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}