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.

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.

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 - 3y = 7 and -x + 4y = -2, we extract the coefficients: equation 1 has coefficients 2 and -3 for variables x and y, giving first row [2 -3], and equation 2 has coefficients -1 and 4, giving second row [-1 4]. The coefficient matrix is A = [[2 -3], [-1 4]], the variable vector is x = [x; y], and the constant vector is b = [7; -2]. Choice A is correct because it properly extracts coefficients into matrix A with the correct signs and places them in the right positions, with constants 7 and -2 forming vector b. Choice D has the wrong sign on the coefficient of y in the first equation, using 3 when the equation has -3y. 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.

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.

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 $2x + 3y = 7$ and $-x + 4y = -1$, we extract the coefficients: equation 1 has coefficients 2 for x and 3 for y, giving first row $[2\ 3]$, and equation 2 has coefficients -1 for x and 4 for y, giving second row $[-1\ 4]$; the coefficient matrix is $A = \begin{bmatrix} 2 & 3 \\ -1 & 4 \end{bmatrix}$, the variable vector is $x = \begin{bmatrix} x \\ y \end{bmatrix}$, and the constant vector is $b = \begin{bmatrix} 7 \\ -1 \end{bmatrix}$. Choice B is correct because it properly extracts coefficients into matrix $A$ and correctly identifies variable and constant vectors. Choice A 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. 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).

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.

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 $$\begin{bmatrix} 0 & -2 & 3 \\ 5 & 1 & -1 \\ \end{bmatrix} \begin{bmatrix} x \\ y \\ z \\ \end{bmatrix} = \begin{bmatrix} 7 \\ -4 \\ \end{bmatrix}$$ expands by performing the matrix-vector multiplication: the first entry of Ax is $0x + (-2)y + 3z = 7$, giving equation $-2y + 3z = 7$; similarly, the second entry gives $5x + 1y + (-1)z = -4$, or $5x + y - z = -4$. Choice B is correct because it accurately represents the system as Ax = b with proper structure, including the missing x in the first equation as coefficient 0. Choice A omits the zero coefficient for the missing variable x in the first equation, incorrectly including x with coefficient -2 instead of recognizing it as missing. 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. 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).

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 $2x + 0y = -4$ and $-x + 3y = 1$, we extract the coefficients: equation 1 has coefficients 2 for x and 0 for y, giving first row $[2\ 0]$, and equation 2 has coefficients -1 for x and 3 for y, giving second row $[-1\ 3]$. Choice A is correct because it properly extracts coefficients into matrix $A$, including the zero for the missing y in the first equation. Choice D has the wrong sign on the coefficient of x in equation 2, using +1 when the equation has -x. 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. 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).

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.

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.