Precalculus Quiz: Representing Linear Systems With Matrices

What this quiz covers

This quiz focuses on Representing Linear Systems With Matrices, giving you a quick way to practice the rules, question types, and explanations that matter most for Precalculus.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

Question 1

1. $-3x + y = 7, \quad 2x - y = -2, \quad x + 2y + z = 5$
2. $x + y - 3z = 7, \quad 2x - y + z = -2, \quad x + 2y = 5$
3. $y - 3z = 7, \quad 2x - y = -2, \quad x + 2y + z = 5$ (correct answer)
4. $x - 3z = 7, \quad 2x - y = -2, \quad x + 2y + z = 5$

Explanation: When you encounter a matrix equation like this, you're seeing a compact way to represent a system of linear equations. The key is understanding how matrix multiplication works to translate between the two forms. To convert the matrix equation to a system, you multiply each row of the coefficient matrix by the variable vector. The first row $\begin{pmatrix} 0 & 1 & -3 \end{pmatrix}$ multiplied by $\begin{pmatrix} x \\ y \\ z \end{pmatrix}$ gives $0x + 1y + (-3)z = y - 3z$, which equals 7. The second row $\begin{pmatrix} 2 & -1 & 0 \end{pmatrix}$ gives $2x - y + 0z = 2x - y$, which equals -2. The third row $\begin{pmatrix} 1 & 2 & 1 \end{pmatrix}$ gives $x + 2y + z$, which equals 5. This produces the system: $y - 3z = 7$, $2x - y = -2$, $x + 2y + z = 5$, which is exactly option C. Option A incorrectly writes the first equation as $-3x + y = 7$, mixing up the coefficients. Option B writes the first equation as $x + y - 3z = 7$, adding an $x$ term that shouldn't exist since the first entry in row 1 is 0. It also incorrectly includes a $z$ term in the second equation and omits $z$ from the third. Option D writes the first equation as $x - 3z = 7$, again adding an $x$ term and omitting the $y$ term. Remember: each row of the coefficient matrix corresponds to one equation, and the position of each number tells you which variable it multiplies. Zero coefficients mean that variable doesn't appear in that equation.

Question 2

1. $d = -5, e = 2, f = -3, q = 7$
2. $d = 2, e = -3, f = 7, q = -5$ (correct answer)
3. $d = 2, e = 7, f = -3, q = -5$
4. $d = -3, e = 2, f = 7, q = -5$

Explanation: When you see a matrix equation like this, remember that matrix multiplication creates a system of linear equations. Each row of the coefficient matrix multiplies the variable vector to produce one equation in the system. To find the correct values, you need to understand how matrix multiplication works. The second row of the coefficient matrix $\begin{pmatrix} d & e & f \end{pmatrix}$ multiplies the variable vector $\begin{pmatrix} s \\ t \\ u \end{pmatrix}$ to give the second component of the result vector. This multiplication yields: $d \cdot s + e \cdot t + f \cdot u = q$ Since you're told the second equation is $2s - 3t + 7u = -5$, you can directly match coefficients:

• The coefficient of $s$ is $2$, so $d = 2$
• The coefficient of $t$ is $-3$, so $e = -3$
• The coefficient of $u$ is $7$, so $f = 7$
• The constant term is $-5$, so $q = -5$

Question 3

1. $\begin{bmatrix}3 & -2\\ 4 & 1\end{bmatrix}$ (correct answer)
2. $\begin{bmatrix}3 & 4\\ -2 & 1\end{bmatrix}$
3. $\begin{bmatrix}3 & -2\\ 4 & -1\end{bmatrix}$
4. $\begin{bmatrix}3 & -2 & 5\\ 4 & 1 & -6\end{bmatrix}$

Explanation: This question tests understanding of how to represent a system of linear equations as a single matrix equation $Ax = b$. To build the coefficient matrix A from a system of equations, create a matrix where entry $a_{ij}$ is the coefficient of variable j in equation i: rows represent equations, columns represent variables, and if a variable doesn't appear in an equation, its coefficient is 0. For the system 3x - 2y = 5, 4x + y = -6, we extract the coefficients: equation 1 has coefficients 3 for x and -2 for y, giving first row $[3 -2]$, and equation 2 has coefficients 4 for x and 1 for y, giving second row $[4 1]$. Choice A is correct because it properly extracts coefficients into matrix A with proper structure. Choice B transposes the coefficient matrix, listing coefficients by column instead of by row, which reverses the structure (variables and equations get mixed up). Key to converting systems to matrix form: create the coefficient matrix A by listing coefficients in order (row i = equation i, column j = variable j), create variable vector $x$ listing unknowns in the same order as columns, and create constant vector $b$ listing the right-hand sides in the same order as equations. Common error: don't put the constants (right-hand side values) into the coefficient matrix A—they belong in the separate vector $b$, while A contains only the coefficients of the variables.

Question 4

1. x-3y=4\\ 2x+5y=-1 \end{cases}$$ (correct answer)
2. x+3y=4\\ 2x+5y=-1 \end{cases}$$
3. x-3y=-1\\ 2x+5y=4 \end{cases}$$
4. x-3y=4\\ 2x-5y=-1 \end{cases}$$

Explanation: This question tests understanding of how to represent a system of linear equations as a single matrix equation $Ax = b$. The matrix equation $Ax = b$ is equivalent to the system of equations: when you multiply the coefficient matrix A by the variable vector x using matrix-vector multiplication, each entry of the result equals the corresponding entry of the constant vector b, giving you the original equations. The matrix equation $\begin{bmatrix} 1 & -3 \\ 2 & 5 \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 4 \\ -1 \end{bmatrix}$ expands by performing the matrix-vector multiplication: the first entry of $Ax$ is $1x + (-3)y$, and the first entry of b is 4, giving equation $x - 3y = 4$; similarly, the second entry gives equation $2x + 5y = -1$. Choice A is correct because it accurately represents the system as $Ax = b$ with proper structure from the matrix multiplication. Choice B has the wrong sign on the coefficient of y in the first equation, using +3 when the matrix has -3. To verify your matrix equation is correct, expand $Ax$ by multiplying: each row of A dotted with x should give the left side of the corresponding equation, and these should equal the entries of b (the right sides). The matrix form $Ax = b$ is just a compact notation for the system—it doesn't solve the system, but it sets up for solution methods like using the inverse (if A is invertible) or row reduction.

Question 5

1. $-1$
2. $4$
3. $1$ (correct answer)
4. $3$

Explanation: This question tests understanding of how to represent a system of linear equations as a single matrix equation Ax = b. To build the coefficient matrix A from a system of equations, create a matrix where entry aᵢⱼ is the coefficient of variable j in equation i: rows represent equations, columns represent variables, and if a variable doesn't appear in an equation, its coefficient is 0. From equation 5x - y + z = -7, the coefficients are 5 for x, -1 for y, and 1 for z, so the corresponding row of the coefficient matrix is [5 -1 1], and thus a_{23} (row 2, column 3) is 1. Choice C is correct because it accurately extracts the coefficient of z in the second equation. Choice A has the wrong sign, using -1 when the actual coefficient is 1 for z in equation 2. Key to converting systems to matrix form: create the coefficient matrix A by listing coefficients in order (row i = equation i, column j = variable j), create variable vector x listing unknowns in the same order as columns, and create constant vector b listing the right-hand sides in the same order as equations. To verify your matrix equation is correct, expand Ax by multiplying: each row of A dotted with x should give the left side of the corresponding equation, and these should equal the entries of b (the right sides).

Question 6

1. $5$ (correct answer)
2. $-2$
3. $4$
4. $-1$

Explanation: This question tests understanding of how to represent a system of linear equations as a single matrix equation Ax = b. To build the coefficient matrix A from a system of equations, create a matrix where entry aᵢⱼ is the coefficient of variable j in equation i: rows represent equations, columns represent variables, and if a variable doesn't appear in an equation, its coefficient is 0. For the system with variable order [x y z], we need to find entry a₂₃, which is the entry in row 2, column 3 of the coefficient matrix A. Row 2 corresponds to equation 2 (3x - 2y + 5z = 0), and column 3 corresponds to variable z (the third variable). From equation 2, the coefficient of z is 5, so a₂₃ = 5. Choice A is correct because the coefficient of z in the second equation 3x - 2y + 5z = 0 is indeed 5. Choice B incorrectly identifies -2 (the coefficient of y) as a₂₃, confusing column positions. Key to converting systems to matrix form: create the coefficient matrix A by listing coefficients in order (row i = equation i, column j = variable j), so aᵢⱼ is the coefficient of variable j in equation i.

Question 7

1. $\begin{pmatrix} 3 & 2 \\ -1 & 4 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 5 \\ 3 \end{pmatrix}$ (correct answer)
2. $\begin{pmatrix} 3 & -1 \\ 2 & 4 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 5 \\ 3 \end{pmatrix}$
3. $\begin{pmatrix} 5 & 3 \\ 3 & 2 \\ -1 & 4 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \end{pmatrix}$
4. $\begin{pmatrix} 3 & 2 \\ -1 & 4 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 3 \\ 5 \end{pmatrix}$

Explanation: The coefficient matrix $A$ is formed by placing coefficients row by row: first row contains coefficients of $3x + 2y = 5$, giving $[3, 2]$, and second row contains coefficients of $-x + 4y = 3$, giving $[-1, 4]$. The vector $\vec{b}$ contains the constants $[5, 3]$. Choice B incorrectly transposes the coefficient matrix. Choice C has wrong dimensions and incorrect constant vector. Choice D swaps the entries in the constant vector.

Question 8

1. $0p + 5q - 1r = -3$
2. $p - 2q + 3r = 8$
3. $4p + 2r = 12$
4. $5q - r = -3$ (correct answer)

Explanation: When you encounter a matrix equation like this, you're looking at a compact way to represent a system of linear equations. The matrix multiplication $A\vec{x} = \vec{b}$ expands into individual equations where each row of the coefficient matrix corresponds to one equation in the system. To find the second equation, focus on the second row of the coefficient matrix and the second element of the result vector. The second row is $\begin{pmatrix} 0 & 5 & -1 \end{pmatrix}$, which means the coefficients are 0 for $p$, 5 for $q$, and -1 for $r$. The second element of the result vector is -3. This gives us: $0p + 5q + (-1)r = -3$, which simplifies to $5q - r = -3$. Choice D correctly represents this equation in its simplest form by omitting the zero coefficient term. Choice A writes out the full form $0p + 5q - 1r = -3$, which is mathematically correct but unnecessarily includes the zero term. Standard form typically omits terms with zero coefficients. Choice B gives you the first equation (first row): $p - 2q + 3r = 8$. This comes from the first row $\begin{pmatrix} 1 & -2 & 3 \end{pmatrix}$. Choice C represents the third equation (third row): $4p + 0q + 2r = 12$, simplified to $4p + 2r = 12$ by dropping the zero term. Study tip: Always match the row number to the equation number. The second row always gives you the second equation, and so on. When writing in standard form, drop any terms with zero coefficients for cleaner presentation.

Question 9

1. $C$ is the inverse of $A$, meaning $C = A^{-1}$
2. $C$ is the transpose of $A$, meaning $C = A^T$ (correct answer)
3. $C$ is the negative of $A$, meaning $C = -A$
4. $C$ is scalar multiple of $A$, meaning $C = kA$ for some constant $k$

Explanation: When you encounter questions comparing matrices, you need to systematically check the fundamental matrix relationships: transpose, inverse, scalar multiplication, and addition/subtraction. Let's examine what happens when we transpose matrix $A$. The transpose operation $A^T$ flips a matrix across its main diagonal, so rows become columns and columns become rows. For matrix $A = \begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix}$, the transpose is $A^T = \begin{pmatrix} 1 & 3 \\ 2 & 4 \end{pmatrix}$. This matches matrix $C$ exactly, confirming that $C = A^T$. Now let's check why the other options fail. Option A claims $C = A^{-1}$. To find $A^{-1}$, we calculate $\frac{1}{\det(A)} \begin{pmatrix} 4 & -2 \\ -3 & 1 \end{pmatrix} = \frac{1}{-2} \begin{pmatrix} 4 & -2 \\ -3 & 1 \end{pmatrix} = \begin{pmatrix} -2 & 1 \\ 1.5 & -0.5 \end{pmatrix}$, which clearly doesn't equal $C$. Option C suggests $C = -A$, but $-A = \begin{pmatrix} -1 & -2 \\ -3 & -4 \end{pmatrix}$, which has all negative entries unlike $C$. Option D proposes $C = kA$ for some scalar $k$. If this were true, then $\frac{1}{1} = \frac{3}{2} = \frac{2}{3} = \frac{4}{4}$, which gives us $1 = 1.5 = 0.67 = 1$ – clearly impossible. Study tip: When comparing matrices, always check transpose first since it's the quickest calculation. Remember that $(i,j)$ entry of $A^T$ equals the $(j,i)$ entry of $A$.

Question 10

1. $\begin{bmatrix}1&-3&2\\4&0&1\\0&-2&3\end{bmatrix}$ (correct answer)
2. $\begin{bmatrix}1&-3&2\\4&1&0\\0&-2&3\end{bmatrix}$
3. $\begin{bmatrix}1&4&0\\-3&0&-2\\2&1&3\end{bmatrix}$
4. $\begin{bmatrix}1&-3&2&0\\4&0&1&5\\0&-2&3&-6\end{bmatrix}$

Explanation: This question tests understanding of how to represent a system of linear equations as a single matrix equation Ax = b. To build the coefficient matrix A from a system of equations, create a matrix where entry aᵢⱼ is the coefficient of variable j in equation i: rows represent equations, columns represent variables, and if a variable doesn't appear in an equation, its coefficient is 0. For the system x - 3y + 2z = 0, 4x + z = 5, and -2y + 3z = -6, we must be careful about missing variables. In equation 2 (4x + z = 5), variable y doesn't appear, which means its coefficient is 0, so the row is [4 0 1] if variables are ordered x, y, z. Similarly, in equation 3 (-2y + 3z = -6), variable x doesn't appear, giving row [0 -2 3]. Choice A is correct because it properly includes zero coefficients for missing variables: [[1 -3 2], [4 0 1], [0 -2 3]]. Choice B makes an error by putting 1 instead of 0 for the coefficient of y in equation 2, incorrectly suggesting the equation is 4x + y + 0z = 5. Remember to include 0 coefficients when a variable doesn't appear in an equation: for example, if equation 2 is '4x + z = 5' in a system with variables x, y, z, the coefficient row is [4 0 1], with the 0 for the missing y.

Question 11

1. -2x+y=8\\ 4x-3y+5z=-1 \end{cases}$$ (correct answer)
2. -2x+y+z=8\\ 4x-3y+5z=-1 \end{cases}$$
3. -2x+y=8\\ 4x+3y+5z=-1 \end{cases}$$
4. 2x-y=8\\ 4x-3y+5z=-1 \end{cases}$$

Explanation: This question tests understanding of how to represent a system of linear equations as a single matrix equation Ax = b. The matrix equation Ax = b is equivalent to the system of equations: when you multiply the coefficient matrix A by the variable vector x using matrix-vector multiplication, each entry of the result equals the corresponding entry of the constant vector b, giving you the original equations. The matrix equation [[-2 1 0]; [4 -3 5]] [x; y; z] = [8; -1] expands by performing the matrix-vector multiplication: the first entry of Ax is -2x + 1y + 0z, and the first entry of b is 8, giving equation -2x + y = 8; similarly, the second entry gives equation 4x - 3y + 5z = -1. Choice A is correct because it accurately represents the system as Ax = b with proper structure, including the missing z in the first equation. Choice B includes an extra z term in the first equation, using +z when the matrix has 0z. To verify your matrix equation is correct, expand Ax by multiplying: each row of A dotted with x should give the left side of the corresponding equation, and these should equal the entries of b (the right sides). Remember to include 0 coefficients when a variable doesn't appear in an equation: for example, if equation 2 is 'x + 3z = 5' in a system with variables x, y, z, the coefficient row is [1 0 3], with the 0 for the missing y.

Question 12

1. $-3$
2. $0$
3. $1$ (correct answer)
4. $2$

Explanation: This question tests understanding of how to represent a system of linear equations as a single matrix equation $Ax = b$. To build the coefficient matrix $A$ from a system of equations, create a matrix where entry $a_{ij}$ is the coefficient of variable j in equation i: rows represent equations, columns represent variables, and if a variable doesn't appear in an equation, its coefficient is 0. From equation $4x - y + z = -2$, the coefficients are 4 for x, -1 for y, and 1 for z, so the corresponding row of the coefficient matrix is $[4, -1, 1]$, and $a_{23}$ is the entry in row 2, column 3, which is 1 for z. Choice C is correct because it properly extracts the coefficient for z in the second equation as 1. Choice A has the wrong sign on the coefficient, using -3 for a different variable when the actual coefficient for z in equation 2 is +1. Key to converting systems to matrix form: create the coefficient matrix $A$ by listing coefficients in order (row i = equation i, column j = variable j), create variable vector x listing unknowns in the same order as columns, and create constant vector b listing the right-hand sides in the same order as equations. Remember to include 0 coefficients when a variable doesn't appear in an equation: for example, if equation 2 is $x + 3z = 5$ in a system with variables x, y, z, the coefficient row is $[1, 0, 3]$, with the 0 for the missing y.

Question 13

1. $-2$ (correct answer)
2. $-1$
3. $4$
4. $1$

Explanation: In matrix form $A\vec{x} = \vec{b}$, matrix $A$ contains the coefficients of the variables. The second row corresponds to the second equation $x + 4y - 2z = -1$, and the third column corresponds to the coefficient of $z$. Therefore, the entry in the second row, third column is $-2$. Choice B is the constant term from the second equation. Choice C is the coefficient of $y$ in the second equation. Choice D is the coefficient of $x$ in the second equation.

Question 14

1. $\begin{pmatrix} 2 & 1 \\ -1 & 5 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 3 \\ -4 \end{pmatrix}$ (correct answer)
2. $\begin{pmatrix} 2 & -1 \\ 1 & 5 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 3 \\ -4 \end{pmatrix}$
3. $\begin{pmatrix} 2 & 1 \\ -1 & 5 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} -4 \\ 3 \end{pmatrix}$
4. $\begin{pmatrix} 2 & 1 \\ 1 & 5 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 3 \\ -4 \end{pmatrix}$

Explanation: This question tests understanding of how to represent a system of linear equations as a single matrix equation $Ax = b$. A system of linear equations can be written compactly as $Ax = b$, where $A$ is the coefficient matrix (containing all the coefficients of the variables), $x$ is the column vector of variables (the unknowns), and $b$ is the column vector of constants (the right-hand side values from each equation). For the system $2x + y = 3$, $-x + 5y = -4$, we extract the coefficients: equation 1 has coefficients 2 for x and 1 for y, giving first row $[2\ 1]$, and equation 2 has coefficients -1 for x and 5 for y, giving second row $[-1\ 5]$. The coefficient matrix is $A = \begin{pmatrix} 2 & 1 \\ -1 & 5 \end{pmatrix}$, the variable vector is $x = \begin{pmatrix} x \\ y \end{pmatrix}$, and the constant vector is $b = \begin{pmatrix} 3 \\ -4 \end{pmatrix}$. Choice A is correct because it properly extracts coefficients into matrix A and correctly identifies variable and constant vectors. Choice B has the wrong sign on the coefficient of x in equation 2, using 1 when the equation has -1, and swaps signs in the rows. The matrix form $Ax = b$ is just a compact notation for the system—it doesn't solve the system, but it sets up for solution methods like using the inverse (if A is invertible) or row reduction.

Question 15

1. $\begin{bmatrix}3 & -2\\ 4 & 1\end{bmatrix}$ (correct answer)
2. $\begin{bmatrix}3 & 4\\ -2 & 1\end{bmatrix}$
3. $\begin{bmatrix}3 & -2\\ 4 & -1\end{bmatrix}$
4. $\begin{bmatrix}3 & -2 & 5\\ 4 & 1 & -6\end{bmatrix}$

Explanation: This question tests understanding of how to represent a system of linear equations as a single matrix equation $Ax = b$. To build the coefficient matrix A from a system of equations, create a matrix where entry $a_{ij}$ is the coefficient of variable j in equation i: rows represent equations, columns represent variables, and if a variable doesn't appear in an equation, its coefficient is 0. For the system $3x - 2y = 5$ and $4x + y = -6$, we extract the coefficients: equation 1 has coefficients 3 for x and -2 for y, giving first row $[3, -2]$, and equation 2 has coefficients 4 for x and 1 for y, giving second row $[4, 1]$. Choice A is correct because it properly extracts coefficients into matrix A with accurate signs and values. Choice D includes the constants in the coefficient matrix instead of putting them in the separate constant vector b—the coefficient matrix A should only contain coefficients of variables, not the right-hand side values. Key to converting systems to matrix form: create the coefficient matrix A by listing coefficients in order (row i = equation i, column j = variable j), create variable vector x listing unknowns in the same order as columns, and create constant vector b listing the right-hand sides in the same order as equations. Common error: don't put the constants (right-hand side values) into the coefficient matrix A—they belong in the separate vector b, while A contains only the coefficients of the variables.

Question 16

1. $\begin{bmatrix}1&1\\2&-1\end{bmatrix}\begin{bmatrix}x\\y\end{bmatrix}=\begin{bmatrix}3\\1\end{bmatrix}$ (correct answer)
2. $\begin{bmatrix}1&2\\1&-1\end{bmatrix}\begin{bmatrix}x\\y\end{bmatrix}=\begin{bmatrix}3\\1\end{bmatrix}$
3. $\begin{bmatrix}1&1\\2&1\end{bmatrix}\begin{bmatrix}x\\y\end{bmatrix}=\begin{bmatrix}3\\1\end{bmatrix}$
4. $\begin{bmatrix}1&1\\2&-1\end{bmatrix}\begin{bmatrix}x\\y\end{bmatrix}=\begin{bmatrix}1\\3\end{bmatrix}$

Explanation: This question tests understanding of how to represent a system of linear equations as a single matrix equation Ax = b. A system of linear equations can be written compactly as Ax = b, where A is the coefficient matrix (containing all the coefficients of the variables), x is the column vector of variables (the unknowns), and b is the column vector of constants (the right-hand side values from each equation). For the system x + y = 3 and 2x - y = 1, we extract the coefficients: equation 1 has coefficients 1 and 1 for variables x and y, giving first row [1 1], and equation 2 has coefficients 2 and -1, giving second row [2 -1]. The coefficient matrix is A = [[1 1], [2 -1]], the variable vector is x = [x; y], and the constant vector is b = [3; 1]. Choice A is correct because it properly extracts coefficients into matrix A = [[1 1], [2 -1]] and correctly identifies the constant vector b = [3; 1]. Choice C has the wrong sign on the coefficient of y in the second equation, using 1 when the equation has -y (coefficient -1). To verify your matrix equation is correct, expand Ax by multiplying: each row of A dotted with x should give the left side of the corresponding equation, and these should equal the entries of b (the right sides).

Question 17

1. $-1$
2. $4$
3. $1$ (correct answer)
4. $3$

Explanation: This question tests understanding of how to represent a system of linear equations as a single matrix equation $Ax = b$. To build the coefficient matrix $A$ from a system of equations, create a matrix where entry $a_{ij}$ is the coefficient of variable $j$ in equation $i$: rows represent equations, columns represent variables, and if a variable doesn't appear in an equation, its coefficient is 0. From equation $5x - y + z = -7$, the coefficients are 5 for $x$, -1 for $y$, and 1 for $z$, so the corresponding row of the coefficient matrix is [5 -1 1], and thus $a_{23}$ (row 2, column 3) is 1. Choice C is correct because it accurately extracts the coefficient of $z$ in the second equation. Choice A has the wrong sign, using -1 when the actual coefficient is 1 for $z$ in equation 2. Key to converting systems to matrix form: create the coefficient matrix $A$ by listing coefficients in order (row $i$ = equation $i$, column $j$ = variable $j$), create variable vector $x$ listing unknowns in the same order as columns, and create constant vector $b$ listing the right-hand sides in the same order as equations. To verify your matrix equation is correct, expand $Ax$ by multiplying: each row of $A$ dotted with $x$ should give the left side of the corresponding equation, and these should equal the entries of $b$ (the right sides).

Question 18

1. $\begin{bmatrix}1 & 2\\ 3 & -1\end{bmatrix}\begin{bmatrix}x\\ y\end{bmatrix}=\begin{bmatrix}-3\\ 8\end{bmatrix}$ (correct answer)
2. $\begin{bmatrix}1 & 2\\ 3 & 1\end{bmatrix}\begin{bmatrix}x\\ y\end{bmatrix}=\begin{bmatrix}-3\\ 8\end{bmatrix}$
3. $\begin{bmatrix}1 & 3\\ 2 & -1\end{bmatrix}\begin{bmatrix}x\\ y\end{bmatrix}=\begin{bmatrix}-3\\ 8\end{bmatrix}$
4. $\begin{bmatrix}1 & 2\\ 3 & -1\end{bmatrix}\begin{bmatrix}x\\ y\end{bmatrix}=\begin{bmatrix}3\\ 8\end{bmatrix}$

Explanation: This question tests understanding of how to represent a system of linear equations as a single matrix equation $Ax = b$. The matrix equation $Ax = b$ is equivalent to the system of equations: when you multiply the coefficient matrix $A$ by the variable vector $x$ using matrix-vector multiplication, each entry of the result equals the corresponding entry of the constant vector $b$, giving you the original equations. For the system $x + 2y = -3$ and $3x - y = 8$, the coefficients match $\begin{bmatrix} 1 & 2 \\ 3 & -1 \end{bmatrix}$ and $b = \begin{bmatrix} -3 \\ 8 \end{bmatrix}$, as expanding gives $1x + 2y = -3$ and $3x + (-1)y = 8$. Choice A is correct because it accurately represents the system as $Ax = b$ with proper structure. Choice C transposes the coefficient matrix, listing coefficients by column instead of by row, which reverses the structure (variables and equations get mixed up). Key to converting systems to matrix form: create the coefficient matrix $A$ by listing coefficients in order (row i = equation i, column j = variable j), create variable vector $x$ listing unknowns in the same order as columns, and create constant vector $b$ listing the right-hand sides in the same order as equations. The matrix form $Ax = b$ is just a compact notation for the system—it doesn't solve the system, but it sets up for solution methods like using the inverse (if $A$ is invertible) or row reduction.

Question 19

1. $\begin{pmatrix}x&y\end{pmatrix}$
2. $\begin{pmatrix}y\\x\end{pmatrix}$
3. $\begin{pmatrix}x\\y\end{pmatrix}$ (correct answer)
4. $\begin{pmatrix}0\\10\end{pmatrix}$

Explanation: This question tests understanding of how to represent a system of linear equations as a single matrix equation Ax = b. A system of linear equations can be written compactly as Ax = b, where A is the coefficient matrix (containing all the coefficients of the variables), x is the column vector of variables (the unknowns), and b is the column vector of constants (the right-hand side values from each equation). For the system 3x - 2y = 0, x + 4y = 10, the variable vector x should list the unknowns in the order matching the columns of A, which for variable order x, y is the column vector [x; y]. Choice C is correct because it correctly identifies the variable vector as a column vector in the order x, y. Choice A treats the variable vector x as a row vector instead of a column vector, which would make the dimensions incompatible for the multiplication Ax. Key to converting systems to matrix form: create the coefficient matrix A by listing coefficients in order (row i = equation i, column j = variable j), create variable vector x listing unknowns in the same order as columns, and create constant vector b listing the right-hand sides in the same order as equations. To verify your matrix equation is correct, expand Ax by multiplying: each row of A dotted with x should give the left side of the corresponding equation, and these should equal the entries of b (the right sides).

Question 20

1. $\begin{bmatrix}5\\ -3\end{bmatrix}$
2. $\begin{bmatrix}2\\ 9\end{bmatrix}$ (correct answer)
3. $\begin{bmatrix}2 & 9\end{bmatrix}$
4. $\begin{bmatrix}5 & -1\\ -3 & 4\end{bmatrix}$

Explanation: This question tests understanding of how to represent a system of linear equations as a single matrix equation $Ax = b$. A system of linear equations can be written compactly as $Ax = b$, where A is the coefficient matrix (containing all the coefficients of the variables), x is the column vector of variables (the unknowns), and b is the column vector of constants (the right-hand side values from each equation). For the system $5x - y = 2$ and $-3x + 4y = 9$, the constant vector b takes the right-hand side values in order, so $b = \begin{bmatrix} 2 \\ 9 \end{bmatrix}$. Choice B is correct because it correctly identifies the constant vector as the column vector of right-hand sides. Choice A scrambles the equation order, putting constants from the coefficient matrix instead of the right-hand sides. Key to converting systems to matrix form: create the coefficient matrix A by listing coefficients in order (row i = equation i, column j = variable j), create variable vector x listing unknowns in the same order as columns, and create constant vector b listing the right-hand sides in the same order as equations. Common error: don't put the constants (right-hand side values) into the coefficient matrix A—they belong in the separate vector b, while A contains only the coefficients of the variables.