Representing Linear Systems with Matrices - Precalculus

1. Given the matrix equation $\begin{pmatrix}1&-4\\3&2\end{pmatrix}\begin{pmatrix}x\\y\end{pmatrix}=\begin{pmatrix}-6\\5\end{pmatrix},$ which system of equations does it represent?
   Answer: $\begin{cases} x-4y=-6 \\ 3x+2y=5 \end{cases}$ 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 $\begin{pmatrix} 1 & -4 \\ 3 & 2 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} -6 \\ 5 \end{pmatrix}$ expands by performing the matrix-vector multiplication: the first entry of $Ax$ is $1x + (-4)y$, and the first entry of $b$ is $-6$, giving equation $x - 4y = -6$; similarly, the second entry gives equation $3x + 2y = 5$. Choice A is correct because it accurately represents the system as $Ax = b$ with proper structure. Choice C has the wrong sign on the constant, using 5 when the matrix has -6 for the first equation and -6 when it has 5 for the second. 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.
2. Consider the matrix equation: $\begin{pmatrix} a & b & c \\ d & e & f \\ g & h & i \end{pmatrix} \begin{pmatrix} s \\ t \\ u \end{pmatrix} = \begin{pmatrix} p \\ q \\ r \end{pmatrix}$ If the corresponding system has the equation $2s - 3t + 7u = -5$ as its second equation, what are the values of $d$, $e$, $f$, and $q$?
   Answer: $d = 2, e = -3, f = 7, q = -5$ 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 $q = -5$ This gives us $d = 2, e = -3, f = 7, q = -5$, which is answer choice B. Looking at the wrong answers: Choice A incorrectly places $-5$ as the value of $d$ instead of $q$. Choice C swaps the values of $e$ and $f$, confusing the coefficients of $t$ and $u$. Choice D mixes up multiple coefficients, placing $-3$ in the $d$ position. Study tip: Always remember that in matrix-to-system conversions, coefficients stay in their exact positions—first matrix entry becomes the first variable's coefficient, second becomes the second variable's coefficient, and so on. The result vector gives you the constants on the right side of each equation.
3. Given the matrix equation $\begin{bmatrix}1 & -3\\ 2 & 5\end{bmatrix}\begin{bmatrix}x\\ y\end{bmatrix}=\begin{bmatrix}4\\ -1\end{bmatrix},$ which system of equations does it represent?​
   Answer: $\begin{cases} x-3y=4\\ 2x+5y=-1 \end{cases}$ 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 [[1 -3]; [2 5]] [x; y] = [4; -1] 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. Choice B has the wrong sign on the coefficient of y in equation 1, using +3 instead of -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.
4. For the system of equations $\begin{cases} 3x+y=-5\\ 2x-4y=6 \end{cases}$ what is the coefficient matrix $A$ in $Ax=b$ (using variable order $[x\; y]^T$)?
   Answer: $\begin{bmatrix}3&1\\2&-4\end{bmatrix}$ 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 3x + y = -5, the coefficients are 3 for x and 1 for y, so the corresponding row of the coefficient matrix is [3 1]. From equation 2x - 4y = 6, the coefficients are 2 for x and -4 for y, giving the second row [2 -4]. The constant -5 and 6 go in the constant vector b, not in the coefficient matrix A. Choice A is correct because it properly extracts coefficients into matrix A = [[3 1], [2 -4]], maintaining the correct order of equations as rows and variables as columns. Choice B transposes the coefficient matrix, listing coefficients by column instead of by row, which reverses the structure (variables and equations get mixed up). 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).
5. Given the matrix equation $\begin{pmatrix}-2&1&0\\4&-3&5\end{pmatrix}\begin{pmatrix}x\\y\\z\end{pmatrix}=\begin{pmatrix}8\\-1\end{pmatrix},$ which system of equations does it represent?​
   Answer: $\begin{cases} -2x+y=8\\ 4x-3y+5z=-1 \end{cases}$ 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.
6. For the system of equations $\begin{cases} 4x + y = -2\\ -3x + 2y = 9 \end{cases}$ what is the constant vector $b$ in $Ax=b$ (using variable order $x, y$)?
   Answer: $\begin{bmatrix}-2\\ 9\end{bmatrix}$ 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). From the system 4x + y = -2, -3x + 2y = 9, the constants are -2 from the first equation and 9 from the second, so the constant vector is $b = \begin{bmatrix} -2 \\ 9 \end{bmatrix}$. Choice A is correct because it correctly identifies the constant vector with the right-hand side values in the order of the equations. Choice B includes coefficients from the equations instead of the constants, using $\begin{bmatrix} 4 \\ -3 \end{bmatrix}$ when those are left-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.
7. For the system of equations $\begin{cases} x+2y-z=4\\ -2x+y+3z=-1\\ 5x-z=6 \end{cases}$ what is the constant vector $b$ in $Ax=b$?
   Answer: $\begin{bmatrix}4\\-1\\6\end{bmatrix}$ 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 + 2y - z = 4, -2x + y + 3z = -1, and 5x - z = 6, we need to identify the constant vector b, which contains the right-hand side values from each equation in order. The first equation has right-hand side 4, the second has -1, and the third has 6, so b = [4; -1; 6]. Choice A is correct because it properly lists the constants from the right-hand sides of the equations in the correct order: [4; -1; 6]. Choice D has the wrong sign on the last constant, using -6 when the equation has positive 6. 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.
8. What system of equations does the matrix equation represent? $\begin{bmatrix}0 & -2 & 3\\ 5 & 1 & -1\end{bmatrix}\begin{bmatrix}x\\ y\\ z\end{bmatrix}=\begin{bmatrix}7\\ -4\end{bmatrix}$
   Answer: $\begin{cases} -2y+3z=7\\ 5x+y-z=-4 \end{cases}$ 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).
9. A system of linear equations is given by: $2x - 3y + z = 7$ $x + 4y - 2z = -1$ $-3x + y + 5z = 8$ When this system is written in the matrix form $A\vec{x} = \vec{b}$, what is the entry in the second row, third column of matrix $A$?
   Answer: $-2$ 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.
10. Verify the matrix equation is equivalent to the system by choosing the matching matrix form. System: $\begin{cases} x+y=3\\ 2x-y=1 \end{cases}$ Which matrix equation correctly represents it as $Ax=b$?
    Answer: $\begin{bmatrix}1&1\\2&-1\end{bmatrix}\begin{bmatrix}x\\y\end{bmatrix}=\begin{bmatrix}3\\1\end{bmatrix}$ 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).
11. For the system of equations $\begin{cases} 5x - y = 2 \\ -3x + 4y = 9 \end{cases}$ what is the constant vector $b$ in $Ax=b$?
    Answer: $\begin{bmatrix}2\\ 9\end{bmatrix}$ 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.
12. Represent the system as a matrix equation $Ax=b$ using variable order $x, y, z$: $\begin{cases} 2x - y = 3\\ -x + 4z = -2\\ 3y + z = 5 \end{cases}$
    Answer: $\begin{bmatrix}2 & -1 & 0\\ -1 & 0 & 4\\ 0 & 3 & 1\end{bmatrix}\begin{bmatrix}x\\ y\\ z\end{bmatrix}=\begin{bmatrix}3\\ -2\\ 5\end{bmatrix}$ 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 $2x - y = 3$, $-x + 4z = -2$, $3y + z = 5$, we extract the coefficients: equation 1 has 2 for $x$, -1 for $y$, 0 for $z$; equation 2 has -1 for $x$, 0 for $y$, 4 for $z$; equation 3 has 0 for $x$, 3 for $y$, 1 for $z$; the coefficient matrix is $A = \begin{bmatrix} 2 & -1 & 0 \\ -1 & 0 & 4 \\ 0 & 3 & 1 \end{bmatrix}$, the variable vector is $\mathbf{x} = \begin{bmatrix} x \\ y \\ z \end{bmatrix}$, and the constant vector is $\mathbf{b} = \begin{bmatrix} 3 \\ -2 \\ 5 \end{bmatrix}$. Choice A is correct because it properly extracts coefficients into matrix $A$, including zeros for missing variables. Choice D scrambles the equation order, putting coefficients in wrong positions, such as 4 in the wrong row and column. Remember to include 0 coefficients when a variable doesn't appear in an equation: for example, if equation 2 is '$-x + 4z = -2$' in a system with variables $x, y, z$, the coefficient row is $[-1 \ 0 \ 4]$, with the 0 for the missing $y$. 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.
13. Given the matrix equation $\begin{bmatrix}1 & -3\\ 2 & 5\end{bmatrix}\begin{bmatrix}x\\ y\end{bmatrix}=\begin{bmatrix}4\\ -1\end{bmatrix},$ which system of equations does it represent?
    Answer: $\begin{cases} x-3y=4\\ 2x+5y=-1 \end{cases}$ 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.
14. Write the system in matrix form $Ax=b$: $\begin{cases} x+2y-z=4\\ -3x+y+2z=-1\\ 2x-5y+z=0 \end{cases}$​
    Answer: $\begin{pmatrix}1&2&-1\\-3&1&2\\2&-5&1\end{pmatrix}\begin{pmatrix}x\\y\\z\end{pmatrix}=\begin{pmatrix}4\\-1\\0\end{pmatrix}$ 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 + 2y - z = 4, -3x + y + 2z = -1, 2x - 5y + z = 0, we extract the coefficients: equation 1 has coefficients [1 2 -1] for variables x, y, z, giving first row [1 2 -1], equation 2 has [-3 1 2], giving second row [-3 1 2], and equation 3 has [2 -5 1], giving third row [2 -5 1]. The coefficient matrix is A = [[1 2 -1]; [-3 1 2]; [2 -5 1]], the variable vector is x = [x; y; z], and the constant vector is b = [4; -1; 0]. 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 3 when the equation has -3. 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.
15. For the system of equations $\begin{cases} 3x-2y=0\\ x+4y=10 \end{cases}$ which variable vector $\mathbf{x}$ should be used in $A\mathbf{x}=\mathbf{b}$ (using variable order $x,y$)?​
    Answer: $\begin{pmatrix}x\\y\end{pmatrix}$ 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).
16. For the system of equations $\begin{cases} 2x-3y=5\\ -x+4y=-2 \end{cases}$ write the system as a matrix equation $Ax=b$.
    Answer: $\begin{pmatrix}2&-3\\-1&4\end{pmatrix}\begin{pmatrix}x\\y\end{pmatrix}=\begin{pmatrix}5\\-2\end{pmatrix}$ 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 $2x - 3y = 5, -x + 4y = -2$, 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 = $\begin{pmatrix} 2 & -3 \\ -1 & 4 \end{pmatrix}$, the variable vector is x = $\begin{pmatrix} x \\ y \end{pmatrix}$, and the constant vector is b = $\begin{pmatrix} 5 \\ -2 \end{pmatrix}$. Choice A is correct because it properly extracts coefficients into matrix A and correctly identifies variable and constant vectors. 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.
17. For the system of equations $\begin{cases} -4x + 2y = 10 \\ x + 5y = -3 \end{cases}$ which matrix correctly represents the coefficients (the matrix $A$ in $Ax=b$ with variable order $(x,y)$)?
    Answer: $\begin{bmatrix}-4 & 2\\ 1 & 5\end{bmatrix}$ 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 $a_{i j}$ 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 -4x + 2y = 10 and x + 5y = -3, we extract the coefficients: equation 1 has -4 for x and 2 for y, giving first row $[-4 \ 2]$, and equation 2 has 1 for x and 5 for y, giving second row $[1 \ 5]$. Choice A is correct because it properly extracts coefficients into matrix A with correct order and signs. 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.
18. Given the matrix equation $\begin{bmatrix}-2&3\\5&1\end{bmatrix}\begin{bmatrix}x\\y\end{bmatrix}=\begin{bmatrix}1\\-9\end{bmatrix},$ what does it represent as a system of equations?
    Answer: $\begin{cases}-2x+3y=1\\5x+y=-9\end{cases}$ 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 3], [5 1]][x; y] = [1; -9] expands by performing the matrix-vector multiplication: the first entry of Ax is -2x + 3y, and the first entry of b is 1, giving equation -2x + 3y = 1. Similarly, the second entry gives equation 5x + 1y = 5x + y = -9. Choice A is correct because it accurately represents the system as -2x + 3y = 1 and 5x + y = -9, matching the matrix multiplication results. Choice B has the wrong sign on the coefficient of y in the first equation, using -3y when the matrix has positive 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).
19. For the system of equations $\begin{cases} -x + 2y = 6\\ 5x - y = -3 \end{cases}$ write the system as a matrix equation $Ax=b$.
    Answer: $\begin{bmatrix}-1 & 2\\ 5 & -1\end{bmatrix}\begin{bmatrix}x\\ y\end{bmatrix}=\begin{bmatrix}6\\ -3\end{bmatrix}$ 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 $-x + 2y = 6$, $5x - y = -3$, we extract the coefficients: equation 1 has -1 for x and 2 for y, giving first row $[-1 \ 2]$, and equation 2 has 5 for x and -1 for y, giving second row $[5 \ -1]$; the coefficient matrix is $A = \begin{bmatrix} -1 & 2 \\ 5 & -1 \end{bmatrix}$, the variable vector is $x = \begin{bmatrix} x \\ y \end{bmatrix}$, and the constant vector is $b = \begin{bmatrix} 6 \\ -3 \end{bmatrix}$. Choice A is correct because it properly extracts coefficients into matrix A and correctly identifies variable and constant vectors. Choice C has the wrong sign on the coefficient of x in equation 1, using 1 instead of -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 the correct signs when extracting coefficients; for example, -y means coefficient -1 for y.
20. Verify that the matrix equation is equivalent to the system by selecting the matching matrix equation. System: $\begin{cases} 2x+y=3\\ -x+5y=-4 \end{cases}$ Which matrix equation $Ax=b$ matches this system (using variable order $x,y$)?​
    Answer: $\begin{pmatrix}2&1\\-1&5\end{pmatrix}\begin{pmatrix}x\\y\end{pmatrix}=\begin{pmatrix}3\\-4\end{pmatrix}$ 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 = [[2 1]; [-1 5]], the variable vector is x = [x; y], and the constant vector is b = [3; -4]. 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.

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