Linear Algebra

1. Check the origin: the point $(0,0)$ moves to $(2,0)$.
2. Since the origin does not stay fixed, the rule is violated.

Answer: No

1. Any two parallel lines have the same slope.
2. Stretching one axis uniformly changes the slope of all lines by the same ratio.
3. Therefore, lines that were parallel remain parallel.

Answer: Yes

1. A line has zero width.
2. Any 2D shape flattened into a 1D line loses its area.
3. The resulting area is zero.

Answer: The area becomes zero

1. The vector points along the $x$-axis, so the $y$-component must be $0$.
2. The vector points in the negative direction, so the $x$-component must be negative.
3. Since the length is $5$, the $x$-component must be $-5$.
4. The resulting column vector is $\begin{bmatrix} -5 \\ 0 \end{bmatrix}$.

Answer: v = \begin{bmatrix} -5 \\ 0 \end{bmatrix}

1. Identify the components: $x = 8$ and $y = 6$.
2. Apply the length formula: $\sqrt{8^2 + 6^2}$.
3. Calculate the squares: $64 + 36 = 100$.
4. Take the square root: $\sqrt{100} = 10$.

Answer: 10

1. The original length is $\sqrt{2^2 + 2^2} = \sqrt{8}$.
2. To double the length while keeping the direction, we must multiply each component by $2$.
3. The new $x$-component is $2 \times 2 = 4$.
4. The new $y$-component is $2 \times 2 = 4$.
5. The new vector is $\begin{bmatrix} 4 \\ 4 \end{bmatrix}$.

Answer: u_{new} = \begin{bmatrix} 4 \\ 4 \end{bmatrix}

1. Add the first components: $4 + (-2) = 2$.
2. Add the second components: $-1 + 3 = 2$.

Answer: \begin{bmatrix} 2 \\ 2 \end{bmatrix}

1. Multiply the first component by $-3$: $-3 \cdot 3 = -9$.
2. Multiply the second component by $-3$: $-3 \cdot (-5) = 15$.

Answer: \begin{bmatrix} -9 \\ 15 \end{bmatrix}

1. First, scale $a$ by $2$: $2 \cdot \begin{bmatrix} 1 \\ 2 \end{bmatrix} = \begin{bmatrix} 2 \\ 4 \end{bmatrix}$.
2. Subtract $b$ from the result: $\begin{bmatrix} 2 \\ 4 \end{bmatrix} - \begin{bmatrix} 0 \\ 4 \end{bmatrix} = \begin{bmatrix} 2 - 0 \\ 4 - 4 \end{bmatrix} = \begin{bmatrix} 2 \\ 0 \end{bmatrix}$.

Answer: \begin{bmatrix} 2 \\ 0 \end{bmatrix}

1. Scale $\mathbf{v}$ by 3: $3 \cdot \begin{bmatrix} 2 \\ -1 \end{bmatrix} = \begin{bmatrix} 6 \\ -3 \end{bmatrix}$.
2. Scale $\mathbf{w}$ by 4: $4 \cdot \begin{bmatrix} 1 \\ 3 \end{bmatrix} = \begin{bmatrix} 4 \\ 12 \end{bmatrix}$.
3. Add the results: $\begin{bmatrix} 6 \\ -3 \end{bmatrix} + \begin{bmatrix} 4 \\ 12 \end{bmatrix} = \begin{bmatrix} 10 \\ 9 \end{bmatrix}$.

Answer: \begin{bmatrix} 10 \\ 9 \end{bmatrix}

1. Scale $\mathbf{v}$ by -2: $-2 \cdot \begin{bmatrix} -1 \\ 4 \end{bmatrix} = \begin{bmatrix} 2 \\ -8 \end{bmatrix}$.
2. Scale $\mathbf{w}$ by 3: $3 \cdot \begin{bmatrix} 2 \\ 0 \end{bmatrix} = \begin{bmatrix} 6 \\ 0 \end{bmatrix}$.
3. Add the results: $\begin{bmatrix} 2 \\ -8 \end{bmatrix} + \begin{bmatrix} 6 \\ 0 \end{bmatrix} = \begin{bmatrix} 8 \\ -8 \end{bmatrix}$.

Answer: \begin{bmatrix} 8 \\ -8 \end{bmatrix}

1. Set up the equation: $c_1 \begin{bmatrix} 1 \\ 1 \end{bmatrix} + c_2 \begin{bmatrix} 0 \\ 1 \end{bmatrix} = \begin{bmatrix} 3 \\ 5 \end{bmatrix}$.
2. From the first component: $c_1(1) + c_2(0) = 3$, so $c_1 = 3$.
3. From the second component: $c_1(1) + c_2(1) = 5$.
4. Substitute $c_1 = 3$ into the second equation: $3 + c_2 = 5$, so $c_2 = 2$.

Answer: c_1 = 3, c_2 = 2

1. Set up the equation $c_1 \begin{bmatrix} 1 \\ 0 \end{bmatrix} + c_2 \begin{bmatrix} 1 \\ 1 \end{bmatrix} = \begin{bmatrix} 5 \\ 1 \end{bmatrix}$.
2. From the second component, $0 \cdot c_1 + 1 \cdot c_2 = 1$, so $c_2 = 1$.
3. Substitute $c_2 = 1$ into the first component: $1 \cdot c_1 + 1 \cdot 1 = 5$.
4. Solve for $c_1$: $c_1 = 5 - 1 = 4$.
5. Since we found weights $c_1 = 4$ and $c_2 = 1$, the vector is in the span.

Answer: Yes, $c_1 = 4, c_2 = 1$

1. Check if $\mathbf{w}$ is a scalar multiple of $\mathbf{v}$.
2. Observe that $-2 \cdot \begin{bmatrix} 3 \\ -2 \end{bmatrix} = \begin{bmatrix} -6 \\ 4 \end{bmatrix}$.
3. Since $\mathbf{w} = -2\mathbf{v}$, the vectors are collinear.
4. The span of collinear vectors is a line passing through the origin.

Answer: A line

1. Notice that $\mathbf{w} = 2\mathbf{v}$, so the span is the line $y = 2x$.
2. Any vector where the second component is not twice the first component will be outside the span.
3. For example, choose $\begin{bmatrix} 1 \\ 1 \end{bmatrix}$.
4. Check: $c_1 \begin{bmatrix} 1 \\ 2 \end{bmatrix} + c_2 \begin{bmatrix} 2 \\ 4 \end{bmatrix} = \begin{bmatrix} c_1 + 2c_2 \\ 2c_1 + 4c_2 \end{bmatrix}$.
5. The second component is always twice the first. Since $1 \neq 2(1)$, the vector $\begin{bmatrix} 1 \\ 1 \end{bmatrix}$ is not in the span.

Answer: Any vector where $y \neq 2x$, e.g., $\begin{bmatrix} 1 \\ 1 \end{bmatrix}$

1. Multiply the first components: $5 \times (-3) = -15$.
2. Multiply the second components: $2 \times 4 = 8$.
3. Sum the results: $-15 + 8 = -7$.

Answer: -7

1. Compute the dot product: $(2)(-6) + (3)(4)$.
2. Calculate the products: $-12 + 12$.
3. The sum is $0$. Since the dot product is zero, the vectors are orthogonal.

Answer: Yes, they are orthogonal

1. Multiply first components: $1 \times 3 = 3$.
2. Multiply second components: $-1 \times 4 = -4$.
3. Multiply third components: $2 \times 0 = 0$.
4. Sum the results: $3 + (-4) + 0 = -1$.

Answer: -1

1. Compute the dot product: $\mathbf{u} \cdot \mathbf{v} = (2)(-6) + (3)(4)$
2. Calculate the values: $-12 + 12 = 0$
3. Since the dot product is $0$, the vectors are perpendicular.

Answer: Yes, they are perpendicular.

1. Compute the dot product: $\mathbf{u} \cdot \mathbf{v} = (1)(1) + (0)(1) = 1$
2. Compute magnitudes: $|\mathbf{u}| = \sqrt{1^2 + 0^2} = 1$ and $|\mathbf{v}| = \sqrt{1^2 + 1^2} = \sqrt{2}$
3. Use the formula: $\cos(\theta) = \frac{1}{1 \cdot \sqrt{2}} = \frac{1}{\sqrt{2}}$
4. Find the angle: $\theta = \arccos(\frac{1}{\sqrt{2}}) = 45^\circ$

Answer: 45^\circ

1. For unit vectors, $\mathbf{u} \cdot \mathbf{v} = \cos(\theta)$
2. Set $\cos(\theta) = -0.5$
3. Find the angle: $\theta = \arccos(-0.5) = 120^\circ$

Answer: 120^\circ

1. Square the components: $8^2 = 64$ and $6^2 = 36$.
2. Sum the squares: $64 + 36 = 100$.
3. Take the square root: $\sqrt{100} = 10$.

Answer: 10

1. Calculate the length of $\mathbf{v}$: $|\mathbf{v}| = \sqrt{1^2 + 1^2} = \sqrt{2}$.
2. Divide each component by the length: $u_1 = 1/\sqrt{2}$ and $u_2 = 1/\sqrt{2}$.

Answer: \begin{bmatrix} 1/\sqrt{2} \\ 1/\sqrt{2} \end{bmatrix}

1. Find the difference vector $\mathbf{b} - \mathbf{a}$: $\begin{bmatrix} -1 - 2 \\ 1 - (-3) \end{bmatrix} = \begin{bmatrix} -3 \\ 4 \end{bmatrix}$.
2. Calculate the length of this vector: $\sqrt{(-3)^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5$.

Answer: 5

1. The first component is $(2)(3) - (0)(1) = 6$.
2. The second component is $(0)(0) - (1)(3) = -3$.
3. The third component is $(1)(1) - (2)(0) = 1$.

Answer: \begin{pmatrix} 6 \\ -3 \\ 1 \end{pmatrix}

1. Compute the cross product: $\mathbf{u} \times \mathbf{v} = \begin{pmatrix} (0)(0) - (0)(3) \\ (0)(0) - (2)(0) \\ (2)(3) - (0)(0) \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 6 \end{pmatrix}$.
2. The area is the magnitude of this vector: $\sqrt{0^2 + 0^2 + 6^2} = 6$.

Answer: 6

1. Compute the first component: $(-1)(-4) - (2)(2) = 4 - 4 = 0$.
2. Compute the second component: $(2)(-2) - (1)(-4) = -4 + 4 = 0$.
3. Compute the third component: $(1)(2) - (-1)(-2) = 2 - 2 = 0$.
4. Since the cross product is the zero vector, the vectors are parallel.

Answer: Yes, they are parallel

1. Identify the starting point $\mathbf{p} = \begin{pmatrix} -2 \\ 5 \end{pmatrix}$.
2. Identify the direction vector $\mathbf{v} = \begin{pmatrix} 4 \\ 3 \end{pmatrix}$.
3. Substitute into the formula $\mathbf{r}(t) = \mathbf{p} + t\mathbf{v}$.

Answer: \mathbf{r}(t) = \begin{pmatrix} -2 \\ 5 \end{pmatrix} + t \begin{pmatrix} 4 \\ 3 \end{pmatrix}

1. Calculate the direction vector $\mathbf{v} = \mathbf{B} - \mathbf{A} = \begin{pmatrix} 3-1 \\ -2-0 \\ 5-2 \end{pmatrix} = \begin{pmatrix} 2 \\ -2 \\ 3 \end{pmatrix}$.
2. Use point $A$ as the starting point $\mathbf{p} = \begin{pmatrix} 1 \\ 0 \\ 2 \end{pmatrix}$.
3. Write the components: $x = 1 + 2t$, $y = 0 - 2t$, $z = 2 + 3t$.

Answer: x = 1 + 2t, y = -2t, z = 2 + 3t

1. Find a point on the line: if $x = 0$, then $y = 4$, so $\mathbf{p} = \begin{pmatrix} 0 \\ 4 \end{pmatrix}$.
2. Identify the slope $m = -3$, which means for $\Delta x = 1$, $\Delta y = -3$.
3. The direction vector is $\mathbf{v} = \begin{pmatrix} 1 \\ -3 \end{pmatrix}$.
4. Combine into the vector form.

Answer: \mathbf{r}(t) = \begin{pmatrix} 0 \\ 4 \end{pmatrix} + t \begin{pmatrix} 1 \\ -3 \end{pmatrix}

1. Use the point-normal form: $2(x - 4) - 3(y - 1) + 1(z - 5) = 0$.
2. Expand the terms: $2x - 8 - 3y + 3 + z - 5 = 0$.
3. Combine constants: $2x - 3y + z - 10 = 0$.
4. Move the constant to the right side: $2x - 3y + z = 10$.

Answer: 2x - 3y + z = 10

1. Identify the normal vector of the plane: $\mathbf{n} = \begin{pmatrix} 3 \\ -1 \\ 2 \end{pmatrix}$.
2. A vector is parallel to the plane if it is perpendicular to the normal vector.
3. Compute the dot product: $\mathbf{n} \cdot \mathbf{v} = (3)(1) + (-1)(2) + (2)(1)$.
4. Calculate the sum: $3 - 2 + 2 = 3$.
5. Since the dot product is not zero, the vector is not parallel to the plane.

Answer: No, it is not parallel.

1. Find two vectors in the plane: $\mathbf{PQ} = Q - P = \begin{pmatrix} -2 \\ 3 \\ 0 \end{pmatrix}$ and $\mathbf{PR} = R - P = \begin{pmatrix} -2 \\ 0 \\ 6 \end{pmatrix}$.
2. Find the normal vector using the cross product: $\mathbf{n} = \mathbf{PQ} \times \mathbf{PR} = \begin{pmatrix} (3)(6) - (0)(0) \\ (0)(-2) - (-2)(6) \\ (-2)(0) - (3)(-2) \end{pmatrix} = \begin{pmatrix} 18 \\ 12 \\ 6 \end{pmatrix}$.
3. Simplify the normal vector by dividing by 6: $\mathbf{n}_{simple} = \begin{pmatrix} 3 \\ 2 \\ 1 \end{pmatrix}$.
4. Use point $P(2, 0, 0)$ in the equation: $3(x - 2) + 2(y - 0) + 1(z - 0) = 0$.
5. Simplify: $3x - 6 + 2y + z = 0 \implies 3x + 2y + z = 6$.

Answer: 3x + 2y + z = 6

1. Identify the normal vector $\mathbf{n} = [1, 1, 1]$.
2. Find a point $A$ on the plane, e.g., $A(1, 0, 0)$.
3. Compute the vector $\mathbf{AP} = [2-1, 3-0, 4-0] = [1, 3, 4]$.
4. Compute the unit normal $\mathbf{\hat{n}} = \frac{1}{\sqrt{3}}[1, 1, 1]$.
5. Calculate the distance $d = |\mathbf{AP} \cdot \mathbf{\hat{n}}| = \frac{|1(1) + 3(1) + 4(1)|}{\sqrt{3}} = \frac{8}{\sqrt{3}}$.

Answer: $\frac{8}{\sqrt{3}}$

1. Sum the x-coordinates: $1 + 5 + 3 = 9$. Divide by 3 to get $3$.
2. Sum the y-coordinates: $2 + 0 + 6 = 8$. Divide by 3 to get $8/3$.
3. Sum the z-coordinates: $0 + 2 + 4 = 6$. Divide by 3 to get $2$.

Answer: $(3, 8/3, 2)$

1. Find the vector $\mathbf{AP} = [1, 1, 1]$.
2. The unit direction vector is $\mathbf{\hat{v}} = [0, 1, 0]$.
3. Compute the cross product $\mathbf{AP} \times \mathbf{\hat{v}} = [1(0)-1(1), 1(0)-1(0), 1(1)-1(0)] = [-1, 0, 1]$.
4. Calculate the magnitude $d = \sqrt{(-1)^2 + 0^2 + 1^2} = \sqrt{2}$.

Answer: $\sqrt{2}$

1. Set up the dot product: $0.3 \times 10 + 0.7 \times 50$
2. Calculate the first term: $0.3 \times 10 = 3$
3. Calculate the second term: $0.7 \times 50 = 35$
4. Sum the results: $3 + 35 = 38$

Answer: 38

1. Find the length of $\mathbf{d_1}$: $|\mathbf{d_1}| = \sqrt{3^2 + 0^2} = 3$
2. Find the length of $\mathbf{d_2}$: $|\mathbf{d_2}| = \sqrt{2^2 + 2^2} = \sqrt{8} = 2\sqrt{2}$
3. Compute the dot product: $\mathbf{d_1} \cdot \mathbf{d_2} = (3 \times 2) + (0 \times 2) = 6$
4. Divide by the product of lengths: $6 / (3 \times 2\sqrt{2}) = 1 / \sqrt{2} \approx 0.707$

Answer: \frac{1}{\sqrt{2}} \approx 0.707

1. Calculate distance $AB$: $\sqrt{(2-1)^2 + (2-2)^2 + (2-1)^2} = \sqrt{1^2 + 0^2 + 1^2} = \sqrt{2} \approx 1.414$
2. Calculate distance $AC$: $\sqrt{(1-1)^2 + (0-2)^2 + (1-1)^2} = \sqrt{0^2 + (-2)^2 + 0^2} = \sqrt{4} = 2$
3. Compare the distances: $\sqrt{2} < 2$

Answer: Point B is closer to A

1. Compute the dot product: $\mathbf{u} \cdot \mathbf{v} = (1)(0) + (1)(1) + (0)(1) = 1$.
2. Compute the lengths: $|\mathbf{u}| = \sqrt{1^2 + 1^2 + 0^2} = \sqrt{2}$ and $|\mathbf{v}| = \sqrt{0^2 + 1^2 + 1^2} = \sqrt{2}$.
3. Use the formula $\cos(\theta) = \frac{\mathbf{u} \cdot \mathbf{v}}{|\mathbf{u}||\mathbf{v}|} = \frac{1}{\sqrt{2}\sqrt{2}} = \frac{1}{2}$.
4. Find the angle: $\theta = \arccos(1/2) = 60^\circ$.

Answer: 60^\circ

1. Find vectors $\vec{AB} = [2, 0, 0]$ and $\vec{AC} = [0, 3, 0]$.
2. Compute the cross product: $\vec{AB} \times \vec{AC} = [0, 0, (2)(3) - (0)(0)] = [0, 0, 6]$.
3. The length of the cross product is $\sqrt{0^2 + 0^2 + 6^2} = 6$.
4. The area of the triangle is half the area of the parallelogram: $6 / 2 = 3$.

Answer: 3

1. Identify the normal vector $\mathbf{n} = [2, -1, 2]$.
2. Find the length of the normal: $|\mathbf{n}| = \sqrt{2^2 + (-1)^2 + 2^2} = \sqrt{9} = 3$.
3. Find a point $A$ on the plane. If $y=0, z=0$, then $2x=4 \implies x=2$. So $A(2, 0, 0)$.
4. Compute the vector $\vec{AP} = P - A = [1-2, 2-0, 1-0] = [-1, 2, 1]$.
5. The distance is $d = \frac{|\vec{AP} \cdot \mathbf{n}|}{|\mathbf{n}|} = \frac{|(-1)(2) + (2)(-1) + (1)(2)|}{3} = \frac{|-2 - 2 + 2|}{3} = \frac{2}{3}$.

Answer: 2/3

1. Identify the coefficients of $x$ as the first column vector: $\mathbf{v}_1 = \begin{pmatrix} 3 \\ 4 \end{pmatrix}$.
2. Identify the coefficients of $y$ as the second column vector: $\mathbf{v}_2 = \begin{pmatrix} 2 \\ -1 \end{pmatrix}$.
3. Identify the constants on the right side as the target vector: $\mathbf{b} = \begin{pmatrix} 7 \\ 2 \end{pmatrix}$.
4. Combine them into the form $x\mathbf{v}_1 + y\mathbf{v}_2 = \mathbf{b}$.

Answer: x \begin{pmatrix} 3 \\ 4 \end{pmatrix} + y \begin{pmatrix} 2 \\ -1 \end{pmatrix} = \begin{pmatrix} 7 \\ 2 \end{pmatrix}

1. In the row picture, the second equation is just twice the first, so both equations describe the same line $x + y = 2$.
2. In the column picture, the columns are $\begin{pmatrix} 1 \\ 2 \end{pmatrix}$ and $\begin{pmatrix} 1 \\ 2 \end{pmatrix}$.
3. The target vector is $\begin{pmatrix} 2 \\ 4 \end{pmatrix}$, which is exactly twice the first column.
4. Since the columns are dependent and the target is in their span, there are infinite solutions.

Answer: Both rows describe the same line; both columns are the same vector and the target is a multiple of them.

1. Write as $x \begin{pmatrix} 2 \\ 1 \end{pmatrix} + y \begin{pmatrix} 1 \\ -1 \end{pmatrix} = \begin{pmatrix} 5 \\ 1 \end{pmatrix}$.
2. Try small integers: if $x = 2$, then $2 \begin{pmatrix} 2 \\ 1 \end{pmatrix} = \begin{pmatrix} 4 \\ 2 \end{pmatrix}$.
3. We need $\begin{pmatrix} 5 \\ 1 \end{pmatrix} - \begin{pmatrix} 4 \\ 2 \end{pmatrix} = \begin{pmatrix} 1 \\ -1 \end{pmatrix}$.
4. This is exactly $1$ times the second column vector, so $y = 1$.
5. Check: $2(2) + 1 = 5$ and $2 - 1 = 1$.

Answer: x = 2, y = 1

1. Observe the coefficients of the first equation: $3$ and $-2$.
2. Observe the coefficients of the second equation: $6$ and $-4$.
3. Notice that $6 = 2 \times 3$ and $-4 = 2 \times (-2)$.
4. Check the constants: $10 = 2 \times 5$.
5. Since the entire second equation is exactly twice the first, the lines are coincident.

Answer: Infinitely many solutions

1. The normal vectors for both lines are $[1, 1]$.
2. Since the normal vectors are identical, the lines are parallel.
3. The constants $2$ and $5$ are different, so the lines are distinct.
4. Parallel distinct lines never intersect.

Answer: No solution

1. A solution must be a point that lies on all three planes simultaneously.
2. If plane 1 and plane 2 are parallel and distinct, they have no points in common.
3. Therefore, there can be no point that lies on all three planes.

Answer: No solution

1. Write the augmented matrix: $\left( \begin{array}{cc|c} 2 & 3 & 8 \\ 1 & -1 & -3 \end{array} \right)$
2. Swap rows to get a pivot of 1: $\left( \begin{array}{cc|c} 1 & -1 & -3 \\ 2 & 3 & 8 \end{array} \right)$
3. Eliminate below pivot: $R_2 \leftarrow R_2 - 2R_1$ gives $\left( \begin{array}{cc|c} 1 & -1 & -3 \\ 0 & 5 & 14 \end{array} \right)$
4. Solve for $y$: $5y = 14 \implies y = 2.8$
5. Back-substitute for $x$: $x - 2.8 = -3 \implies x = -0.2$

Answer: x = -0.2, y = 2.8

1. Augmented matrix: $\left( \begin{array}{ccc|c} 1 & 1 & 1 & 6 \\ 2 & -1 & 1 & 3 \\ 1 & 2 & -1 & 2 \end{array} \right)$
2. Clear first column: $R_2 \leftarrow R_2 - 2R_1$ and $R_3 \leftarrow R_3 - R_1$ gives $\left( \begin{array}{ccc|c} 1 & 1 & 1 & 6 \\ 0 & -3 & -1 & -9 \\ 0 & 1 & -2 & -4 \end{array} \right)$
3. Swap $R_2$ and $R_3$ for easier pivot: $\left( \begin{array}{ccc|c} 1 & 1 & 1 & 6 \\ 0 & 1 & -2 & -4 \\ 0 & -3 & -1 & -9 \end{array} \right)$
4. Clear second column: $R_3 \leftarrow R_3 + 3R_2$ gives $\left( \begin{array}{ccc|c} 1 & 1 & 1 & 6 \\ 0 & 1 & -2 & -4 \\ 0 & 0 & -7 & -21 \end{array} \right)$

Answer: \left( \begin{array}{ccc|c} 1 & 1 & 1 & 6 \\ 0 & 1 & -2 & -4 \\ 0 & 0 & -7 & -21 \end{array} \right)

1. From the last row: $-7z = -21 \implies z = 3$
2. From the second row: $y - 2(3) = -4 \implies y - 6 = -4 \implies y = 2$
3. From the first row: $x + 2 + 3 = 6 \implies x + 5 = 6 \implies x = 1$

Answer: x = 1, y = 2, z = 3

1. Write the augmented matrix: $[[1, 2, -1, 3], [0, 1, 4, 2], [2, 5, 2, 10]]$.
2. Eliminate $x$ from row 3: $R_3 \leftarrow R_3 - 2R_1$ gives $[[1, 2, -1, 3], [0, 1, 4, 2], [0, 1, 4, 4]]$.
3. Eliminate $y$ from row 3: $R_3 \leftarrow R_3 - R_2$ gives $[[1, 2, -1, 3], [0, 1, 4, 2], [0, 0, 0, 2]]$.
4. Observe the third row: $0x + 0y + 0z = 2$, which is impossible.

Answer: No solution (Inconsistent system)

1. Swap $R_1$ and $R_2$ to get a pivot in the first column: $[[1, 2, 1, 4], [0, 1, 1, 2], [2, 7, 1, 19]]$.
2. Eliminate $x$ from row 3: $R_3 \leftarrow R_3 - 2R_1$ gives $[[1, 2, 1, 4], [0, 1, 1, 2], [0, 3, -1, 11]]$.
3. Eliminate $y$ from row 3: $R_3 \leftarrow R_3 - 3R_2$ gives $[[1, 2, 1, 4], [0, 1, 1, 2], [0, 0, -4, 5]]$.
4. Back-substitute: $-4z = 5 \implies z = -1.25$.
5. Substitute $z$ into $R_2$: $y - 1.25 = 2 \implies y = 3.25$.
6. Substitute $y, z$ into $R_1$: $x + 6.5 - 1.25 = 4 \implies x = -1.25$.

Answer: x = -1.25, y = 3.25, z = -1.25

1. Augmented matrix: $[[1, 1, 1, 6], [2, 4, -2, 2], [3, 5, 0, 18]]$.
2. Clear Col 1: $R_2 \leftarrow R_2 - 2R_1$ and $R_3 \leftarrow R_3 - 3R_1$ gives $[[1, 1, 1, 6], [0, 2, -4, -10], [0, 2, -3, 0]]$.
3. Clear Col 2: $R_3 \leftarrow R_3 - R_2$ gives $[[1, 1, 1, 6], [0, 2, -4, -10], [0, 0, 1, 10]]$.
4. Back-substitute: $z = 10$.
5. Substitute $z$ into $R_2$: $2y - 40 = -10 \implies 2y = 30 \implies y = 15$.
6. Substitute $y, z$ into $R_1$: $x + 15 + 10 = 6 \implies x + 25 = 6 \implies x = -19$.

Answer: x = -19, y = 15, z = 10

1. Step 1: Eliminate the entry below the first pivot. $R_2 \leftarrow R_2 - 2R_1$ gives $\begin{pmatrix} 1 & 3 & | & 7 \\ 0 & 2 & | & 3 \end{pmatrix}$.
2. Step 2: Scale the second row to make the pivot one. $R_2 \leftarrow 0.5R_2$ gives $\begin{pmatrix} 1 & 3 & | & 7 \\ 0 & 1 & | & 1.5 \end{pmatrix}$.
3. Step 3: Eliminate the entry above the second pivot. $R_1 \leftarrow R_1 - 3R_2$ gives $\begin{pmatrix} 1 & 0 & | & 2.5 \\ 0 & 1 & | & 1.5 \end{pmatrix}$.

Answer: \begin{pmatrix} 1 & 0 & | & 2.5 \\ 0 & 1 & | & 1.5 \end{pmatrix}

1. Step 1: Check REF conditions. All zero rows are at the bottom. The leading entry of row 2 is to the right of row 1. All entries below pivots are zero. It is in REF.
2. Step 2: Check RREF conditions. All pivots are one. The pivot in column 1 is the only non-zero entry in column 1. The pivot in column 2 is the only non-zero entry in column 2.

Answer: RREF

1. Step 1: Identify pivots. Pivots are in columns 1 and 2. Column 3 has no pivot, so $z$ is a free variable.
2. Step 2: Write the equations from the rows. Row 1: $x + 2z = 4$. Row 2: $y - z = 3$.
3. Step 3: Solve for pivot variables. $x = 4 - 2z$ and $y = 3 + z$.

Answer: x = 4 - 2z, y = 3 + z, z \text{ is free}

1. Identify that there is one equation and two variables, so one variable must be free.
2. Let $y = t$.
3. Solve for $x$: $2x = 12 - 3t$, so $x = 6 - 1.5t$.
4. Write in vector form: $\begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 6 \\ 0 \end{pmatrix} + t \begin{pmatrix} -1.5 \\ 1 \end{pmatrix}$.

Answer: \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 6 \\ 0 \end{pmatrix} + t \begin{pmatrix} -1.5 \\ 1 \end{pmatrix}

1. Pivots are in columns 1 and 3. Column 2 is free.
2. Let $x_2 = t$.
3. From row 2: $x_3 = 5$.
4. From row 1: $x_1 + 2t = 4$, so $x_1 = 4 - 2t$.
5. The solution vector is $\begin{pmatrix} 4-2t \\ t \\ 5 \end{pmatrix} = \begin{pmatrix} 4 \\ 0 \\ 5 \end{pmatrix} + t \begin{pmatrix} -2 \\ 1 \\ 0 \end{pmatrix}$.

Answer: \begin{pmatrix} x_1 \\ x_2 \\ x_3 \end{pmatrix} = \begin{pmatrix} 4 \\ 0 \\ 5 \end{pmatrix} + t \begin{pmatrix} -2 \\ 1 \\ 0 \end{pmatrix}

1. The number of variables is 5.
2. The number of pivots is 2.
3. The number of free variables is $5 - 2 = 3$.
4. The dimension of the solution set is equal to the number of free variables.

Answer: Dimension is 3; 3 parameters are needed.

1. Write the augmented matrix: $\left[ \begin{array}{cc|c} 1 & 3 & 4 \\ 2 & 6 & 10 \end{array} \right]$
2. Perform $R_2 \leftarrow R_2 - 2R_1$: $2 - 2(1) = 0$, $6 - 2(3) = 0$, $10 - 2(4) = 2$.
3. The resulting matrix is $\left[ \begin{array}{cc|c} 1 & 3 & 4 \\ 0 & 0 & 2 \end{array} \right]$.
4. The second row represents $0 = 2$, which is a contradiction.

Answer: Inconsistent

1. The row $[0 \ 0 \ 0 \ | \ 1]$ translates to the equation $0x + 0y + 0z = 1$.
2. This simplifies to $0 = 1$.
3. Since $0$ can never equal $1$, there are no values of $x, y, z$ that satisfy this equation.

Answer: Inconsistent

1. Augmented matrix: $\left[ \begin{array}{ccc|c} 1 & 1 & 0 & 2 \\ 0 & 1 & 1 & 2 \\ 1 & 0 & 1 & 2 \end{array} \right]$
2. Perform $R_3 \leftarrow R_3 - R_1$: $\left[ \begin{array}{ccc|c} 1 & 1 & 0 & 2 \\ 0 & 1 & 1 & 2 \\ 0 & -1 & 1 & 0 \end{array} \right]$
3. Perform $R_3 \leftarrow R_3 + R_2$: $\left[ \begin{array}{ccc|c} 1 & 1 & 0 & 2 \\ 0 & 1 & 1 & 2 \\ 0 & 0 & 2 & 2 \end{array} \right]$
4. The system is consistent because there is no row of the form $[0 \ 0 \ 0 \ | \ k]$.
5. Back-substitute: $2z = 2 \implies z = 1$. Then $y + 1 = 2 \implies y = 1$. Then $x + 1 = 2 \implies x = 1$.

Answer: Consistent; solution is $(1, 1, 1)$

1. Write the augmented matrix and perform row reduction.
2. Row 2 minus 2 times Row 1 gives $[0 \ 0 \ 0]$.
3. Row 3 minus Row 1 gives $[0 \ -1 \ -1]$.
4. The RREF is $\begin{bmatrix} 1 & 0 & -1 \\ 0 & 1 & 1 \\ 0 & 0 & 0 \end{bmatrix}$.
5. There is a free variable $z$, so there are infinitely many solutions.

Answer: Infinitely many solutions

1. Use the superposition principle: any linear combination of solutions is a solution.
2. Let $\mathbf{v}_3 = \mathbf{v}_1 + \mathbf{v}_2$.
3. Compute the sum: $[1+0, 2+1, -1+3]^T = [1, 3, 2]^T$.

Answer: $[1, 3, 2]^T$ (or any other linear combination)

1. The second equation is a multiple of the first, so we have only one independent equation: $x + 2y - z = 0$.
2. Identify $x$ as the pivot variable and $y, z$ as free variables.
3. Solve for $x$: $x = -2y + z$.
4. Write in vector form: $\mathbf{x} = y[-2, 1, 0]^T + z[1, 0, 1]^T$.

Answer: $\mathbf{x} = y \begin{bmatrix} -2 \\ 1 \\ 0 \end{bmatrix} + z \begin{bmatrix} 1 \\ 0 \\ 1 \end{bmatrix}$

1. Set up the augmented matrix $[[1, 2, 1, 4], [3, -1, 2, 1], [4, 1, 3, 5]]$.
2. Perform $R2 \leftarrow R2 - 3R1$ and $R3 \leftarrow R3 - 4R1$ to get $[[1, 2, 1, 4], [0, -7, -1, -11], [0, -7, -1, -11]]$.
3. Perform $R3 \leftarrow R3 - R2$ to get $[[1, 2, 1, 4], [0, -7, -1, -11], [0, 0, 0, 0]]$.
4. Identify $z$ as a free variable. Let $z = t$.
5. From $R2$, $-7y - t = -11$, so $y = (11 - t)/7$.
6. From $R1$, $x = 4 - 2y - z = 4 - 2(11-t)/7 - t = (28 - 22 + 2t - 7t)/7 = (6 - 5t)/7$.

Answer: \mathbf{x} = \begin{bmatrix} 6/7 \\ 11/7 \\ 0 \end{bmatrix} + t \begin{bmatrix} -5/7 \\ -1/7 \\ 1 \end{bmatrix}

1. Set up the augmented matrix $[[1, 1, 2], [2, 2, 5]]$.
2. Perform $R2 \leftarrow R2 - 2R1$.
3. The resulting matrix is $[[1, 1, 2], [0, 0, 1]]$.
4. The second row represents the equation $0 = 1$, which is a contradiction.

Answer: Inconsistent (No solution)

1. Set up the coefficient matrix $[[1, -2, 1], [2, -4, 2]]$.
2. Perform $R2 \leftarrow R2 - 2R1$ to get $[[1, -2, 1], [0, 0, 0]]$.
3. Variables $y$ and $z$ are free. Let $y = s$ and $z = t$.
4. Then $x = 2s - t$.

Answer: \mathbf{x} = s \begin{bmatrix} 2 \\ 1 \\ 0 \end{bmatrix} + t \begin{bmatrix} -1 \\ 0 \\ 1 \end{bmatrix}

1. Node 1: $x_1 + x_2 = 8$
2. Node 2: $x_1 = x_3$
3. Node 3: $x_2 + x_3 = 8$
4. The third equation is redundant. From node 2, $x_1 = x_3$. Substituting into node 1, $x_3 + x_2 = 8$, which matches node 3.
5. Let $x_3 = t$. Then $x_1 = t$ and $x_2 = 8 - t$.

Answer: $x_1 = t, x_2 = 8 - t, x_3 = t$

1. Conservation at node: $I_1 + I_2 = I_3 + I_4$
2. Substitute knowns: $5 + 3 = I_3 + I_4 \implies I_3 + I_4 = 8$
3. Substitute constraint: $2I_4 + I_4 = 8 \implies 3I_4 = 8$
4. Solve for $I_4$: $I_4 = 8/3\text{A}$
5. Solve for $I_3$: $I_3 = 2(8/3) = 16/3\text{A}$

Answer: $I_3 = 16/3\text{A}, I_4 = 8/3\text{A}$

1. Eq 1: $x_1 + x_2 = 10$
2. Eq 2: $x_1 - x_3 - x_4 = 0$
3. Eq 3: $x_2 + x_3 - x_5 = 0$
4. Eq 4: $x_4 + x_5 = 10$
5. Matrix: $[[1, 1, 0, 0, 0, 10], [1, 0, -1, -1, 0, 0], [0, 1, 1, 0, -1, 0], [0, 0, 0, 1, 1, 10]]$.
6. Row reduction shows the 4th equation is a linear combination of the first three.
7. With 5 variables and 3 independent equations, there are $5 - 3 = 2$ free variables.

Answer: The system has infinitely many solutions with 2 free variables.

1. Plug in $(0,1)$: $a(0)^2 + b(0) + c = 1 \implies c = 1$.
2. Plug in $(1,3)$: $a(1)^2 + b(1) + 1 = 3 \implies a + b = 2$.
3. Plug in $(2,7)$: $a(2)^2 + b(2) + 1 = 7 \implies 4a + 2b = 6$.
4. Divide the last equation by 2: $2a + b = 3$.
5. Subtract $(a + b = 2)$ from $(2a + b = 3)$ to get $a = 1$.
6. Substitute $a = 1$ into $a + b = 2$ to get $b = 1$.

Answer: y = x^2 + x + 1

1. Carbon: $3x_1 = 3x_3 \implies 3x_1 - 3x_3 = 0$.
2. Hydrogen: $8x_1 = 6x_3 + 2x_4 \implies 8x_1 - 6x_3 - 2x_4 = 0$.
3. Oxygen: $2x_2 = 3x_3 + x_4 \implies 2x_2 - 3x_3 - x_4 = 0$.

Answer: 3x_1 - 3x_3 = 0, 8x_1 - 6x_3 - 2x_4 = 0, 2x_2 - 3x_3 - x_4 = 0

1. The points $(1,2)$ and $(1,5)$ have the same x-coordinate but different y-coordinates.
2. This means for $x=1$, the system requires $a+b+c=2$ and $a+b+c=5$.
3. Subtracting these equations gives $0 = 3$, which is a contradiction.
4. The system is inconsistent, meaning no such parabola exists.

Answer: No, the system is inconsistent because the points $(1,2)$ and $(1,5)$ violate the vertical line test for functions.

1. Start with the $3 \times 3$ identity matrix $I$.
2. The operation affects row 3 using row 2.
3. Place the multiplier $-4$ in the position corresponding to row 3, column 2.
4. Keep all other entries of the identity matrix the same.

Answer: \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & -4 & 1 \end{pmatrix}

1. Recognize that $E$ is the elementary matrix for swapping row 1 and row 2.
2. Apply the swap to the rows of $A$.
3. The first row of $A$ becomes the second row of the result.
4. The second row of $A$ becomes the first row of the result.

Answer: \begin{pmatrix} 3 & 4 \\ 1 & 2 \end{pmatrix}

1. Write $E_1 = \begin{pmatrix} 5 & 0 \\ 0 & 1 \end{pmatrix}$.
2. Write $E_2 = \begin{pmatrix} 1 & 0 \\ 2 & 1 \end{pmatrix}$.
3. Compute the product $E_2 E_1$.
4. Top row: $(1 \cdot 5) + (0 \cdot 0) = 5$ and $(1 \cdot 0) + (0 \cdot 1) = 0$.
5. Bottom row: $(2 \cdot 5) + (1 \cdot 0) = 10$ and $(2 \cdot 0) + (1 \cdot 1) = 1$.

Answer: \begin{pmatrix} 5 & 0 \\ 10 & 1 \end{pmatrix}

1. Step 1: Clear the first column. $R_3 \leftarrow R_3 - 2R_1$ gives $\begin{bmatrix} 1 & 2 & 1 & | & 4 \\ 0 & 1 & -1 & | & 1 \\ 0 & 1 & 0 & | & 1 \end{bmatrix}$.
2. Step 2: Clear the second column. $R_3 \leftarrow R_3 - R_2$ gives $\begin{bmatrix} 1 & 2 & 1 & | & 4 \\ 0 & 1 & -1 & | & 1 \\ 0 & 0 & 1 & | & 0 \end{bmatrix}$.
3. Step 3: Back substitute. From $R_3$, $z = 0$.
4. Step 4: From $R_2$, $y - 0 = 1$, so $y = 1$.
5. Step 5: From $R_1$, $x + 2(1) + 0 = 4$, so $x = 2$.

Answer: (2, 1, 0)

1. Step 1: Perform $R_2 \leftarrow R_2 - 2R_1$.
2. Step 2: The resulting matrix is $\begin{bmatrix} 1 & 1 & | & 2 \\ 0 & 0 & | & 1 \end{bmatrix}$.
3. Step 3: The second row implies $0 = 1$, which is a contradiction.

Answer: Inconsistent (no solution)

1. Step 1: Identify pivots. The only pivot is in the first column.
2. Step 2: Identify free variables. The second column has no pivot, so $y$ is free.
3. Step 3: Express $x$ in terms of $y$. $x + 3y = 5 \implies x = 5 - 3y$.

Answer: x = 5 - 3y, y is free

1. Locate the second row of the matrix, which is $[4, 5, 6]$.
2. Locate the third column of that row, which is the value $6$.

Answer: 6

1. Start with a $3 \times 3$ matrix of zeros.
2. Place $2$ at row $1$, column $1$.
3. Place $-1$ at row $2$, column $2$.
4. Place $4$ at row $3$, column $3$.

Answer: \begin{bmatrix} 2 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & 4 \end{bmatrix}

1. The total number of entries is the product of the number of rows and columns: $4 \times 2 = 8$.
2. A square matrix must have the same number of rows and columns. Since $4 \neq 2$, it is not square.

Answer: 8 entries; not square

1. Using the row view, the first component is $1(2) + 0(1) + 2(3) = 2 + 0 + 6 = 8$.
2. The second component is $-1(2) + 3(1) + 1(3) = -2 + 3 + 3 = 4$.
3. The resulting vector is $\begin{bmatrix} 8 \\ 4 \end{bmatrix}$.

Answer: \begin{bmatrix} 8 \\ 4 \end{bmatrix}

1. The column view is $1 \begin{bmatrix} 2 \\ 0 \\ 3 \end{bmatrix} + 2 \begin{bmatrix} 1 \\ -2 \\ 4 \end{bmatrix}$.
2. This equals $\begin{bmatrix} 2 \\ 0 \\ 3 \end{bmatrix} + \begin{bmatrix} 2 \\ -4 \\ 8 \end{bmatrix}$.
3. Summing the components gives $\begin{bmatrix} 4 \\ -4 \\ 11 \end{bmatrix}$.

Answer: \begin{bmatrix} 4 \\ -4 \\ 11 \end{bmatrix}

1. The product of a $4 \times 3$ matrix and a $3 \times 1$ vector results in a $4 \times 1$ vector.
2. For multiplication to be possible, the number of columns in the matrix must match the number of components in the vector.
3. Since the matrix has 3 columns, a vector with 4 components would not match, making the multiplication impossible.

Answer: 4 \times 1; No

1. Check dimensions: $(2 \times 2) \times (2 \times 2)$ is defined and results in a $2 \times 2$ matrix.
2. Compute $c_{11} = (2)(1) + (-1)(2) = 2 - 2 = 0$.
3. Compute $c_{12} = (2)(4) + (-1)(5) = 8 - 5 = 3$.
4. Compute $c_{21} = (0)(1) + (3)(2) = 0 + 6 = 6$.
5. Compute $c_{22} = (0)(4) + (3)(5) = 0 + 15 = 15$.

Answer: \begin{bmatrix} 0 & 3 \\ 6 & 15 \end{bmatrix}

1. Compute $AB$: $c_{11} = 1(1)+2(0)=1$, $c_{12} = 1(1)+2(2)=5$, $c_{21} = 3(1)+0(0)=3$, $c_{22} = 3(1)+0(2)=3$. So $AB = \begin{bmatrix} 1 & 5 \\ 3 & 3 \end{bmatrix}$.
2. Compute $BA$: $c_{11} = 1(1)+1(3)=4$, $c_{12} = 1(2)+1(0)=2$, $c_{21} = 0(1)+2(3)=6$, $c_{22} = 0(2)+2(0)=0$. So $BA = \begin{bmatrix} 4 & 2 \\ 6 & 0 \end{bmatrix}$.
3. Compare the results: $\begin{bmatrix} 1 & 5 \\ 3 & 3 \end{bmatrix} \neq \begin{bmatrix} 4 & 2 \\ 6 & 0 \end{bmatrix}$.

Answer: AB = \begin{bmatrix} 1 & 5 \\ 3 & 3 \end{bmatrix}, BA = \begin{bmatrix} 4 & 2 \\ 6 & 0 \end{bmatrix}

1. For $C = AB$, the dimensions are $(3 \times 2) \times (2 \times 4)$. The inner dimensions match (2), so the product is defined.
2. The resulting dimensions are the outer dimensions: $3 \times 4$.
3. For $BA$, the dimensions are $(2 \times 4) \times (3 \times 2)$.
4. The inner dimensions are 4 and 3. Since $4 \neq 3$, the product $BA$ is undefined.

Answer: C is $3 \times 4$; BA is undefined.

1. The vector is rotated first, so $R$ is the rightmost matrix.
2. The vector is then scaled, so $S$ is the leftmost matrix.
3. We compute $C = SR$.
4. The first row of $S$ dotted with the first column of $R$ is $3(0) + 0(1) = 0$.
5. The first row of $S$ dotted with the second column of $R$ is $3(-1) + 0(0) = -3$.
6. The second row of $S$ dotted with the first column of $R$ is $0(0) + 3(1) = 3$.
7. The second row of $S$ dotted with the second column of $R$ is $0(-1) + 3(0) = 0$.

Answer: \begin{bmatrix} 0 & -3 \\ 3 & 0 \end{bmatrix}

1. Compute $AB$: row 1 of $A$ dot col 1 of $B$ is $1(1) + 2(3) = 7$; row 1 of $A$ dot col 2 of $B$ is $1(0) + 2(1) = 2$; row 2 of $A$ dot col 1 of $B$ is $0(1) + 1(3) = 3$; row 2 of $A$ dot col 2 of $B$ is $0(0) + 1(1) = 1$. So $AB = \begin{bmatrix} 7 & 2 \\ 3 & 1 \end{bmatrix}$.
2. Compute $BA$: row 1 of $B$ dot col 1 of $A$ is $1(1) + 0(0) = 1$; row 1 of $B$ dot col 2 of $A$ is $1(2) + 0(1) = 2$; row 2 of $B$ dot col 1 of $A$ is $3(1) + 1(0) = 3$; row 2 of $B$ dot col 2 of $A$ is $3(2) + 1(1) = 7$. So $BA = \begin{bmatrix} 1 & 2 \\ 3 & 7 \end{bmatrix}$.
3. Since $AB \neq BA$, the matrices do not commute.

Answer: AB = \begin{bmatrix} 7 & 2 \\ 3 & 1 \end{bmatrix}, BA = \begin{bmatrix} 1 & 2 \\ 3 & 7 \end{bmatrix}, \text{ No they do not commute.}

1. The sequence is $T_1$ then $T_2$, so the composition matrix is $C = M_2 M_1$.
2. Compute $C = \begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix} \begin{bmatrix} 2 & 0 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 1(2)+1(0) & 1(0)+1(1) \\ 0(2)+1(0) & 0(0)+1(1) \end{bmatrix} = \begin{bmatrix} 2 & 1 \\ 0 & 1 \end{bmatrix}$.
3. Now apply $C$ to $\mathbf{v}$: $\begin{bmatrix} 2 & 1 \\ 0 & 1 \end{bmatrix} \begin{bmatrix} 1 \\ 1 \end{bmatrix} = \begin{bmatrix} 2(1) + 1(1) \\ 0(1) + 1(1) \end{bmatrix} = \begin{bmatrix} 3 \\ 1 \end{bmatrix}$.

Answer: \begin{bmatrix} 3 \\ 1 \end{bmatrix}

1. Compute $AB$: the first row is $(1\cdot 0 + 2\cdot 1, 1\cdot 1 + 2\cdot 0) = (2, 1)$, the second row is $(3\cdot 0 + 4\cdot 1, 3\cdot 1 + 4\cdot 0) = (4, 3)$. So $AB = \begin{pmatrix} 2 & 1 \ 4 & 3 \end{pmatrix}$.
2. Compute $BA$: the first row is $(0\cdot 1 + 1\cdot 3, 0\cdot 2 + 1\cdot 4) = (3, 4)$, the second row is $(1\cdot 1 + 0\cdot 3, 1\cdot 2 + 0\cdot 4) = (1, 2)$. So $BA = \begin{pmatrix} 3 & 4 \ 1 & 2 \end{pmatrix}$.
3. Compare the results: $\begin{pmatrix} 2 & 1 \ 4 & 3 \end{pmatrix} \neq \begin{pmatrix} 3 & 4 \ 1 & 2 \end{pmatrix}$.

Answer: AB = \begin{pmatrix} 2 & 1 \ 4 & 3 \end{pmatrix}, BA = \begin{pmatrix} 3 & 4 \ 1 & 2 \end{pmatrix}

1. Compute $AB$: the first row is $(1\cdot 5 + 0\cdot 7, 1\cdot 6 + 0\cdot 8) = (5, 6)$, the second row is $(0\cdot 5 + 1\cdot 7, 0\cdot 6 + 1\cdot 8) = (7, 8)$. So $AB = \begin{pmatrix} 5 & 6 \ 7 & 8 \end{pmatrix}$.
2. Compute $BA$: the first row is $(5\cdot 1 + 6\cdot 0, 5\cdot 0 + 6\cdot 1) = (5, 6)$, the second row is $(7\cdot 1 + 8\cdot 0, 7\cdot 0 + 8\cdot 1) = (7, 8)$. So $BA = \begin{pmatrix} 5 & 6 \ 7 & 8 \end{pmatrix}$.
3. Compare the results: $AB = BA$.

Answer: Yes, AB = BA = \begin{pmatrix} 5 & 6 \ 7 & 8 \end{pmatrix}

1. Apply the distributive law: $A(B + C) = AB + AC$.
2. Substitute this back into the expression: $(AB + AC) - AB$.
3. Rearrange the terms: $AB - AB + AC$.
4. The $AB$ terms cancel out, leaving $AC$.

Answer: AC

1. The top left element is the dot product of the first row $(1, 2)$ and first column $(1, 3)$, which is $1(1) + 2(3) = 7$.
2. The top right element is the dot product of the first row $(1, 2)$ and second column $(2, 4)$, which is $1(2) + 2(4) = 10$.
3. The bottom left element is the dot product of the second row $(3, 4)$ and first column $(1, 3)$, which is $3(1) + 4(3) = 15$.
4. The bottom right element is the dot product of the second row $(3, 4)$ and second column $(2, 4)$, which is $3(2) + 4(4) = 22$.

Answer: \begin{pmatrix} 7 & 10 \\ 15 & 22 \end{pmatrix}

1. Compute $A^2$: the top left is $0(0) + (-1)(1) = -1$; top right is $0(-1) + (-1)(0) = 0$; bottom left is $1(0) + 0(1) = 0$; bottom right is $1(-1) + 0(0) = -1$. So $A^2 = -I$.
2. Compute $A^4$ as $(A^2)^2$. Since $A^2 = -I$, then $A^4 = (-I)(-I)$.
3. Multiplying $-I$ by $-I$ is the same as $(-1)(-1) I^2$, which is $1 \times I = I$.

Answer: A^2 = \begin{pmatrix} -1 & 0 \\ 0 & -1 \end{pmatrix}, A^4 = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}

1. Substitute $A^2 = I$ into the expression: $A^3 + 2A^2 + A = A(A^2) + 2(I) + A$.
2. Since $A^2 = I$, the first term $A(A^2)$ becomes $A(I) = A$.
3. The expression is now $A + 2I + A$.
4. Combine the $A$ terms to get $2A + 2I$.

Answer: 2A + 2I

1. The first row $(1, 2)$ becomes the first column.
2. The second row $(3, 4)$ becomes the second column.
3. The third row $(5, 6)$ becomes the third column.

Answer: \begin{pmatrix} 1 & 3 & 5 \\ 2 & 4 & 6 \end{pmatrix}

1. First, find $A+B = \begin{pmatrix} 1+0 & 0+1 \\ 2+1 & 1+0 \end{pmatrix} = \begin{pmatrix} 1 & 1 \\ 3 & 1 \end{pmatrix}$.
2. Now, transpose the result by swapping rows and columns.

Answer: \begin{pmatrix} 1 & 3 \\ 1 & 1 \end{pmatrix}

1. The product $AB$ has dimensions $2 \times 4$.
2. The transpose of a $2 \times 4$ matrix swaps the number of rows and columns.

Answer: 4 \times 2

1. Calculate the determinant: $\text{det}(A) = (3)(2) - (1)(5) = 6 - 5 = 1$.
2. Swap the main diagonal elements (3 and 2) and negate the off-diagonal elements (1 and 5).
3. The resulting matrix is $\begin{pmatrix} 2 & -1 \\ -5 & 3 \end{pmatrix}$.
4. Multiply by $1/\text{det}(A) = 1/1$.

Answer: \begin{pmatrix} 2 & -1 \\ -5 & 3 \end{pmatrix}

1. Calculate the determinant: $\text{det}(B) = (2)(2) - (4)(1) = 4 - 4 = 0$.
2. Since the determinant is zero, the matrix is singular.

Answer: Not invertible (singular)

1. Calculate the determinant: $\text{det}(C) = (4)(2) - (3)(3) = 8 - 9 = -1$.
2. Apply the formula: $C^{-1} = \frac{1}{-1} \begin{pmatrix} 2 & -3 \\ -3 & 4 \end{pmatrix} = \begin{pmatrix} -2 & 3 \\ 3 & -4 \end{pmatrix}$.
3. Verify: $\begin{pmatrix} 4 & 3 \\ 3 & 2 \end{pmatrix} \begin{pmatrix} -2 & 3 \\ 3 & -4 \end{pmatrix} = \begin{pmatrix} -8+9 & 12-12 \\ -6+6 & 9-8 \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}$.

Answer: \begin{pmatrix} -2 & 3 \\ 3 & -4 \end{pmatrix}

1. Set up augmented matrix $[A | I]$.
2. R3 <- R3 - 2*R1 results in $[\begin{smallmatrix} 1 & 2 & 1 & | & 1 & 0 & 0 \\ 0 & 1 & 3 & | & 0 & 1 & 0 \\ 0 & 1 & 0 & | & -2 & 0 & 1 \end{smallmatrix}]$.
3. R3 <- R3 - R2 results in $[\begin{smallmatrix} 1 & 2 & 1 & | & 1 & 0 & 0 \\ 0 & 1 & 3 & | & 0 & 1 & 0 \\ 0 & 0 & -3 & | & -2 & -1 & 1 \end{smallmatrix}]$.
4. R3 <- -1/3 * R3 results in $[\begin{smallmatrix} 1 & 2 & 1 & | & 1 & 0 & 0 \\ 0 & 1 & 3 & | & 0 & 1 & 0 \\ 0 & 0 & 1 & | & 2/3 & 1/3 & -1/3 \end{smallmatrix}]$.
5. R2 <- R2 - 3*R3 and R1 <- R1 - R3 results in $[\begin{smallmatrix} 1 & 2 & 0 & | & 1/3 & -1/3 & 1/3 \\ 0 & 1 & 0 & | & -2 & 0 & 1 \\ 0 & 0 & 1 & | & 2/3 & 1/3 & -1/3 \end{smallmatrix}]$.
6. R1 <- R1 - 2*R2 results in $[\begin{smallmatrix} 1 & 0 & 0 & | & 13/3 & -1/3 & -5/3 \\ 0 & 1 & 0 & | & -2 & 0 & 1 \\ 0 & 0 & 1 & | & 2/3 & 1/3 & -1/3 \end{smallmatrix}]$.

Answer: \begin{pmatrix} 13/3 & -1/3 & -5/3 \\ -2 & 0 & 1 \\ 2/3 & 1/3 & -1/3 \end{pmatrix}

1. Set up augmented matrix $[B | I]$.
2. R2 <- R2 - 4*R1 and R3 <- R3 - 7*R1 results in $[\begin{smallmatrix} 1 & 2 & 3 & | & 1 & 0 & 0 \\ 0 & -3 & -6 & | & -4 & 1 & 0 \\ 0 & -6 & -12 & | & -7 & 0 & 1 \end{smallmatrix}]$.
3. R3 <- R3 - 2*R2 results in $[\begin{smallmatrix} 1 & 2 & 3 & | & 1 & 0 & 0 \\ 0 & -3 & -6 & | & -4 & 1 & 0 \\ 0 & 0 & 0 & | & 1 & -2 & 1 \end{smallmatrix}]$.
4. The left side contains a row of zeros, meaning the matrix is singular.

Answer: Not invertible (singular)

1. Formula: $\text{det}(C) = (2)(3) - (1)(5) = 1$. $C^{-1} = \frac{1}{1} \begin{pmatrix} 3 & -1 \\ -5 & 2 \end{pmatrix} = \begin{pmatrix} 3 & -1 \\ -5 & 2 \end{pmatrix}$.
2. Gauss-Jordan: $[\begin{smallmatrix} 2 & 1 & | & 1 & 0 \\ 5 & 3 & | & 0 & 1 \end{smallmatrix}]$.
3. R1 <- 0.5*R1 results in $[\begin{smallmatrix} 1 & 0.5 & | & 0.5 & 0 \\ 5 & 3 & | & 0 & 1 \end{smallmatrix}]$.
4. R2 <- R2 - 5*R1 results in $[\begin{smallmatrix} 1 & 0.5 & | & 0.5 & 0 \\ 0 & 0.5 & | & -2.5 & 1 \end{smallmatrix}]$.
5. R2 <- 2*R2 results in $[\begin{smallmatrix} 1 & 0.5 & | & 0.5 & 0 \\ 0 & 1 & | & -5 & 2 \end{smallmatrix}]$.
6. R1 <- R1 - 0.5*R2 results in $[\begin{smallmatrix} 1 & 0 & | & 3 & -1 \\ 0 & 1 & | & -5 & 2 \end{smallmatrix}]$.

Answer: \begin{pmatrix} 3 & -1 \\ -5 & 2 \end{pmatrix}

1. Observe the first column $c_1 = \begin{pmatrix} 3 \ 2 \end{pmatrix}$ and the second column $c_2 = \begin{pmatrix} 6 \ 4 \end{pmatrix}$.
2. Notice that $c_2 = 2 \cdot c_1$.
3. Since the columns are linearly dependent, the matrix is singular.

Answer: Singular

1. Set up the system $x + y = 0$ and $2x + 2y = 0$.
2. Both equations simplify to $y = -x$.
3. Pick $x = 1$, then $y = -1$.
4. The vector is $\begin{pmatrix} 1 \ -1 \end{pmatrix}$.

Answer: $\begin{pmatrix} 1 \ -1 \end{pmatrix}$

1. A $3 \times 3$ matrix is invertible if and only if it has three pivots (one in every row/column).
2. Since $M$ only has two pivots, the third row of the echelon form must be a row of zeros.
3. A row of zeros in the echelon form indicates that the matrix is singular.

Answer: No, it is singular.

1. Apply $R_2 \leftarrow R_2 - 4 R_1$ to $I = \begin{pmatrix} 1 & 0 \ 0 & 1 \end{pmatrix}$.
2. The first row remains $\begin{pmatrix} 1 & 0 \end{pmatrix}$. The second row becomes $\begin{pmatrix} 0 - 4(1) & 1 - 4(0) \end{pmatrix} = \begin{pmatrix} -4 & 1 \end{pmatrix}$.
3. So $E = \begin{pmatrix} 1 & 0 \ -4 & 1 \end{pmatrix}$.
4. Compute $E A = \begin{pmatrix} 1 & 0 \ -4 & 1 \end{pmatrix} \begin{pmatrix} 1 & 1 \ 4 & 5 \end{pmatrix} = \begin{pmatrix} 1(1)+0(4) & 1(1)+0(5) \\ -4(1)+1(4) & -4(1)+1(5) \end{pmatrix} = \begin{pmatrix} 1 & 1 \ 0 & 1 \end{pmatrix}$.

Answer: E = \begin{pmatrix} 1 & 0 \ -4 & 1 \end{pmatrix}, E A = \begin{pmatrix} 1 & 1 \ 0 & 1 \end{pmatrix}

1. Compare $E$ to the identity matrix $I = \begin{pmatrix} 1 & 0 \ 0 & 1 \end{pmatrix}$.
2. The rows of $I$ have been swapped. Thus, $E$ performs the operation $R_1 \leftrightarrow R_2$.
3. To undo a swap, you swap the rows back. The operation $R_1 \leftrightarrow R_2$ is its own inverse.
4. Therefore, $E^{-1} = E = \begin{pmatrix} 0 & 1 \ 1 & 0 \end{pmatrix}$.

Answer: Operation: R_1 \leftrightarrow R_2, E^{-1} = \begin{pmatrix} 0 & 1 \ 1 & 0 \end{pmatrix}

1. Step 1: $R_1 \leftarrow 2 R_1$. The elementary matrix is $E_1 = \begin{pmatrix} 2 & 0 \ 0 & 1 \end{pmatrix}$.
2. Step 2: $R_2 \leftarrow R_2 + 3 R_1$. The elementary matrix is $E_2 = \begin{pmatrix} 1 & 0 \ 3 & 1 \end{pmatrix}$.
3. The total transformation is $R = E_2 (E_1 A) = (E_2 E_1) A$.
4. Compute $E_2 E_1 = \begin{pmatrix} 1 & 0 \ 3 & 1 \end{pmatrix} \begin{pmatrix} 2 & 0 \ 0 & 1 \end{pmatrix} = \begin{pmatrix} 1(2)+0(0) & 1(0)+0(1) \\ 3(2)+1(0) & 3(0)+1(1) \end{pmatrix} = \begin{pmatrix} 2 & 0 \ 6 & 1 \end{pmatrix}$.

Answer: E_2 E_1 = \begin{pmatrix} 2 & 0 \ 6 & 1 \end{pmatrix}

1. Perform elimination on A to find U. Subtract 4 times row 1 from row 2: $8 - 4(2) = 0$ and $7 - 4(1) = 3$.
2. The resulting upper triangular matrix is $U = \begin{pmatrix} 2 & 1 \\ 0 & 3 \end{pmatrix}$.
3. The multiplier used was 4. Place this in the second row, first column of L.
4. The lower triangular matrix is $L = \begin{pmatrix} 1 & 0 \\ 4 & 1 \end{pmatrix}$.

Answer: L = \begin{pmatrix} 1 & 0 \\ 4 & 1 \end{pmatrix}, U = \begin{pmatrix} 2 & 1 \\ 0 & 3 \end{pmatrix}

1. Compute the product $L U$.
2. First row: $1(3) + 0(0) = 3$ and $1(5) + 0(4) = 5$.
3. Second row: $2(3) + 1(0) = 6$ and $2(5) + 1(4) = 14$.

Answer: A = \begin{pmatrix} 3 & 5 \\ 6 & 14 \end{pmatrix}

1. Step 1: Eliminate column 1. $R_2 \leftarrow R_2 - 2R_1$ and $R_3 \leftarrow R_3 - 1R_1$. Multipliers are $L_{21}=2, L_{31}=1$.
2. Matrix becomes $\begin{pmatrix} 1 & 2 & 1 \\ 0 & 1 & 1 \\ 0 & -2 & 0 \end{pmatrix}$.
3. Step 2: Eliminate column 2. $R_3 \leftarrow R_3 - (-2)R_2$, which is $R_3 + 2R_2$. Multiplier is $L_{32}=-2$.
4. Matrix becomes $U = \begin{pmatrix} 1 & 2 & 1 \\ 0 & 1 & 1 \\ 0 & 0 & 2 \end{pmatrix}$.
5. Step 3: Construct $L$ using multipliers: $L = \begin{pmatrix} 1 & 0 & 0 \\ 2 & 1 & 0 \\ 1 & -2 & 1 \end{pmatrix}$.

Answer: L = \begin{pmatrix} 1 & 0 & 0 \\ 2 & 1 & 0 \\ 1 & -2 & 1 \end{pmatrix}, U = \begin{pmatrix} 1 & 2 & 1 \\ 0 & 1 & 1 \\ 0 & 0 & 2 \end{pmatrix}

1. Step 1: Solve $L y = b$. $y_1 = 5$. $2(5) + y_2 = 10 \implies y_2 = 0$. $3(5) + 4(0) + y_3 = 23 \implies y_3 = 8$. So $y = [5, 0, 8]^T$.
2. Step 2: Solve $U x = y$. $1 x_3 = 8 \implies x_3 = 8$. $3 x_2 + 2(8) = 0 \implies 3 x_2 = -16 \implies x_2 = -16/3$. $2 x_1 + (-16/3) + 8 = 5 \implies 2 x_1 = 5 - 8 + 16/3 = -3 + 16/3 = 7/3 \implies x_1 = 7/6$.

Answer: x = [7/6, -16/3, 8]^T

1. The determinant of $A$ is the product of the diagonal elements of $U$.
2. $\text{det}(A) = 2 \times 4 = 8$.

Answer: 8

1. The determinant of $A$ is the product of the diagonal elements of $U$.
2. $\text{det}(A) = 2 \times 0 \times 5 = 0$.
3. Since the determinant is zero, the matrix $A$ is singular.

Answer: The matrix A is singular, meaning the system may have no solution or infinitely many solutions.

1. Check if it is diagonal: The entry $a_{13} = 2$ is not on the main diagonal, so it is not diagonal.
2. Check if it is triangular: There are non-zero entries both above ($a_{13}=2$) and below ($a_{31}=2$) the diagonal, so it is neither upper nor lower triangular.
3. Check if it is symmetric: The transpose $A^T$ has $a_{13}$ becoming $a_{31}$ and vice versa. Since $a_{13} = 2$ and $a_{31} = 2$, and all other off-diagonal pairs match, $A = A^T$.

Answer: Symmetric

1. The first row of $P$ is $[0, 0, 1]$, so the first entry of $Pv$ is $0(10) + 0(20) + 1(30) = 30$.
2. The second row of $P$ is $[1, 0, 0]$, so the second entry of $Pv$ is $1(10) + 0(20) + 0(30) = 10$.
3. The third row of $P$ is $[0, 1, 0]$, so the third entry of $Pv$ is $0(10) + 1(20) + 0(30) = 20$.

Answer: \begin{bmatrix} 30 \\ 10 \\ 20 \end{bmatrix}

1. Compute $D_1 D_2$ by multiplying diagonal entries: $\begin{bmatrix} 2 \cdot 4 & 0 \\ 0 & -3 \cdot 5 \end{bmatrix} = \begin{bmatrix} 8 & 0 \\ 0 & -15 \end{bmatrix}$.
2. Add $D_1$ to the result: $\begin{bmatrix} 8 & 0 \\ 0 & -15 \end{bmatrix} + \begin{bmatrix} 2 & 0 \\ 0 & -3 \end{bmatrix}$.
3. Sum the corresponding entries: $\begin{bmatrix} 8+2 & 0 \\ 0 & -15-3 \end{bmatrix} = \begin{bmatrix} 10 & 0 \\ 0 & -18 \end{bmatrix}$.

Answer: \begin{bmatrix} 10 & 0 \\ 0 & -18 \end{bmatrix}

1. Identify $A_{11} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$, $A_{12} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$, $B_{11} = \begin{bmatrix} 2 & 0 \\ 0 & 2 \end{bmatrix}$, and $B_{21} = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix}$.
2. Apply the block formula: $C_{11} = A_{11}B_{11} + A_{12}B_{21}$.
3. Compute $A_{11}B_{11} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \begin{bmatrix} 2 & 0 \\ 0 & 2 \end{bmatrix} = \begin{bmatrix} 2 & 0 \\ 0 & 2 \end{bmatrix}$.
4. Compute $A_{12}B_{21} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix} = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix}$.
5. Sum the results: $C_{11} = \begin{bmatrix} 2 & 0 \\ 0 & 2 \end{bmatrix} + \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix} = \begin{bmatrix} 2 & 0 \\ 0 & 2 \end{bmatrix}$.

Answer: \begin{bmatrix} 2 & 0 \\ 0 & 2 \end{bmatrix}

1. A block diagonal matrix has the form $\begin{bmatrix} A_{11} & 0 \\ 0 & A_{22} \end{bmatrix}$.
2. The product of two such matrices is $\begin{bmatrix} A_{11} & 0 \\ 0 & A_{22} \end{bmatrix} \begin{bmatrix} B_{11} & 0 \\ 0 & B_{22} \end{bmatrix}$.
3. Using block multiplication: $C_{11} = A_{11}B_{11} + 0 \cdot 0 = A_{11}B_{11}$.
4. Similarly, $C_{12} = A_{11} \cdot 0 + 0 \cdot B_{22} = 0$.
5. And $C_{21} = 0 \cdot B_{11} + A_{22} \cdot 0 = 0$.
6. Finally, $C_{22} = 0 \cdot 0 + A_{22}B_{22} = A_{22}B_{22}$.

Answer: \begin{bmatrix} A_{11}B_{11} & 0 \\ 0 & A_{22}B_{22} \end{bmatrix}

1. The formula for this product is $C = A_{11}B_{11} + A_{12}B_{21}$.
2. Substitute the given blocks: $C = I \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix} + I \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}$.
3. Since $I$ is the identity, $I M = M$ for any matrix $M$.
4. So, $C = \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix} + \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}$.
5. Perform the addition: $C = \begin{bmatrix} 1+0 & 2+1 \\ 3+1 & 4+0 \end{bmatrix} = \begin{bmatrix} 1 & 3 \\ 4 & 4 \end{bmatrix}$.

Answer: \begin{bmatrix} 1 & 3 \\ 4 & 4 \end{bmatrix}

1. Compute the determinant: $\text{det}(A) = (4)(2) - (3)(3) = 8 - 9 = -1$.
2. Find the inverse: $A^{-1} = \frac{1}{-1} \begin{bmatrix} 2 & -3 \\ -3 & 4 \end{bmatrix} = \begin{bmatrix} -2 & 3 \\ 3 & -4 \end{bmatrix}$.
3. Solve for $x$: $x = A^{-1}b = \begin{bmatrix} -2 & 3 \\ 3 & -4 \end{bmatrix} \begin{bmatrix} 1 \\ 0 \end{bmatrix} = \begin{bmatrix} -2 \\ 3 \end{bmatrix}$.

Answer: x = \begin{bmatrix} -2 \\ 3 \end{bmatrix}

1. Use block multiplication: $MN = \begin{bmatrix} AC & 0 \\ 0 & BD \end{bmatrix}$.
2. Compute $AC = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \begin{bmatrix} 3 & 0 \\ 0 & 3 \end{bmatrix} = \begin{bmatrix} 3 & 0 \\ 0 & 3 \end{bmatrix}$.
3. Compute $BD = \begin{bmatrix} 2 & 0 \\ 0 & 2 \end{bmatrix} \begin{bmatrix} 4 & 0 \\ 0 & 4 \end{bmatrix} = \begin{bmatrix} 8 & 0 \\ 0 & 8 \end{bmatrix}$.
4. Assemble the result: $MN = \begin{bmatrix} 3 & 0 & 0 & 0 \\ 0 & 3 & 0 & 0 \\ 0 & 0 & 8 & 0 \\ 0 & 0 & 0 & 8 \end{bmatrix}$.

Answer: \begin{bmatrix} 3 & 0 & 0 & 0 \\ 0 & 3 & 0 & 0 \\ 0 & 0 & 8 & 0 \\ 0 & 0 & 0 & 8 \end{bmatrix}

1. Eliminate the 8 in the second row: $R_2 \leftarrow R_2 - 4 R_1$.
2. The resulting upper triangular matrix is $U = \begin{bmatrix} 2 & 1 \\ 0 & 3 \end{bmatrix}$.
3. The multiplier was 4, so the lower triangular matrix is $L = \begin{bmatrix} 1 & 0 \\ 4 & 1 \end{bmatrix}$.
4. Verify: $LU = \begin{bmatrix} 1 & 0 \\ 4 & 1 \end{bmatrix} \begin{bmatrix} 2 & 1 \\ 0 & 3 \end{bmatrix} = \begin{bmatrix} (1)(2)+(0)(0) & (1)(1)+(0)(3) \\ (4)(2)+(1)(0) & (4)(1)+(1)(3) \end{bmatrix} = \begin{bmatrix} 2 & 1 \\ 8 & 7 \end{bmatrix}$.

Answer: L = \begin{bmatrix} 1 & 0 \\ 4 & 1 \end{bmatrix}, U = \begin{bmatrix} 2 & 1 \\ 0 & 3 \end{bmatrix}

1. First, compute the determinant of $A$: $\det(A) = (4 \times 1) - (1 \times 2) = 4 - 2 = 2$.
2. The area scaling factor is the absolute value of the determinant, which is $|2| = 2$.
3. Multiply the original area by the scaling factor: $7 \times 2 = 14$.

Answer: 14

1. Compute the determinant: $\det(M) = (2 \times 3) - (5 \times 1) = 6 - 5 = 1$.
2. Since the determinant is positive, the orientation is preserved.
3. Since the absolute value of the determinant is $1$, the area is preserved.

Answer: Yes to both

1. The area scaling factor is $|\det(A)| = 4$.
2. Since the orientation is reversed, $\det(A)$ must be $-4$.
3. Set up the equation: $(a \times 1) - (1 \times 2) = -4$.
4. Solve for $a$: $a - 2 = -4$, so $a = -2$.

Answer: -2

1. Identify the entries: $a=5, b=2, c=3, d=4$.
2. Apply the 2x2 formula: $\det(A) = ad - bc$.
3. Calculate the products: $5 \times 4 = 20$ and $2 \times 3 = 6$.
4. Subtract the results: $20 - 6 = 14$.

Answer: 14

1. Calculate downward diagonals: $(1 \times 1 \times 1) + (2 \times 3 \times 2) + (1 \times 0 \times 0) = 1 + 12 + 0 = 13$.
2. Calculate upward diagonals: $(2 \times 1 \times 1) + (0 \times 3 \times 1) + (1 \times 0 \times 2) = 2 + 0 + 0 = 2$.
3. Subtract the upward sum from the downward sum: $13 - 2 = 11$.

Answer: 11

1. Compute the determinant: $\det(C) = (0 \times 0) - (1 \times 1)$.
2. The result is $0 - 1 = -1$.
3. Since the determinant is negative, the transformation reverses the orientation of the plane.

Answer: Reverses orientation because $\det(C) = -1$.

1. Observe that the second row $[5, 10]$ is $2.5$ times the first row $[2, 4]$.
2. Since the rows are linearly dependent, the vectors are collinear.
3. The area of the resulting parallelogram is zero.

Answer: 0

1. Starting determinant is $5$.
2. Swapping two rows multiplies the determinant by $-1$, giving $-5$.
3. Multiplying one row by $3$ multiplies the determinant by $3$, giving $-5 \times 3 = -15$.

Answer: -15

1. Perform the operation: $R_2$ becomes $[3 - 3(1), 7 - 3(2)] = [0, 1]$.
2. The new matrix is $\begin{bmatrix} 1 & 2 \\ 0 & 1 \end{bmatrix}$.
3. The row addition property states the determinant remains unchanged.
4. The determinant of the resulting matrix is $1 \times 1 - 2 \times 0 = 1$.

Answer: 1

1. Choose the second column for expansion because it has two zeros.
2. The only non-zero element is $a_{32} = 5$.
3. The cofactor sign for row 3, column 2 is $(-1)^{3+2} = -1$.
4. The minor $M_{32}$ is the determinant of $\begin{bmatrix} 2 & 1 \\ 3 & 4 \end{bmatrix}$, which is $(2 \cdot 4) - (1 \cdot 3) = 8 - 3 = 5$.
5. The determinant is $a_{32} \cdot C_{32} = 5 \cdot (-1 \cdot 5) = -25$.

Answer: -25

1. Expand along row 1: elements are 1 and 2.
2. First term: $1 \cdot C_{11} = 1 \cdot (-1)^{1+1} \cdot \det[4] = 1 \cdot 1 \cdot 4 = 4$.
3. Second term: $2 \cdot C_{12} = 2 \cdot (-1)^{1+2} \cdot \det[3] = 2 \cdot (-1) \cdot 3 = -6$.
4. Sum: $4 - 6 = -2$.

Answer: -2

1. Expand along column 1: elements are 1, 2, 0.
2. Term 1: $1 \cdot C_{11} = 1 \cdot (-1)^{1+1} \cdot \det \begin{bmatrix} 1 & 0 \\ 4 & 5 \end{bmatrix} = 1 \cdot 1 \cdot (5 - 0) = 5$.
3. Term 2: $2 \cdot C_{21} = 2 \cdot (-1)^{2+1} \cdot \det \begin{bmatrix} 0 & 3 \\ 4 & 5 \end{bmatrix} = 2 \cdot (-1) \cdot (0 - 12) = -2 \cdot (-12) = 24$.
4. Term 3: $0 \cdot C_{31} = 0$.
5. Sum: $5 + 24 + 0 = 29$.

Answer: 29

1. Step 1: Replace $R_2$ with $R_2 - 0.5 R_1$. The matrix becomes $\begin{bmatrix} 2 & 1 & 0 \\ 0 & 1.5 & 1 \\ 0 & 1 & 2 \end{bmatrix}$. Determinant is unchanged.
2. Step 2: Replace $R_3$ with $R_3 - (1/1.5) R_2$, which is $R_3 - (2/3) R_2$. The matrix becomes $\begin{bmatrix} 2 & 1 & 0 \\ 0 & 1.5 & 1 \\ 0 & 0 & 2 - 2/3 \end{bmatrix} = \begin{bmatrix} 2 & 1 & 0 \\ 0 & 1.5 & 1 \\ 0 & 0 & 4/3 \end{bmatrix}$. Determinant is unchanged.
3. Step 3: Multiply the diagonal elements: $2 \cdot 1.5 \cdot (4/3) = 2 \cdot (3/2) \cdot (4/3) = 4$.

Answer: 4

1. Step 1: Swap $R_1$ and $R_2$ to get a non-zero pivot. The matrix becomes $\begin{bmatrix} 3 & 4 \\ 0 & 2 \end{bmatrix}$. This introduces a factor of $-1$.
2. Step 2: The matrix is now upper triangular. The product of the pivots is $3 \cdot 2 = 6$.
3. Step 3: Apply the swap factor: $(-1) \cdot 6 = -6$.

Answer: -6

1. Step 1: Replace $R_2$ with $R_2 - 2 R_1$. The matrix becomes $\begin{bmatrix} 1 & 2 & 3 \\ 0 & 0 & 0 \\ 1 & 0 & 1 \end{bmatrix}$.
2. Step 2: We observe that the second row has become all zeros.
3. Step 3: Any triangular matrix with a zero on the diagonal has a determinant of $0$. Therefore, $\det(C) = 0$ and the matrix is singular.

Answer: 0 (Singular)

1. Compute $\det(A) = (2)(3) - (1)(0) = 6$.
2. Compute $\det(B) = (1)(2) - (0)(4) = 2$.
3. Apply the property $\det(AB) = \det(A)\det(B) = 6 \times 2 = 12$.

Answer: 12

1. Use the power property $\det(M^k) = (\det(M))^k$.
2. Substitute $k = 3$ and $\det(M) = -5$.
3. Calculate $(-5)^3 = (-5) \times (-5) \times (-5) = -125$.

Answer: -125

1. Use the property $\det(A^2 B^{-1}) = \det(A)^2 \times \det(B^{-1})$.
2. Calculate $\det(A)^2 = 2^2 = 4$.
3. Calculate $\det(B^{-1}) = 1/\det(B) = 1/0.5 = 2$.
4. Multiply the results: $4 \times 2 = 8$.

Answer: 8

1. Identify $A = \begin{bmatrix} 3 & 2 \\ 4 & -1 \end{bmatrix}$ and $b = \begin{bmatrix} 7 \\ 2 \end{bmatrix}$.
2. Compute $\det(A) = (3)(-1) - (2)(4) = -3 - 8 = -11$.
3. Construct $A_1 = \begin{bmatrix} 7 & 2 \\ 2 & -1 \end{bmatrix}$.
4. Compute $\det(A_1) = (7)(-1) - (2)(2) = -7 - 4 = -11$.
5. Calculate $x = \frac{-11}{-11} = 1$.

Answer: x = 1

1. $\det(A) = (1)(3) - (1)(2) = 1$.
2. $\det(A_1) = (5)(3) - (1)(13) = 15 - 13 = 2$.
3. $\det(A_2) = (1)(13) - (5)(2) = 13 - 10 = 3$.
4. Calculate $x = 2/1 = 2$ and $y = 3/1 = 3$.

Answer: x = 2, y = 3

1. The coefficient matrix is $A = \begin{bmatrix} a & b \\ c & d \end{bmatrix}$.
2. The determinant is $\det(A) = ad - bc$.
3. The modified matrix for $x$ is $A_1 = \begin{bmatrix} e & b \\ f & d \end{bmatrix}$.
4. The determinant is $\det(A_1) = ed - bf$.
5. The formula is $x = \frac{ed - bf}{ad - bc}$.

Answer: x = \frac{ed - bf}{ad - bc}

1. Compute the determinant: $\det(A) = (4)(3) - (6)(2)$
2. Calculate the values: $12 - 12 = 0$
3. Since the determinant is zero, the matrix is singular.

Answer: Not invertible (singular) because $\det(A) = 0$.

1. The determinant $\det(B) = -5$ is non zero.
2. A non zero determinant implies the matrix $B$ is invertible.
3. If $B$ is invertible, we can multiply $Bx = 0$ by $B^{-1}$ on the left.
4. This gives $x = B^{-1} 0$, which simplifies to $x = 0$.

Answer: Yes, because $\det(B) \neq 0$ implies $B$ is invertible, ensuring a unique solution.

1. A matrix is singular if its determinant is zero.
2. Set up the determinant equation: $(k)(6) - (2)(3) = 0$
3. Simplify the equation: $6k - 6 = 0$
4. Solve for $k$: $6k = 6$, so $k = 1$.

Answer: $k = 1$

1. Notice that the second column has two zeros.
2. Expand along the second column. The only non-zero entry is $4$ at position $(3,2)$.
3. The sign for position $(3,2)$ is negative.
4. The determinant is $-4 \times \det \begin{pmatrix} 2 & 1 \\ 5 & 3 \end{pmatrix}$.
5. Calculate the $2 \times 2$ determinant: $(2 \times 3) - (1 \times 5) = 6 - 5 = 1$.
6. The final result is $-4 \times 1 = -4$.

Answer: -4

1. First, find the determinant of the base matrix $B$.
2. $\det(B) = (1 \times 4) - (2 \times 3) = 4 - 6 = -2$.
3. Use the property $\det(B^k) = (\det(B))^k$.
4. The determinant is $(-2)^3$.
5. Calculate $(-2) \times (-2) \times (-2) = -8$.

Answer: -8

1. Observe the rows of the matrix.
2. Row 2 is exactly $2$ times Row 1.
3. Since the rows are linearly dependent, the determinant must be $0$.
4. Because $\det(C) = 0$, the matrix is singular.
5. Therefore, the matrix is not invertible.

Answer: Not invertible (\det(C) = 0)

1. Notice that the matrix is lower triangular.
2. The determinant of a triangular matrix is the product of its diagonal entries.
3. Calculate $2 \times 3 \times 2 = 12$.

Answer: 12

1. The transpose does not change the determinant: $\det(A^T) = \det(A) = -2$.
2. For a $3 \times 3$ matrix, $\det(kA) = k^3 \det(A)$.
3. Substitute $k=5$ and $\det(A^T)=-2$: $5^3 \times (-2) = 125 \times (-2) = -250$.

Answer: -250

1. Compute the determinant: $\det(M) = (1 \times 6) - (2 \times 3)$.
2. Calculate $6 - 6 = 0$.
3. Since the determinant is zero, the matrix is singular and therefore not invertible.

Answer: Not invertible because $\det(M) = 0$

1. Check closure under addition: if $u = [x_1, y_1, z_1]$ and $v = [x_2, y_2, z_2]$ are in $V$, then $x_1+y_1+z_1=0$ and $x_2+y_2+z_2=0$. Their sum is $[x_1+x_2, y_1+y_2, z_1+z_2]$. The sum of these components is $(x_1+y_1+z_1) + (x_2+y_2+z_2) = 0 + 0 = 0$. So it is closed under addition.
2. Check closure under scalar multiplication: $c u = [cx_1, cy_1, cz_1]$. The sum of components is $c(x_1+y_1+z_1) = c(0) = 0$. So it is closed under scalar multiplication.
3. Check for zero vector: $[0, 0, 0]$ satisfies $0+0+0=0$, so it is in $V$.
4. Since it is a subset of $\mathbb{R}^3$ and satisfies these properties, it is a vector space.

Answer: Yes, it is a vector space.

1. Check closure under addition. Let $A = \begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix}$ and $B = \begin{bmatrix} 0 & 0 \\ 0 & 1 \end{bmatrix}$.
2. Both have $\det(A)=0$ and $\det(B)=0$, so they are in the set.
3. The sum is $A+B = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$.
4. The determinant of the sum is $\det(A+B) = 1 \cdot 1 - 0 \cdot 0 = 1$.
5. Since the sum is not in the set, it is not closed under addition.

Answer: No, it is not a vector space.

1. Check for the zero vector. The zero vector in the space of functions is the function $z(x) = 0$ for all $x$.
2. For this function, $z(0) = 0$.
3. However, the set requires $f(0) = 1$.
4. Since the zero function is not in the set, it cannot be a vector space.

Answer: No, it is not a vector space.

1. Check zero vector: $2(0) - 0 = 0$, so $0 \in S$.
2. Check addition: If $2u_1 - u_2 = 0$ and $2v_1 - v_2 = 0$, then $2(u_1+v_1) - (u_2+v_2) = (2u_1-u_2) + (2v_1-v_2) = 0 + 0 = 0$.
3. Check scalar multiplication: If $2x - y = 0$, then $2(cx) - (cy) = c(2x-y) = c(0) = 0$.

Answer: Yes, it is a subspace.

1. Check zero vector: $0^2 = 0$, so $0 \in S$.
2. Check scalar multiplication: Let $v = \begin{bmatrix} 1 \\ 1 \end{bmatrix}$. This is in $S$ because $1^2 = 1$.
3. Multiply by $c=2$: $2v = \begin{bmatrix} 2 \\ 2 \end{bmatrix}$. Check if $2^2 = 2$. Since $4 \neq 2$, the vector $2v$ is not in $S$.

Answer: No, it is not a subspace.

1. A vector $v$ is in $S$ if $n \cdot v = 0$.
2. Zero vector: $n \cdot 0 = 0$, so $0 \in S$.
3. Addition: If $n \cdot u = 0$ and $n \cdot v = 0$, then $n \cdot (u+v) = n \cdot u + n \cdot v = 0 + 0 = 0$.
4. Scalar multiplication: If $n \cdot v = 0$, then $n \cdot (cv) = c(n \cdot v) = c(0) = 0$.

Answer: Yes, it is a subspace.

1. Perform row reduction: $R_2 \leftarrow R_2 - 2R_1$ gives $\begin{pmatrix} 1 & -3 \\ 0 & 0 \end{pmatrix}$.
2. Identify variables: $x_1$ is a pivot variable, $x_2$ is a free variable.
3. Solve for $x_1$: $x_1 - 3x_2 = 0 \implies x_1 = 3x_2$.
4. Write in vector form: $x = \begin{pmatrix} 3x_2 \\ x_2 \end{pmatrix} = x_2 \begin{pmatrix} 3 \\ 1 \end{pmatrix}$.

Answer: \text{Null}(A) = \text{span} \left\{ \begin{pmatrix} 3 \\ 1 \end{pmatrix} \right\}

1. The matrix is already in RREF.
2. Pivot variables are $x_1$ and $x_2$; the free variable is $x_3$.
3. From row 1: $x_1 + 2x_3 = 0 \implies x_1 = -2x_3$.
4. From row 2: $x_2 - x_3 = 0 \implies x_2 = x_3$.
5. The general solution is $x = x_3 \begin{pmatrix} -2 \\ 1 \\ 1 \end{pmatrix}$.

Answer: \left\{ \begin{pmatrix} -2 \\ 1 \\ 1 \end{pmatrix} \right\}

1. The matrix is in RREF. Pivots are in columns 1 and 3.
2. Free variables are $x_2$ and $x_4$.
3. Row 1: $x_1 + 2x_2 + x_4 = 0 \implies x_1 = -2x_2 - x_4$.
4. Row 2: $x_3 + 3x_4 = 0 \implies x_3 = -3x_4$.
5. Vector form: $x = x_2 \begin{pmatrix} -2 \\ 1 \\ 0 \\ 0 \end{pmatrix} + x_4 \begin{pmatrix} -1 \\ 0 \\ -3 \\ 1 \end{pmatrix}$.

Answer: \text{Null}(A) = \text{span} \left\{ \begin{pmatrix} -2 \\ 1 \\ 0 \\ 0 \end{pmatrix}, \begin{pmatrix} -1 \\ 0 \\ -3 \\ 1 \end{pmatrix} \right\}

1. We check if $x_1 \begin{bmatrix} 1 \\ 3 \\ 5 \end{bmatrix} + x_2 \begin{bmatrix} 2 \\ 4 \\ 6 \end{bmatrix} = \begin{bmatrix} 1 \\ 2 \\ 3 \end{bmatrix}$.
2. From the first row: $x_1 + 2x_2 = 1$.
3. From the second row: $3x_1 + 4x_2 = 2$.
4. Subtracting 3 times the first from the second: $(3-3)x_1 + (4-6)x_2 = 2-3$, so $-2x_2 = -1$, which means $x_2 = 0.5$.
5. Substituting back: $x_1 + 2(0.5) = 1$, so $x_1 = 0$.
6. Check the third row: $5(0) + 6(0.5) = 3$. This matches the third component of $b$.

Answer: Yes, $b$ is in the column space because $b = 0 v_1 + 0.5 v_2$.

1. Perform row reduction: $R_3 \to R_3 - R_1$ gives $\begin{bmatrix} 1 & 2 & 3 \\ 0 & 1 & 1 \\ 0 & 1 & 1 \end{bmatrix}$.
2. Next, $R_3 \to R_3 - R_2$ gives $\begin{bmatrix} 1 & 2 & 3 \\ 0 & 1 & 1 \\ 0 & 0 & 0 \end{bmatrix}$.
3. The pivots are in the first and second columns.
4. The basis for the column space consists of the first and second columns of the original matrix.

Answer: \left\{ \begin{bmatrix} 1 \\ 0 \\ 1 \end{bmatrix}, \begin{bmatrix} 2 \\ 1 \\ 3 \end{bmatrix} \right\}

1. The column space is a subspace of $\mathbb{R}^4$ because the columns have 4 components.
2. The dimension of the column space is equal to the rank of the matrix.
3. Since the rank is 2, the dimension of the column space is 2.

Answer: The column space is a 2-dimensional plane passing through the origin in 4-dimensional space.

1. Check if $v_2$ is a scalar multiple of $v_1$.
2. Observe that $2 \times 2 = 4$ and $2 \times 3 = 6$.
3. Thus, $v_2 = 2 v_1$.
4. Since one vector is a multiple of the other, they are linearly dependent.

Answer: Linearly dependent

1. Stack the vectors as rows of a matrix $A = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 1 & 1 & 0 \end{pmatrix}$.
2. Perform row reduction: $R_3 \leftarrow R_3 - R_1$ gives $\begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 1 & 0 \end{pmatrix}$.
3. Then $R_3 \leftarrow R_3 - R_2$ gives $\begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 0 \end{pmatrix}$.
4. The presence of a zero row indicates linear dependence.

Answer: Linearly dependent

1. Form the matrix $A = \begin{pmatrix} 1 & 2 & 1 \\ 0 & 1 & 1 \\ 1 & 0 & -1 \end{pmatrix}$.
2. Compute the determinant: $1(1(-1) - 1(0)) - 2(0(-1) - 1(1)) + 1(0(0) - 1(1))$.
3. This equals $1(-1) - 2(-1) + 1(-1) = -1 + 2 - 1 = 0$.
4. Since the determinant is zero, the vectors are linearly dependent.

Answer: Linearly dependent

1. Form the matrix $A = \begin{bmatrix} 1 & 1 & 0 \\ 0 & 1 & 1 \\ 1 & 0 & 1 \end{bmatrix}$.
2. Perform $R_3 \leftarrow R_3 - R_1$ to get $\begin{bmatrix} 1 & 1 & 0 \\ 0 & 1 & 1 \\ 0 & -1 & 1 \end{bmatrix}$.
3. Perform $R_3 \leftarrow R_3 + R_2$ to get $\begin{bmatrix} 1 & 1 & 0 \\ 0 & 1 & 1 \\ 0 & 0 & 2 \end{bmatrix}$.
4. There are 3 pivots for 3 vectors.

Answer: Linearly Independent

1. Form the matrix $A = \begin{bmatrix} 1 & 3 \\ 2 & 6 \end{bmatrix}$.
2. Perform $R_2 \leftarrow R_2 - 2R_1$ to get $\begin{bmatrix} 1 & 3 \\ 0 & 0 \end{bmatrix}$.
3. There is only 1 pivot, but there are 2 vectors.

Answer: Linearly Dependent

1. Form the matrix $A = \begin{bmatrix} 1 & 1 & 1 \\ 0 & 1 & 1 \\ 0 & 0 & 1 \end{bmatrix}$.
2. The matrix is already in row echelon form.
3. There are pivots in every column (the diagonal entries are all 1).
4. Since the number of pivots equals the number of vectors, they are independent.

Answer: Linearly Independent

1. Form a matrix $A$ with the vectors as columns: $A = \begin{bmatrix} 2 & 1 \\ 1 & 3 \end{bmatrix}$.
2. Calculate the determinant: $\det(A) = (2)(3) - (1)(1) = 6 - 1 = 5$.
3. Since the determinant is non-zero, the vectors are linearly independent.
4. In $\mathbb{R}^2$, any two linearly independent vectors span the space.
5. Therefore, the set is a basis.

Answer: Yes, it is a basis.

1. Set up the equation $c_1 \begin{bmatrix} 1 \\ 2 \end{bmatrix} + c_2 \begin{bmatrix} 1 \\ 0 \end{bmatrix} = \begin{bmatrix} 4 \\ 10 \end{bmatrix}$.
2. From the second component: $2c_1 + 0c_2 = 10$, which implies $c_1 = 5$.
3. From the first component: $1c_1 + 1c_2 = 4$.
4. Substitute $c_1 = 5$: $5 + c_2 = 4$, which implies $c_2 = -1$.
5. The coordinates are $c_1 = 5$ and $c_2 = -1$.

Answer: \begin{bmatrix} 5 \\ -1 \end{bmatrix}

1. Check for linear independence by calculating the determinant of the matrix $A = \begin{bmatrix} 1 & 1 & 2 \\ 0 & 1 & 1 \\ 0 & 0 & 0 \end{bmatrix}$.
2. The determinant of an upper triangular matrix is the product of the diagonal: $1 \cdot 1 \cdot 0 = 0$.
3. Since the determinant is zero, the vectors are linearly dependent.
4. A basis must be linearly independent.
5. Therefore, the set is not a basis.

Answer: No, it is not a basis.

1. Solve for one variable in terms of the others: $y = 2x + 3z$.
2. Write a general vector in $V$: $(x, 2x + 3z, z)$.
3. Decompose into linear combinations: $x(1, 2, 0) + z(0, 3, 1)$.
4. The vectors $(1, 2, 0)$ and $(0, 3, 1)$ are linearly independent and span $V$.
5. The basis has 2 vectors.

Answer: 2

1. Use the Rank-Nullity Theorem: $\text{rank}(A) + \text{dim}(\text{Null } A) = n$, where $n$ is the number of columns.
2. Substitute the known values: $3 + \text{dim}(\text{Null } A) = 7$.
3. Solve for the dimension of the null space: $7 - 3 = 4$.

Answer: 4

1. Check for linear independence: $v_3 = v_1 + v_2$.
2. Since $v_3$ is a linear combination of $v_1$ and $v_2$, the set is linearly dependent.
3. The spanning set can be reduced to the linearly independent set $\{v_1, v_2\}$.
4. This set forms a basis for the subspace.
5. The number of vectors in the basis is 2.

Answer: 2

1. Perform row reduction: $R2 \to R2 - 2R1$ and $R3 \to R3 - R1$ gives $\begin{pmatrix} 1 & 3 & 1 \\ 0 & 0 & -2 \\ 0 & 0 & -2 \end{pmatrix}$.
2. Further reduce to RREF: $R2 \to -0.5 R2$ and $R3 \to R3 - R2$ and $R1 \to R1 - R2$ gives $\begin{pmatrix} 1 & 3 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{pmatrix}$.
3. Pivots are in columns 1 and 3. The basis for $\text{Col}(A)$ is the 1st and 3rd columns of the original matrix: $\left\{ \begin{pmatrix} 1 \\ 2 \\ 1 \end{pmatrix}, \begin{pmatrix} 1 \\ 0 \\ -1 \end{pmatrix} \right\}$.
4. For the null space, $x_2$ is free. $x_1 + 3x_2 = 0 \implies x_1 = -3x_2$ and $x_3 = 0$.
5. Setting $x_2 = 1$ gives the special solution $\begin{pmatrix} -3 \\ 1 \\ 0 \end{pmatrix}$. This is the basis for $\text{Null}(A)$.

Answer: \text{Col basis: } \left\{ \begin{pmatrix} 1 \\ 2 \\ 1 \end{pmatrix}, \begin{pmatrix} 1 \\ 0 \\ -1 \end{pmatrix} \right\}, \text{Null basis: } \left\{ \begin{pmatrix} -3 \\ 1 \\ 0 \end{pmatrix} \right\}

1. The rank of a matrix is the dimension of its column space, so $\text{dim}(\text{Col } A) = 2$.
2. The rank-nullity theorem states that $\text{dim}(\text{Col } A) + \text{dim}(\text{Null } A) = n$, where $n$ is the number of columns.
3. Here $n = 4$, so $2 + \text{dim}(\text{Null } A) = 4$.
4. Therefore, $\text{dim}(\text{Null } A) = 2$.

Answer: 2 vectors

1. Reduce to RREF: $R2 \to R2 - 2R1$ gives $\begin{pmatrix} 1 & 2 \\ 0 & 0 \end{pmatrix}$.
2. The pivot is in column 1, so $x_2$ is a free variable.
3. The equation is $x_1 + 2x_2 = 0$, which means $x_1 = -2x_2$.
4. Setting $x_2 = 1$ gives the vector $\begin{pmatrix} -2 \\ 1 \end{pmatrix}$.

Answer: \left\{ \begin{pmatrix} -2 \\ 1 \end{pmatrix} \right\}

1. Perform row operations: $R_2 \leftarrow R_2 - 2R_1$ and $R_3 \leftarrow R_3 - 3R_1$.
2. The resulting matrix is $\begin{bmatrix} 1 & 2 & 1 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix}$.
3. There is only one non-zero row, so the rank is 1.
4. The basis for the row space is the set containing the single non-zero row.

Answer: \text{rank}(A) = 1, \text{Basis}(\text{Row } A) = \{ (1, 2, 1) \}

1. The number of pivots is given as 3.
2. The dimension of the column space is equal to the number of pivots.
3. The dimension of the row space is also equal to the number of pivots.
4. Therefore, both dimensions are 3.

Answer: \text{dim}(\text{Row } B) = 3, \text{dim}(\text{Col } B) = 3

1. Perform row operations: $R_3 \leftarrow R_3 - R_1$.
2. The matrix becomes $\begin{bmatrix} 1 & 0 & 1 \\ 0 & 1 & 1 \\ 0 & 1 & 1 \end{bmatrix}$.
3. Perform $R_3 \leftarrow R_3 - R_2$.
4. The RREF is $\begin{bmatrix} 1 & 0 & 1 \\ 0 & 1 & 1 \\ 0 & 0 & 0 \end{bmatrix}$.
5. The non-zero rows are $(1, 0, 1)$ and $(0, 1, 1)$.

Answer: \text{Basis}(\text{Row } C) = \{ (1, 0, 1), (0, 1, 1) \}

1. Identify the number of columns $n = 6$.
2. Identify the rank $\text{rank}(A) = 4$.
3. Apply the Rank-Nullity Theorem: $\text{rank}(A) + \text{nullity}(A) = n$.
4. Substitute the values: $4 + \text{nullity}(A) = 6$.
5. Solve for nullity: $\text{nullity}(A) = 6 - 4 = 2$.

Answer: 2

1. The null space is spanned by $3$ linearly independent vectors, so $\text{nullity}(A) = 3$.
2. The number of columns $n$ is $7$.
3. Apply the Rank-Nullity Theorem: $\text{rank}(A) + 3 = 7$.
4. Solve for rank: $\text{rank}(A) = 7 - 3 = 4$.

Answer: 4

1. If $Ax = 0$ has only the trivial solution, the null space contains only the zero vector.
2. The dimension of the zero subspace is $0$, so $\text{nullity}(A) = 0$.
3. The number of columns $n$ is $3$.
4. Apply the Rank-Nullity Theorem: $\text{rank}(A) + 0 = 3$.
5. Therefore, $\text{rank}(A) = 3$.

Answer: 3

1. Perform row reduction: $R3 \leftarrow R3 - R1$ gives $\begin{bmatrix} 1 & 2 & 0 & 1 \\ 0 & 1 & 1 & 0 \\ 0 & 1 & 1 & 0 \end{bmatrix}$.
2. Further reduction: $R3 \leftarrow R3 - R2$ gives $\begin{bmatrix} 1 & 2 & 0 & 1 \\ 0 & 1 & 1 & 0 \\ 0 & 0 & 0 & 0 \end{bmatrix}$.
3. There are 2 pivots, so $\text{rank}(A) = r = 2$.
4. The matrix has $m=3$ rows and $n=4$ columns.
5. $\text{dim}(C(A)) = r = 2$ and $\text{dim}(C(A^T)) = r = 2$.
6. $\text{dim}(N(A)) = n - r = 4 - 2 = 2$ and $\text{dim}(N(A^T)) = m - r = 3 - 2 = 1$.

Answer: \text{dim}(C(A))=2, \text{dim}(C(A^T))=2, \text{dim}(N(A))=2, \text{dim}(N(A^T))=1

1. We are given $n = 12$, $m = 7$, and $\text{dim}(N(A)) = 5$.
2. Using the rank-nullity theorem: $r + 5 = 12$, so $r = 7$.
3. The dimension of the column space $C(A)$ is $r = 7$.
4. The dimension of the row space $C(A^T)$ is $r = 7$.
5. The dimension of the left null space $N(A^T)$ is $m - r = 7 - 7 = 0$.

Answer: \text{dim}(C(A))=7, \text{dim}(C(A^T))=7, \text{dim}(N(A^T))=0

1. If $N(A^T) = \{0\}$, then $\text{dim}(N(A^T)) = 0$.
2. Since $\text{dim}(N(A^T)) = m - r$ and $m = n$, we have $n - r = 0$, which means $r = n$.
3. Since $r = n$, the column space $C(A)$ has dimension $n$, meaning $C(A) = \mathbb{R}^n$.
4. The row space $C(A^T)$ also has dimension $n$, meaning $C(A^T) = \mathbb{R}^n$.
5. The null space $N(A)$ has dimension $n - r = n - n = 0$, meaning $N(A) = \{0\}$.

Answer: All four subspaces are determined: $C(A)=\mathbb{R}^n, C(A^T)=\mathbb{R}^n, N(A)=\{0\}, N(A^T)=\{0\}$

1. Set up the equation $\begin{bmatrix} 1 & 2 \\ 1 & 0 \end{bmatrix} \begin{bmatrix} c_1 \\ c_2 \end{bmatrix} = \begin{bmatrix} 5 \\ 2 \end{bmatrix}$.
2. From the second row, $1 c_1 + 0 c_2 = 2$, so $c_1 = 2$.
3. Substitute $c_1 = 2$ into the first row: $2 + 2 c_2 = 5$.
4. Subtract 2 from both sides: $2 c_2 = 3$.
5. Divide by 2: $c_2 = 1.5$.

Answer: $[v]_B = \begin{bmatrix} 2 \\ 1.5 \end{bmatrix}$

1. Set up the system $\begin{bmatrix} 1 & 0 & 1 \\ 0 & 1 & 1 \\ 1 & 1 & 0 \end{bmatrix} \begin{bmatrix} c_1 \\ c_2 \\ c_3 \end{bmatrix} = \begin{bmatrix} 2 \\ 2 \\ 3 \end{bmatrix}$.
2. Subtract row 1 from row 3: $\begin{bmatrix} 1 & 0 & 1 \\ 0 & 1 & 1 \\ 0 & 1 & -1 \end{bmatrix} \begin{bmatrix} c_1 \\ c_2 \\ c_3 \end{bmatrix} = \begin{bmatrix} 2 \\ 2 \\ 1 \end{bmatrix}$.
3. Subtract row 2 from row 3: $\begin{bmatrix} 1 & 0 & 1 \\ 0 & 1 & 1 \\ 0 & 0 & -2 \end{bmatrix} \begin{bmatrix} c_1 \\ c_2 \\ c_3 \end{bmatrix} = \begin{bmatrix} 2 \\ 2 \\ -1 \end{bmatrix}$.
4. Solve for $c_3$: $-2 c_3 = -1 \implies c_3 = 0.5$.
5. Solve for $c_2$: $c_2 + 0.5 = 2 \implies c_2 = 1.5$.
6. Solve for $c_1$: $c_1 + 0.5 = 2 \implies c_1 = 1.5$.

Answer: $[v]_B = \begin{bmatrix} 1.5 \\ 1.5 \\ 0.5 \end{bmatrix}$

1. The coordinates relative to the standard basis are simply the vector $v$ itself.
2. Use the formula $v = P [v]_B$.
3. Calculate $v = 3 \begin{bmatrix} 1 \\ 2 \end{bmatrix} + (-2) \begin{bmatrix} 0 \\ 1 \end{bmatrix}$.
4. Compute the components: $x = 3(1) + (-2)(0) = 3$ and $y = 3(2) + (-2)(1) = 6 - 2 = 4$.
5. The vector is $\begin{bmatrix} 3 \\ 4 \end{bmatrix}$.

Answer: $[v]_{std} = \begin{bmatrix} 3 \\ 4 \end{bmatrix}$

1. Find coordinates of $\begin{bmatrix} 1 \\ 2 \end{bmatrix}$ relative to $B_2$: $1\begin{bmatrix} 1 \\ 1 \end{bmatrix} + 1\begin{bmatrix} 0 \\ 1 \end{bmatrix} = \begin{bmatrix} 1 \\ 2 \end{bmatrix}$, so the first column is $\begin{bmatrix} 1 \\ 1 \end{bmatrix}$.
2. Find coordinates of $\begin{bmatrix} 3 \\ 5 \end{bmatrix}$ relative to $B_2$: $3\begin{bmatrix} 1 \\ 1 \end{bmatrix} + 2\begin{bmatrix} 0 \\ 1 \end{bmatrix} = \begin{bmatrix} 3 \\ 5 \end{bmatrix}$, so the second column is $\begin{bmatrix} 3 \\ 2 \end{bmatrix}$.
3. Assemble the matrix.

Answer: \begin{bmatrix} 1 & 3 \\ 1 & 2 \end{bmatrix}

1. Use the formula $[v]_{B_2} = P_{B_2 \leftarrow B_1} [v]_{B_1}$.
2. Compute the product $\begin{bmatrix} 2 & 1 \\ 0 & 3 \end{bmatrix} \begin{bmatrix} 1 \\ -1 \end{bmatrix}$.
3. The first component is $2(1) + 1(-1) = 1$.
4. The second component is $0(1) + 3(-1) = -3$.

Answer: \begin{bmatrix} 1 \\ -3 \end{bmatrix}

1. The matrix from $B_2$ to $B_1$ is $P_{B_1 \leftarrow B_2} = P_1^{-1} P_2$.
2. Compute $P_1^{-1}$: for $\begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix}$, the inverse is $\begin{bmatrix} 1 & -1 \\ 0 & 1 \end{bmatrix}$.
3. Multiply $P_1^{-1}$ by $P_2$: $\begin{bmatrix} 1 & -1 \\ 0 & 1 \end{bmatrix} \begin{bmatrix} 2 & 0 \\ 1 & 1 \end{bmatrix}$.
4. Top left: $1(2) + (-1)(1) = 1$. Top right: $1(0) + (-1)(1) = -1$.
5. Bottom left: $0(2) + 1(1) = 1$. Bottom right: $0(0) + 1(1) = 1$.

Answer: \begin{bmatrix} 1 & -1 \\ 1 & 1 \end{bmatrix}

1. The matrix $P_1$ has columns $[2, 1]^T$ and $[1, 2]^T$.
2. The matrix $P_2$ is the identity matrix $I$.
3. The formula is $P_{B_2 \leftarrow B_1} = P_2^{-1} P_1 = I^{-1} P_1 = P_1$.
4. Therefore, the matrix is simply the matrix of $B_1$ vectors.

Answer: \begin{bmatrix} 2 & 1 \\ 1 & 2 \end{bmatrix}

1. Construct $P_1 = \begin{bmatrix} 1 & 0 \\ 1 & 1 \end{bmatrix}$ and $P_2 = \begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix}$.
2. Compute $P_2^{-1}$. The determinant is $1(1) - 1(0) = 1$. The inverse is $\begin{bmatrix} 1 & -1 \\ 0 & 1 \end{bmatrix}$.
3. Multiply $P_2^{-1} P_1$: $\begin{bmatrix} 1 & -1 \\ 0 & 1 \end{bmatrix} \begin{bmatrix} 1 & 0 \\ 1 & 1 \end{bmatrix} = \begin{bmatrix} 1-1 & 0-1 \\ 0+1 & 0+1 \end{bmatrix} = \begin{bmatrix} 0 & -1 \\ 1 & 1 \end{bmatrix}$.

Answer: \begin{bmatrix} 0 & -1 \\ 1 & 1 \end{bmatrix}

1. The change of basis matrix from $B_1$ to the standard basis $B_2$ is $P_1 = \begin{bmatrix} 1 & 3 \\ 2 & 4 \end{bmatrix}$.
2. The coordinates in $B_2$ are found by $[v]_{B_2} = P_1 [v]_{B_1}$.
3. Compute the product: $\begin{bmatrix} 1 & 3 \\ 2 & 4 \end{bmatrix} \begin{bmatrix} 3 \\ -2 \end{bmatrix} = \begin{bmatrix} 1(3) + 3(-2) \\ 2(3) + 4(-2) \end{bmatrix} = \begin{bmatrix} 3-6 \\ 6-8 \end{bmatrix} = \begin{bmatrix} -3 \\ -2 \end{bmatrix}$.

Answer: \begin{bmatrix} -3 \\ -2 \end{bmatrix}

1. Convert polynomials to coordinate vectors relative to basis $\{1, x, x^2\}$: $[p_1] = [1, 0, 1]^T$, $[p_2] = [0, 1, 1]^T$, $[p_3] = [1, 1, 0]^T$.
2. Form matrix $A = \begin{bmatrix} 1 & 0 & 1 \\ 0 & 1 & 1 \\ 1 & 1 & 0 \end{bmatrix}$.
3. Perform row reduction: $R_3 \leftarrow R_3 - R_1$ gives $\begin{bmatrix} 1 & 0 & 1 \\ 0 & 1 & 1 \\ 0 & 1 & -1 \end{bmatrix}$.
4. Then $R_3 \leftarrow R_3 - R_2$ gives $\begin{bmatrix} 1 & 0 & 1 \\ 0 & 1 & 1 \\ 0 & 0 & -2 \end{bmatrix}$.
5. Since there are 3 pivots, the vectors are linearly independent.

Answer: Yes, they are linearly independent.

1. Set up the equation $4 + 2x - 3x^2 = c_1(1) + c_2(1+x) + c_3(1+x+x^2)$.
2. Equate coefficients of $x^2$: $c_3 = -3$.
3. Equate coefficients of $x$: $c_2 + c_3 = 2 \implies c_2 - 3 = 2 \implies c_2 = 5$.
4. Equate constants: $c_1 + c_2 + c_3 = 4 \implies c_1 + 5 - 3 = 4 \implies c_1 + 2 = 4 \implies c_1 = 2$.
5. The coordinate vector is $[2, 5, -3]^T$.

Answer: [2, 5, -3]^T

1. Convert to coordinate vectors in $\mathbb{R}^4$: $v_1=[1,1,0,0]^T, v_2=[0,1,1,0]^T, v_3=[0,0,1,1]^T, v_4=[1,0,0,1]^T$.
2. Form matrix $A = \begin{bmatrix} 1 & 0 & 0 & 1 \\ 1 & 1 & 0 & 0 \\ 0 & 1 & 1 & 0 \\ 0 & 0 & 1 & 1 \end{bmatrix}$.
3. Row reduce: $R_2 \leftarrow R_2 - R_1$ gives $\begin{bmatrix} 1 & 0 & 0 & 1 \\ 0 & 1 & 0 & -1 \\ 0 & 1 & 1 & 0 \\ 0 & 0 & 1 & 1 \end{bmatrix}$.
4. Then $R_3 \leftarrow R_3 - R_2$ gives $\begin{bmatrix} 1 & 0 & 0 & 1 \\ 0 & 1 & 0 & -1 \\ 0 & 0 & 1 & 1 \\ 0 & 0 & 1 & 1 \end{bmatrix}$.
5. Then $R_4 \leftarrow R_4 - R_3$ gives $\begin{bmatrix} 1 & 0 & 0 & 1 \\ 0 & 1 & 0 & -1 \\ 0 & 0 & 1 & 1 \\ 0 & 0 & 0 & 0 \end{bmatrix}$.
6. There are only 3 pivots for 4 vectors, so they are linearly dependent.

Answer: No, they are linearly dependent.

1. First, check if the vectors are in $S$. For $[-1, 1, 0]^T$, $-1+1+0=0$. For $[-1, 0, 1]^T$, $-1+0+1=0$. Both are in $S$.
2. Check for linear independence. Since neither vector is a scalar multiple of the other, they are linearly independent.
3. Determine the dimension of $S$. $S$ is a plane in $\mathbb{R}^3$ defined by one equation, so $\text{dim}(S) = 3 - 1 = 2$.
4. Since we have 2 linearly independent vectors in a 2-dimensional space, they must form a basis.

Answer: Yes, it is a basis.

1. Perform row reduction. $R_3 \leftarrow R_3 - R_1$ gives $\begin{bmatrix} 1 & 2 & 1 \\ 0 & 1 & 1 \\ 0 & 1 & 1 \end{bmatrix}$.
2. Then $R_3 \leftarrow R_3 - R_2$ gives $\begin{bmatrix} 1 & 2 & 1 \\ 0 & 1 & 1 \\ 0 & 0 & 0 \end{bmatrix}$.
3. There are 2 pivots, so $\text{rank}(A) = 2$.
4. The matrix has 3 columns, so $\text{nullity}(A) = 3 - 2 = 1$.

Answer: \text{rank}=2, \text{nullity}=1

1. We need to express vectors of $B_1$ in terms of $B_2$.
2. For $[1, 0]^T$: $c_1[1, 1]^T + c_2[1, -1]^T = [1, 0]^T$. This gives $c_1+c_2=1$ and $c_1-c_2=0$, so $c_1=0.5, c_2=0.5$.
3. For $[0, 1]^T$: $c_1[1, 1]^T + c_2[1, -1]^T = [0, 1]^T$. This gives $c_1+c_2=0$ and $c_1-c_2=1$, so $c_1=0.5, c_2=-0.5$.
4. The matrix $P$ is formed by these coordinate vectors as columns.

Answer: \begin{bmatrix} 0.5 & 0.5 \\ 0.5 & -0.5 \end{bmatrix}

1. Multiply the matrix $A$ by the vector $\mathbf{v}$ using the column picture.
2. The result is $5 \begin{bmatrix} 2 \\ 1 \end{bmatrix} + 2 \begin{bmatrix} -3 \\ 4 \end{bmatrix}$.
3. Compute the components: $5(2) + 2(-3) = 10 - 6 = 4$ and $5(1) + 2(4) = 5 + 8 = 13$.

Answer: \begin{bmatrix} 4 \\ 13 \end{bmatrix}

1. Construct the matrix $A$ using the images of the basis vectors as columns: $A = \begin{bmatrix} 0 & -1 \\ 2 & 3 \end{bmatrix}$.
2. Multiply $A$ by $\mathbf{w}$: $\begin{bmatrix} 0 & -1 \\ 2 & 3 \end{bmatrix} \begin{bmatrix} 3 \\ -2 \end{bmatrix}$.
3. Compute the components: $0(3) + (-1)(-2) = 2$ and $2(3) + 3(-2) = 6 - 6 = 0$.

Answer: \begin{bmatrix} 2 \\ 0 \end{bmatrix}

1. Test the homogeneity rule $T(c\mathbf{v}) = cT(\mathbf{v})$.
2. Let $\mathbf{v} = \begin{bmatrix} 1 \\ 0 \end{bmatrix}$ and $c = 2$.
3. Compute $T(2\mathbf{v}) = T\begin{bmatrix} 2 \\ 0 \end{bmatrix} = \begin{bmatrix} 2^2 \\ 0 \end{bmatrix} = \begin{bmatrix} 4 \\ 0 \end{bmatrix}$.
4. Compute $2T(\mathbf{v}) = 2 T\begin{bmatrix} 1 \\ 0 \end{bmatrix} = 2 \begin{bmatrix} 1^2 \\ 0 \end{bmatrix} = \begin{bmatrix} 2 \\ 0 \end{bmatrix}$.
5. Since $\begin{bmatrix} 4 \\ 0 \end{bmatrix} \neq \begin{bmatrix} 2 \\ 0 \end{bmatrix}$, the rule is violated.

Answer: No, it is not a linear transformation.

1. Find the image of the first basis vector $\mathbf{i} = [1, 0]^T$. Scaling $x$ by $2$ gives $T(\mathbf{i}) = [2, 0]^T$.
2. Find the image of the second basis vector $\mathbf{j} = [0, 1]^T$. Scaling $y$ by $3$ gives $T(\mathbf{j}) = [0, 3]^T$.
3. Place these images as the columns of the matrix $A$.

Answer: A = \begin{bmatrix} 2 & 0 \\ 0 & 3 \end{bmatrix}

1. Evaluate $T(1, 0)$ to find the first column: $T(1, 0) = (1 + 2(0), 3(1) - 4(0)) = (1, 3)$.
2. Evaluate $T(0, 1)$ to find the second column: $T(0, 1) = (0 + 2(1), 3(0) - 4(1)) = (2, -4)$.
3. Assemble the columns into matrix $A$.

Answer: A = \begin{bmatrix} 1 & 2 \\ 3 & -4 \end{bmatrix}

1. The images of the basis vectors are given directly as the columns of the matrix.
2. Column 1 is $[1, 1]^T$.
3. Column 2 is $[0, 1]^T$.
4. Column 3 is $[-1, 0]^T$.
5. Combine these into a $2 \times 3$ matrix.

Answer: A = \begin{bmatrix} 1 & 0 & -1 \\ 1 & 1 & 0 \end{bmatrix}

1. The reflection across the $y$-axis is $Ref_y = \begin{bmatrix} -1 & 0 \\ 0 & 1 \end{bmatrix}$.
2. The scaling of the $x$-axis by $5$ is $S = \begin{bmatrix} 5 & 0 \\ 0 & 1 \end{bmatrix}$.
3. The combined transformation is $A = S \cdot Ref_y = \begin{bmatrix} 5 & 0 \\ 0 & 1 \end{bmatrix} \begin{bmatrix} -1 & 0 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} -5 & 0 \\ 0 & 1 \end{bmatrix}$.

Answer: \begin{bmatrix} -5 & 0 \\ 0 & 1 \end{bmatrix}

1. Using $R_\theta$ with $\theta = 180^\circ$, we have $\cos(180^\circ) = -1$ and $\sin(180^\circ) = 0$.
2. The matrix is $R_{180} = \begin{bmatrix} -1 & 0 \\ 0 & -1 \end{bmatrix}$.
3. Applying this to $\begin{bmatrix} x \\ y \end{bmatrix}$ gives $\begin{bmatrix} -x \\ -y \end{bmatrix}$, which is the negation of the original vector.

Answer: \begin{bmatrix} -1 & 0 \\ 0 & -1 \end{bmatrix}

1. The matrix has ones on the diagonal and a non-zero entry in the bottom-left. This is a vertical shear $H_y$ with factor $k=2$.
2. The image is $A\mathbf{x} = \begin{bmatrix} 1 & 0 \\ 2 & 1 \end{bmatrix} \begin{bmatrix} 3 \\ 4 \end{bmatrix}$.
3. The new $x$ is $1(3) + 0(4) = 3$.
4. The new $y$ is $2(3) + 1(4) = 6 + 4 = 10$.

Answer: \begin{bmatrix} 3 \\ 10 \end{bmatrix}

1. The reflection matrix is $A = \begin{bmatrix} -1 & 0 \\ 0 & 1 \end{bmatrix}$.
2. The scaling matrix is $B = \begin{bmatrix} 5 & 0 \\ 0 & 1 \end{bmatrix}$.
3. Since reflection is first, the product is $C = BA$.
4. Compute $C = \begin{bmatrix} 5 & 0 \\ 0 & 1 \end{bmatrix} \begin{bmatrix} -1 & 0 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} -5 & 0 \\ 0 & 1 \end{bmatrix}$.

Answer: \begin{bmatrix} -5 & 0 \\ 0 & 1 \end{bmatrix}

1. The shear matrix is $H = \begin{bmatrix} 1 & 3 \\ 0 & 1 \end{bmatrix}$.
2. The rotation matrix is $R = \begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix}$.
3. The composition is $C = RH$.
4. Compute $C = \begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix} \begin{bmatrix} 1 & 3 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 0 & -1 \\ 1 & 3 \end{bmatrix}$.

Answer: \begin{bmatrix} 0 & -1 \\ 1 & 3 \end{bmatrix}

1. Scaling matrix $S = \begin{bmatrix} 2 & 0 \\ 0 & 2 \end{bmatrix}$.
2. Reflection across $y=x$ matrix $A = \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}$.
3. Composition $M = AS = \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix} \begin{bmatrix} 2 & 0 \\ 0 & 2 \end{bmatrix} = \begin{bmatrix} 0 & 2 \\ 2 & 0 \end{bmatrix}$.
4. Check reverse order: $SA = \begin{bmatrix} 2 & 0 \\ 0 & 2 \end{bmatrix} \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix} = \begin{bmatrix} 0 & 2 \\ 2 & 0 \end{bmatrix}$.
5. In this specific case, the order does not matter because $S$ is a scalar multiple of the identity matrix.

Answer: \begin{bmatrix} 0 & 2 \\ 2 & 0 \end{bmatrix}, \text{ order does not matter}

1. Solve $A\mathbf{x} = \mathbf{0}$ using row reduction. $R_3 \leftarrow R_3 - R_1$ gives $\begin{bmatrix} 1 & 2 & 1 \\ 0 & 1 & 1 \\ 0 & 1 & 1 \end{bmatrix}$, then $R_3 \leftarrow R_3 - R_2$ gives $\begin{bmatrix} 1 & 2 & 1 \\ 0 & 1 & 1 \\ 0 & 0 & 0 \end{bmatrix}$.
2. From $R_2$, $x_2 = -x_3$. From $R_1$, $x_1 = -2x_2 - x_3 = -2(-x_3) - x_3 = x_3$.
3. The kernel is $\text{span} \{ [1, -1, 1]^T \}$. Since it is non-trivial, the map is not injective.
4. The rank is 2 (two pivots). Since the target space is $\mathbb{R}^3$ and $2 < 3$, the map is not surjective.

Answer: \text{ker}(T) = \text{span} \{ [1, -1, 1]^T \}, \text{im}(T) = \text{span} \{ [1, 0, 1]^T, [2, 1, 3]^T \}, \text{not injective, not surjective}

1. Set $T([x, y]^T) = [0, 0, 0]^T$.
2. This gives the system: $x+y=0$, $x-y=0$, and $2x=0$.
3. From $2x=0$, we have $x=0$. Substituting into $x+y=0$ gives $y=0$.
4. The only solution is the zero vector $[0, 0]^T$.
5. Since the kernel is trivial, the map is injective.

Answer: \text{Yes, it is injective.}

1. Use the Rank-Nullity Theorem: $\dim(V) = \text{rank}(T) + \dim(\text{ker}(T))$.
2. Substitute the known values: $4 = \text{rank}(T) + 1$.
3. This implies $\text{rank}(T) = 3$.
4. The dimension of the image is 3, and the dimension of the target space $W$ is also 3.
5. Since $\text{im}(T) = W$, the map is surjective.

Answer: \text{Yes, it is surjective.}

1. Calculate the determinant: $\det(A) = (1)(7) - (3)(2) = 7 - 6 = 1$.
2. Since $\det(A) \neq 0$, the matrix is invertible.
3. Apply the 2x2 inverse formula: swap diagonal elements (7 and 1), negate off-diagonal elements (-3 and -2), and divide by the determinant (1).
4. The result is $A^{-1} = \begin{bmatrix} 7 & -3 \\ -2 & 1 \end{bmatrix}$.

Answer: A^{-1} = \begin{bmatrix} 7 & -3 \\ -2 & 1 \end{bmatrix}

1. Calculate the determinant: $\det(A) = (2)(2) - (4)(1) = 4 - 4 = 0$.
2. Because the determinant is zero, the transformation collapses the plane, making it non-invertible.
3. Observe that the second column $[4, 2]^T$ is exactly twice the first column $[2, 1]^T$.
4. Therefore, the image of $T$ is the line spanned by the vector $[2, 1]^T$.

Answer: Not invertible because $\det(A)=0$; the image is a line.

1. Identify $A$ as a 90-degree counter-clockwise rotation. Its inverse $A^{-1}$ should be a 90-degree clockwise rotation.
2. Compute $\det(A) = (0)(0) - (-1)(1) = 1$.
3. Find $A^{-1} = \frac{1}{1} \begin{bmatrix} 0 & 1 \\ -1 & 0 \end{bmatrix} = \begin{bmatrix} 0 & 1 \\ -1 & 0 \end{bmatrix}$.
4. Multiply $A^{-1}$ by the output vector: $\mathbf{x} = \begin{bmatrix} 0 & 1 \\ -1 & 0 \end{bmatrix} \begin{bmatrix} 3 \\ -2 \end{bmatrix} = \begin{bmatrix} (0)(3) + (1)(-2) \\ (-1)(3) + (0)(-2) \end{bmatrix} = \begin{bmatrix} -2 \\ -3 \end{bmatrix}$.

Answer: \mathbf{x} = \begin{bmatrix} -2 \\ -3 \end{bmatrix}

1. The normal vector to the $xz$-plane is $\mathbf{n} = [0, 1, 0]^T$.
2. The projection of $\mathbf{e}_1 = [1, 0, 0]^T$ onto the plane is $[1, 0, 0]^T$ because it already lies in the plane.
3. The projection of $\mathbf{e}_2 = [0, 1, 0]^T$ onto the plane is the zero vector because it is the normal vector.
4. The projection of $\mathbf{e}_3 = [0, 0, 1]^T$ onto the plane is $[0, 0, 1]^T$ because it already lies in the plane.
5. Assembling these as columns gives the matrix.

Answer: \begin{bmatrix} 1 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 1 \end{bmatrix}

1. Compute $\mathbf{x} \cdot \mathbf{n} = (2)(0) + (3)(1) + (4)(1) = 7$.
2. Compute $\mathbf{n} \cdot \mathbf{n} = 0^2 + 1^2 + 1^2 = 2$.
3. The projection onto the normal is $\text{Proj}_n(\mathbf{x}) = \frac{7}{2}[0, 1, 1]^T = [0, 3.5, 3.5]^T$.
4. The projection onto the plane is $\mathbf{x} - \text{Proj}_n(\mathbf{x}) = [2, 3, 4]^T - [0, 3.5, 3.5]^T = [2, -0.5, 0.5]^T$.

Answer: [2, -0.5, 0.5]^T

1. Geometrically, $P\mathbf{x}$ is the projection of $\mathbf{x}$ onto the plane $W$.
2. The resulting vector $P\mathbf{x}$ already lies within the plane $W$.
3. Projecting a vector that is already in the plane does not change it, so $P(P\mathbf{x}) = P\mathbf{x}$.
4. Since this holds for all $\mathbf{x}$, we have $P^2 = P$.

Answer: P^2 = P

1. Construct $P = \begin{pmatrix} 1 & 0 \\ 1 & 1 \end{pmatrix}$.
2. Compute $P^{-1} = \begin{pmatrix} 1 & 0 \\ -1 & 1 \end{pmatrix}$.
3. Compute $AP = \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix} \begin{pmatrix} 1 & 0 \\ 1 & 1 \end{pmatrix} = \begin{pmatrix} 2 & 1 \\ 1 & 1 \end{pmatrix}$.
4. Compute $B' = P^{-1}(AP) = \begin{pmatrix} 1 & 0 \\ -1 & 1 \end{pmatrix} \begin{pmatrix} 2 & 1 \\ 1 & 1 \end{pmatrix} = \begin{pmatrix} 2 & 1 \\ -1 & 0 \end{pmatrix}$.

Answer: \begin{pmatrix} 2 & 1 \\ -1 & 0 \end{pmatrix}

1. Since $A$ and $B'$ are similar, $B' = P^{-1} A P$.
2. The determinant of a product is the product of the determinants: $\det(B') = \det(P^{-1}) \det(A) \det(P)$.
3. Since $\det(P^{-1}) = 1/\det(P)$, these terms cancel out.
4. Therefore, $\det(B') = \det(A) = 5$.

Answer: 5

1. We need $P$ such that $AP = PB'$.
2. This means $A v_1 = 1 v_1$ and $A v_2 = -1 v_2$, where $v_1$ and $v_2$ are the columns of $P$.
3. Solving $A v_1 = v_1$ for $v_1 = [x, y]^T$ gives $y = x$, so $v_1 = [1, 1]^T$ is a solution.
4. Solving $A v_2 = -v_2$ for $v_2 = [x, y]^T$ gives $y = -x$, so $v_2 = [1, -1]^T$ is a solution.
5. One such basis is $B = \left\{ \begin{pmatrix} 1 \\ 1 \end{pmatrix}, \begin{pmatrix} 1 \\ -1 \end{pmatrix} \right\}$.

Answer: \left\{ \begin{pmatrix} 1 \\ 1 \end{pmatrix}, \begin{pmatrix} 1 \\ -1 \end{pmatrix} \right\}

1. Compute $\det(A) = (2)(2) - (1)(1) = 3$.
2. Compute $\det(B) = (3)(1) - (0)(0) = 3$.
3. Compute $\text{tr}(A) = 2 + 2 = 4$.
4. Compute $\text{tr}(B) = 3 + 1 = 4$.
5. Since both the determinant and trace match, the matrices can be similar.

Answer: Yes, they share the same invariants ($\det=3, \text{tr}=4$).

1. Recall that the determinant is invariant under similarity, so $\det(B) = \det(A) = 5$.
2. Recall that the trace is invariant under similarity, so $\text{tr}(B) = \text{tr}(A) = 2$.

Answer: \det(B) = 5, \text{tr}(B) = 2

1. Compute $\det(A) = 1$ and $\det(B) = 1$.
2. Compute $\text{tr}(A) = 2$ and $\text{tr}(B) = 2$.
3. The invariants match, but similarity requires $B = P^{-1} A P$.
4. For $A = I$, $P^{-1} I P = P^{-1} P = I$.
5. Since $B \neq I$, $A$ and $B$ cannot be similar despite having the same invariants.

Answer: No, because the only matrix similar to the identity matrix is the identity matrix itself.

1. Step 1: Write the polynomial as a coordinate vector $\mathbf{v} = \begin{bmatrix} 2 \\ 7 \\ 5 \end{bmatrix}$.
2. Step 2: Multiply the matrix $D$ by the vector $\mathbf{v}$.
3. Step 3: The first component is $0(2) + 1(7) + 0(5) = 7$.
4. Step 4: The second component is $0(2) + 0(7) + 2(5) = 10$.
5. Step 5: The third component is $0(2) + 0(7) + 0(5) = 0$.
6. Step 6: The resulting vector $\begin{bmatrix} 7 \\ 10 \\ 0 \end{bmatrix}$ corresponds to the polynomial $7 + 10x$.

Answer: 7 + 10x

1. Step 1: Multiply $D$ by itself: $\begin{bmatrix} 0 & 1 & 0 \\ 0 & 0 & 2 \\ 0 & 0 & 0 \end{bmatrix} \begin{bmatrix} 0 & 1 & 0 \\ 0 & 0 & 2 \\ 0 & 0 & 0 \end{bmatrix}$.
2. Step 2: Row 1 times Col 1 is 0, Row 1 times Col 2 is 0, Row 1 times Col 3 is $1(2) = 2$.
3. Step 3: Row 2 times all columns is 0, Row 3 times all columns is 0.
4. Step 4: The result is $\begin{bmatrix} 0 & 0 & 2 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix}$.
5. Step 5: This matrix represents the second derivative operator.

Answer: \begin{bmatrix} 0 & 0 & 2 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix}

1. Step 1: Find the derivative of each basis element: $D(1)=0, D(x)=1, D(x^2)=2x, D(x^3)=3x^2$.
2. Step 2: Convert these to coordinate vectors: $\begin{bmatrix} 0 \\ 0 \\ 0 \\ 0 \end{bmatrix}, \begin{bmatrix} 1 \\ 0 \\ 0 \\ 0 \end{bmatrix}, \begin{bmatrix} 0 \\ 2 \\ 0 \\ 0 \end{bmatrix}, \begin{bmatrix} 0 \\ 0 \\ 3 \\ 0 \end{bmatrix}$.
3. Step 3: Place these vectors as columns in a $4 \times 4$ matrix.

Answer: \begin{bmatrix} 0 & 1 & 0 & 0 \\ 0 & 0 & 2 & 0 \\ 0 & 0 & 0 & 3 \\ 0 & 0 & 0 & 0 \end{bmatrix}

1. The first column of the matrix is the image of $e_1$, which is $[2, 0]^T$.
2. The second column of the matrix is the image of $e_2$, which is $[1, 1]^T$.
3. Placing these into a matrix gives $A = \begin{bmatrix} 2 & 1 \\ 0 & 1 \end{bmatrix}$.

Answer: \begin{bmatrix} 2 & 1 \\ 0 & 1 \end{bmatrix}

1. A $90^\circ$ rotation maps $[1, 0]^T$ to $[0, 1]^T$ and $[0, 1]^T$ to $[-1, 0]^T$.
2. Scaling by $3$ multiplies these resulting vectors by $3$.
3. The new image of $e_1$ is $[0, 3]^T$ and the new image of $e_2$ is $[-3, 0]^T$.
4. The matrix is $A = \begin{bmatrix} 0 & -3 \\ 3 & 0 \end{bmatrix}$.

Answer: \begin{bmatrix} 0 & -3 \\ 3 & 0 \end{bmatrix}

1. The first column is $T(e_1) = [1, -1]^T$.
2. The second column is $T(e_2) = [-2, 2]^T$.
3. The matrix is $A = \begin{bmatrix} 1 & -2 \\ -1 & 2 \end{bmatrix}$.
4. The determinant is $(1)(2) - (-2)(-1) = 2 - 2 = 0$.

Answer: \begin{bmatrix} 1 & -2 \\ -1 & 2 \end{bmatrix}

1. The reflection across $y=x$ maps $e_1 = [1, 0]^T$ to $[0, 1]^T$ and $e_2 = [0, 1]^T$ to $[1, 0]^T$.
2. Applying the scaling factor of $2$ to these results gives $T(e_1) = [0, 2]^T$ and $T(e_2) = [2, 0]^T$.
3. The standard matrix $A$ is formed by these images as columns.

Answer: A = \begin{bmatrix} 0 & 2 \\ 2 & 0 \end{bmatrix}

1. The columns are identical, so the image is the span of $[1, 1]^T$, which has dimension $1$.
2. Using the Rank-Nullity Theorem, $\dim(\ker(A)) + \dim(\text{im}(A)) = 2$.
3. Therefore, $\dim(\ker(A)) = 2 - 1 = 1$.

Answer: \dim(\ker(A)) = 1, \dim(\text{im}(A)) = 1

1. Compute the determinant of $A$: $\det(A) = 1 \times 2 - 0 \times 0 = 2$.
2. Compute the determinant of $B$: $\det(B) = 1 \times 2 - 1 \times 0 = 2$.
3. Compute the trace of $A$: $\text{tr}(A) = 1 + 2 = 3$.
4. Compute the trace of $B$: $\text{tr}(B) = 1 + 2 = 3$.
5. Since both the determinant and trace match, the matrices are candidates for similarity, although matching invariants are necessary but not sufficient to prove similarity.

Answer: The invariants match, so they may be similar.

1. Identify the basis for $V$ as $v_1 = [1, 2, 1]^T$.
2. Identify the basis for $W$ as $w_1 = [1, 0, -1]^T$ and $w_2 = [0, 1, -2]^T$.
3. Compute $v_1 \cdot w_1 = (1)(1) + (2)(0) + (1)(-1) = 1 + 0 - 1 = 0$.
4. Compute $v_1 \cdot w_2 = (1)(0) + (2)(1) + (1)(-2) = 0 + 2 - 2 = 0$.
5. Since all pairs of basis vectors are orthogonal, the subspaces are orthogonal.

Answer: Yes, they are orthogonal.

1. Recall the dimension constraint for orthogonal subspaces: $\dim(V) + \dim(W) \le n$.
2. Here, $n = 4$.
3. The sum of dimensions is $3 + 3 = 6$.
4. Since $6 > 4$, the condition is violated.

Answer: No, because the sum of their dimensions would exceed the dimension of the space.

1. The row space is spanned by $[1, 2]^T$ since the second row is a multiple of the first.
2. To find the null space, solve $x + 2y = 0$, which gives $x = -2y$. A basis vector is $[-2, 1]^T$.
3. Compute the dot product: $(1)(-2) + (2)(1) = -2 + 2 = 0$.
4. The basis vectors are orthogonal, so the subspaces are orthogonal.

Answer: Row space basis: $\{\begin{bmatrix} 1 \\ 2 \end{bmatrix}\}$, Null space basis: $\{\begin{bmatrix} -2 \\ 1 \end{bmatrix}\}$. Dot product is 0.

1. A vector $\begin{bmatrix} x \\ y \\ z \end{bmatrix}$ is in $V^{\perp}$ if its dot product with both spanning vectors is zero.
2. This gives the system: $x + 2y + z = 0$ and $y - z = 0$.
3. From the second equation, $y = z$.
4. Substitute into the first: $x + 2z + z = 0$, so $x + 3z = 0$, which means $x = -3z$.
5. The vectors are of the form $\begin{bmatrix} -3z \\ z \\ z \end{bmatrix} = z\begin{bmatrix} -3 \\ 1 \\ 1 \end{bmatrix}$.

Answer: \text{span}\{\begin{bmatrix} -3 \\ 1 \\ 1 \end{bmatrix}\}

1. The row space of $A$ is a subspace of $\mathbb{R}^5$ because the rows have 5 components.
2. The dimension of the row space is equal to the rank of the matrix, which is 2.
3. The dimension of the orthogonal complement $V^{\perp}$ is $n - \dim(V)$.
4. Here $n = 5$ and $\dim(Row(A)) = 2$.
5. Therefore, $\dim(Row(A)^{\perp}) = 5 - 2 = 3$.

Answer: 3

1. The subspace $V$ is defined as the set of all vectors $\begin{bmatrix} x_1 \\ x_2 \\ x_3 \\ x_4 \end{bmatrix}$ such that $\begin{bmatrix} 1 \\ 1 \\ 1 \\ 1 \end{bmatrix} \cdot \begin{bmatrix} x_1 \\ x_2 \\ x_3 \\ x_4 \end{bmatrix} = 0$.
2. This means $V$ is the null space of the matrix $A = \begin{bmatrix} 1 & 1 & 1 & 1 \end{bmatrix}$.
3. We know that $Null(A)^{\perp} = Row(A)$.
4. The row space of $A$ is spanned by the single row vector $\begin{bmatrix} 1 \\ 1 \\ 1 \\ 1 \end{bmatrix}$.
5. Thus, $V^{\perp}$ is spanned by the vector $\begin{bmatrix} 1 \\ 1 \\ 1 \\ 1 \end{bmatrix}$.

Answer: \text{span}\{\begin{bmatrix} 1 \\ 1 \\ 1 \\ 1 \end{bmatrix}\}

1. Compute $a \cdot b = (3)(5) + (4)(2) = 15 + 8 = 23$.
2. Compute $a \cdot a = 3^2 + 4^2 = 9 + 16 = 25$.
3. Calculate the scalar $c = 23/25 = 0.92$.
4. Multiply $c$ by $a$: $0.92 \begin{bmatrix} 3 \\ 4 \end{bmatrix} = \begin{bmatrix} 2.76 \\ 3.68 \end{bmatrix}$.

Answer: \begin{bmatrix} 2.76 \\ 3.68 \end{bmatrix}

1. Compute $a \cdot b = (0)(1) + (1)(1) + (1)(0) = 1$.
2. Compute $a \cdot a = 0^2 + 1^2 + 1^2 = 2$.
3. The scalar is $c = 1/2$.
4. The projection is $p = \frac{1}{2} \begin{bmatrix} 0 \\ 1 \\ 1 \end{bmatrix} = \begin{bmatrix} 0 \\ 0.5 \\ 0.5 \end{bmatrix}$.

Answer: \begin{bmatrix} 0 \\ 0.5 \\ 0.5 \end{bmatrix}

1. Compute $a \cdot b = (1)(-1) + (2)(1) = 1$.
2. Compute $a \cdot a = 1^2 + 2^2 = 5$.
3. Find $p = \frac{1}{5} \begin{bmatrix} 1 \\ 2 \end{bmatrix} = \begin{bmatrix} 0.2 \\ 0.4 \end{bmatrix}$.
4. Find $e = b - p = \begin{bmatrix} -1 - 0.2 \\ 1 - 0.4 \end{bmatrix} = \begin{bmatrix} -1.2 \\ 0.6 \end{bmatrix}$.
5. Check $p \cdot e = (0.2)(-1.2) + (0.4)(0.6) = -0.24 + 0.24 = 0$.

Answer: p = \begin{bmatrix} 0.2 \\ 0.4 \end{bmatrix}, e = \begin{bmatrix} -1.2 \\ 0.6 \end{bmatrix}

1. Set $A = \begin{bmatrix} 1 \\ 1 \end{bmatrix}$.
2. Compute $A^T A = [1 \ 1] \begin{bmatrix} 1 \\ 1 \end{bmatrix} = [2]$.
3. The inverse is $(A^T A)^{-1} = [1/2]$.
4. Compute $P = A (A^T A)^{-1} A^T = \begin{bmatrix} 1 \\ 1 \end{bmatrix} [1/2] [1 \ 1] = \begin{bmatrix} 0.5 & 0.5 \\ 0.5 & 0.5 \end{bmatrix}$.

Answer: P = \begin{bmatrix} 0.5 & 0.5 \\ 0.5 & 0.5 \end{bmatrix}

1. The projection is $p = P b$.
2. Compute $\begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix} \begin{bmatrix} 3 \\ 7 \end{bmatrix} = \begin{bmatrix} 1(3) + 0(7) \\ 0(3) + 0(7) \end{bmatrix} = \begin{bmatrix} 3 \\ 0 \end{bmatrix}$.

Answer: p = [3, 0]^T

1. A projection matrix must satisfy $M^2 = M$.
2. Compute $M^2 = \begin{bmatrix} 0.6 & 0.8 \\ 0.8 & 0.6 \end{bmatrix} \begin{bmatrix} 0.6 & 0.8 \\ 0.8 & 0.6 \end{bmatrix}$.
3. Top left: $0.6(0.6) + 0.8(0.8) = 0.36 + 0.64 = 1.0$.
4. Since the top left entry of $M^2$ is $1.0$ and the top left entry of $M$ is $0.6$, $M^2 \neq M$.

Answer: No, it is not a projection matrix.

1. Compute $A^T A = [1 \quad 2] \begin{bmatrix} 1 \\ 2 \end{bmatrix} = [1^2 + 2^2] = [5]$.
2. Compute $A^T b = [1 \quad 2] \begin{bmatrix} 1 \\ 4 \end{bmatrix} = [1(1) + 2(4)] = [9]$.
3. Solve the normal equation $5\hat{x} = 9$.
4. Divide by 5 to get $\hat{x} = 1.8$.

Answer: \hat{x} = 1.8

1. Compute $A^T A = \begin{bmatrix} 1 & 0 & 1 \\ 0 & 1 & 1 \end{bmatrix} \begin{bmatrix} 1 & 0 \\ 0 & 1 \\ 1 & 1 \end{bmatrix} = \begin{bmatrix} 2 & 1 \\ 1 & 2 \end{bmatrix}$.
2. Compute $A^T b = \begin{bmatrix} 1 & 0 & 1 \\ 0 & 1 & 1 \end{bmatrix} \begin{bmatrix} 1 \\ 1 \\ 0 \end{bmatrix} = \begin{bmatrix} 1 \\ 1 \end{bmatrix}$.
3. Solve $\begin{bmatrix} 2 & 1 \\ 1 & 2 \end{bmatrix} \begin{bmatrix} \hat{x}_1 \\ \hat{x}_2 \end{bmatrix} = \begin{bmatrix} 1 \\ 1 \end{bmatrix}$.
4. Subtracting the equations or using the inverse gives $\hat{x}_1 = 1/3$ and $\hat{x}_2 = 1/3$.

Answer: \hat{x} = \begin{bmatrix} 1/3 \\ 1/3 \end{bmatrix}

1. The system $Ax = b$ is inconsistent if $b$ is not in the column space of $A$.
2. The normal equations seek a vector $p = A\hat{x}$ such that $p$ is the projection of $b$ onto $Col(A)$.
3. The projection $p$ always exists for any vector $b$ and any subspace $Col(A)$.
4. Since $p$ is by definition in $Col(A)$, there must exist some $\hat{x}$ such that $A\hat{x} = p$.
5. This $\hat{x}$ satisfies $A^T(b - A\hat{x}) = 0$, which is the normal equation.

Answer: Because the projection of $b$ onto the column space of $A$ always exists and is always reachable by a linear combination of the columns of $A$.

1. Construct $A = \begin{pmatrix} 1 & 1 \\ 2 & 1 \\ 3 & 1 \end{pmatrix}$ and $b = \begin{pmatrix} 2 \\ 3 \\ 5 \end{pmatrix}$.
2. Compute $A^T A = \begin{pmatrix} 1+4+9 & 1+2+3 \\ 1+2+3 & 1+1+1 \end{pmatrix} = \begin{pmatrix} 14 & 6 \\ 6 & 3 \end{pmatrix}$.
3. Compute $A^T b = \begin{pmatrix} 1(2)+2(3)+3(5) \\ 2+3+5 \end{pmatrix} = \begin{pmatrix} 23 \\ 10 \end{pmatrix}$.
4. Solve $\begin{pmatrix} 14 & 6 \\ 6 & 3 \end{pmatrix} \begin{pmatrix} m \\ c \end{pmatrix} = \begin{pmatrix} 23 \\ 10 \end{pmatrix}$.
5. From the second equation, $3c = 10 - 6m$, so $c = 10/3 - 2m$.
6. Substitute into first: $14m + 6(10/3 - 2m) = 23 \implies 14m + 20 - 12m = 23 \implies 2m = 3 \implies m = 1.5$.
7. Then $c = 10/3 - 3 = 1/3$.

Answer: y = 1.5x + 1/3

1. The points are horizontal. The line $y = 1$ should be the best fit.
2. Construct $A = \begin{pmatrix} 0 & 1 \\ 1 & 1 \\ 2 & 1 \end{pmatrix}$ and $b = \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix}$.
3. Compute $A^T A = \begin{pmatrix} 5 & 3 \\ 3 & 3 \end{pmatrix}$.
4. Compute $A^T b = \begin{pmatrix} 0+1+2 \\ 1+1+1 \end{pmatrix} = \begin{pmatrix} 3 \\ 3 \end{pmatrix}$.
5. Solve $\begin{pmatrix} 5 & 3 \\ 3 & 3 \end{pmatrix} \begin{pmatrix} m \\ c \end{pmatrix} = \begin{pmatrix} 3 \\ 3 \end{pmatrix}$.
6. Subtracting equations gives $2m = 0 \implies m = 0$.
7. Then $3(0) + 3c = 3 \implies c = 1$.

Answer: y = 1

1. The equation becomes $y = mx$.
2. The system is now $m x_i = y_i$ for each $i$.
3. The matrix $A$ no longer needs the column of ones.
4. A becomes a single column vector containing the x values: $A = \begin{pmatrix} x_1 \\ \vdots \\ x_n \end{pmatrix}$.

Answer: A becomes a single column vector of x-coordinates.