This question tests your ability to represent a system of linear equations as a single matrix equation in the form AX = B, where A is the coefficient matrix, X is the variable vector, and B is the constant vector. Matrix form AX = B is a compact way to write entire systems: the coefficient matrix A contains all the coefficients from the left sides of equations (organized by rows for equations and columns for variables), the variable vector X lists the unknowns as a column, and the constant vector B lists the right-hand side values. When you multiply matrix A times vector X, you get the left sides of all equations, which equals vector B (the right sides). For example, the system 2x + 3y = 7 and x - y = 1 becomes [[2, 3], [1, -1]] times [x, y] = [7, 1]. Matrix multiplication recovers the original equations! To convert the system 3x + 2y = 8 and x - 5y = -3 to matrix form: (1) Write coefficient matrix A by using equation 1 coefficients as row 1 and equation 2 coefficients as row 2, with columns for each variable in order: A = [[3, 2], [1, -5]] (first column is x-coefficients: 3, 1; second column is y-coefficients: 2, -5). (2) Write variable vector X = [x, y] as column (order matches columns of A). (3) Write constant vector B = [8, -3] as column (first equation's constant 8, second equation's constant -3, matching equation order). (4) Combine: [[3, 2], [1, -5]] times [x, y] = [8, -3]. This matrix equation is equivalent to the original system! Choice A correctly constructs the coefficient matrix A with proper row-column organization, variable vector X in correct order, and constant vector B matching the equations. Choice B has the rows and columns of matrix A transposed (swapped), putting coefficients in wrong positions. Remember: rows correspond to equations (equation 1 is row 1, equation 2 is row 2), columns correspond to variables (x-column, then y-column). Don't flip this relationship! Matrix equation construction recipe: (1) Label equations (first, second, third, etc.) and variables (x, y, z in consistent order), (2) Build matrix A: make rows from equations (row i = coefficients from equation i), make columns from variables (column j = coefficient of variable j in each equation), include 0 for missing variables, (3) Write X as column vector with variables in same order as A's columns, (4) Write B as column vector with constants in same order as equations, (5) Combine as AX = B. Example: x + 2y = 5 and 3x - y = 7 gives A = [[1, 2], [3, -1]], X = [x, y], B = [5, 7]. Systematic! To verify your matrix equation represents the system correctly, perform the matrix multiplication AX mentally or on paper: the first row of A times X should give the left side of equation 1, the second row times X should give equation 2's left side, etc. If [[2, 3], [1, -1]] times [x, y] gives [2x + 3y, x - y], and B = [7, 1], then matrix equation represents 2x + 3y = 7 and x - y = 1. Matrix multiplication recovers the system—this verification prevents errors!

This question tests your ability to identify the correct coefficient matrix A for a system of linear equations in the form AX = B, including zeros for missing variables. Matrix form AX = B is a compact way to write entire systems: the coefficient matrix A contains all the coefficients from the left sides of equations (organized by rows for equations and columns for variables), the variable vector X lists the unknowns as a column, and the constant vector B lists the right-hand side values. When you multiply matrix A times vector X, you get the left sides of all equations, which equals vector B (the right sides). For example, the system $2x + 3y = 7$ and $x - y = 1$ becomes $$\begin{bmatrix} 2 & 3 \\ 1 & -1 \end{bmatrix}$$ times $$\begin{bmatrix} x \\ y \end{bmatrix}$$ = $$\begin{bmatrix} 7 \\ 1 \end{bmatrix}$$. Matrix multiplication recovers the original equations! For the system $5x - 2y + z = 0$, $3x - 4z = 9$, $-x + y + 2z = -6$ (order x, y, z), $$A = \begin{bmatrix} 5 & -2 & 1 \\ 3 & 0 & -4 \\ -1 & 1 & 2 \end{bmatrix}$$—note the 0 for missing y in the second equation. Choice A correctly places all coefficients, including the 0 for y in row 2 and proper signs like -4 for z. Choice B incorrectly moves the -4 to the y-column in row 2, confusing the variables—always align columns with the specified variable order! Matrix equation construction recipe: (1) Label equations (first, second, third, etc.) and variables (x, y, z in consistent order), (2) Build matrix A: make rows from equations (row i = coefficients from equation i), make columns from variables (column j = coefficient of variable j in each equation), include 0 for missing variables, (3) Write X as column vector with variables in same order as A's columns, (4) Write B as column vector with constants in same order as equations, (5) Combine as AX = B. Example: $x + 2y = 5$ and $3x - y = 7$ gives $$A = \begin{bmatrix} 1 & 2 \\ 3 & -1 \end{bmatrix}$$, $X = \begin{bmatrix} x \\ y \end{bmatrix}$, $B = \begin{bmatrix} 5 \\ 7 \end{bmatrix}$. Systematic! To verify your matrix equation represents the system correctly, perform the matrix multiplication AX mentally or on paper: the first row of A times X should give the left side of equation 1, the second row times X should give equation 2's left side, etc. If $$\begin{bmatrix} 2 & 3 \\ 1 & -1 \end{bmatrix}$$ times $$\begin{bmatrix} x \\ y \end{bmatrix}$$ gives $$\begin{bmatrix} 2x + 3y \\ x - y \end{bmatrix}$$, and $B = \begin{bmatrix} 7 \\ 1 \end{bmatrix}$, then matrix equation represents $2x + 3y = 7$ and $x - y = 1$. Matrix multiplication recovers the system—this verification prevents errors!

This question tests your ability to represent a system of linear equations as a single matrix equation in the form $AX = B$, where A is the coefficient matrix, X is the variable vector, and B is the constant vector. Matrix form $AX = B$ is a compact way to write entire systems: the coefficient matrix A contains all the coefficients from the left sides of equations (organized by rows for equations and columns for variables), the variable vector X lists the unknowns as a column, and the constant vector B lists the right-hand side values. When you multiply matrix A times vector X, you get the left sides of all equations, which equals vector B (the right sides). For example, the system $2x + 3y = 7$ and $x - y = 1$ becomes $$\begin{bmatrix} 2 & 3 \\ 1 & -1 \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 7 \\ 1 \end{bmatrix}$$. Matrix multiplication recovers the original equations! To convert the system $-4x + y = 6$ and $2x + 3y = -1$ to matrix form: A = $$\begin{bmatrix} -4 & 1 \\ 2 & 3 \end{bmatrix}$$, X = $$\begin{bmatrix} x \\ y \end{bmatrix}$$, B = $$\begin{bmatrix} 6 \\ -1 \end{bmatrix}$$—note the positive 3y and negative constant -1 are correctly placed. Choice A correctly constructs the coefficient matrix A with all signs preserved, variable vector X in order, and constant vector B matching the right sides. Choice B incorrectly changes the first constant to -6, perhaps from misapplying a sign—always copy constants directly from the system! Matrix equation construction recipe: (1) Label equations (first, second, third, etc.) and variables (x, y, z in consistent order), (2) Build matrix A: make rows from equations (row i = coefficients from equation i), make columns from variables (column j = coefficient of variable j in each equation), include 0 for missing variables, (3) Write X as column vector with variables in same order as A's columns, (4) Write B as column vector with constants in same order as equations, (5) Combine as $AX = B$. Example: $x + 2y = 5$ and $3x - y = 7$ gives A = $$\begin{bmatrix} 1 & 2 \\ 3 & -1 \end{bmatrix}$$, X = $$\begin{bmatrix} x \\ y \end{bmatrix}$$, B = $$\begin{bmatrix} 5 \\ 7 \end{bmatrix}$$. Systematic! To verify your matrix equation represents the system correctly, perform the matrix multiplication $AX$ mentally or on paper: the first row of A times X should give the left side of equation 1, the second row times X should give equation 2's left side, etc. If $$\begin{bmatrix} 2 & 3 \\ 1 & -1 \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix}$$ gives $[2x + 3y, x - y]$, and B = $$\begin{bmatrix} 7 \\ 1 \end{bmatrix}$$, then matrix equation represents $2x + 3y = 7$ and $x - y = 1$. Matrix multiplication recovers the system—this verification prevents errors!

This question tests your ability to represent a system of linear equations as a single matrix equation in the form $AX = B$, where $A$ is the coefficient matrix, $X$ is the variable vector, and $B$ is the constant vector, including zeros for missing variables. Matrix form $AX = B$ is a compact way to write entire systems: the coefficient matrix $A$ contains all the coefficients from the left sides of equations (organized by rows for equations and columns for variables), the variable vector $X$ lists the unknowns as a column, and the constant vector $B$ lists the right-hand side values. When you multiply matrix $A$ times vector $X$, you get the left sides of all equations, which equals vector $B$ (the right sides). For example, the system $2x + 3y = 7$ and $x - y = 1$ becomes $$\begin{bmatrix} 2 & 3 \\ 1 & -1 \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 7 \\ 1 \end{bmatrix}$$. Matrix multiplication recovers the original equations! To convert the system $-x + 2y + z = 3$, $4x - y = 0$, $2x + 5y - 3z = -7$ to matrix form (order $x, y, z$): $A = \begin{bmatrix} -1 & 2 & 1 \\ 4 & -1 & 0 \\ 2 & 5 & -3 \end{bmatrix}$, $X = \begin{bmatrix} x \\ y \\ z \end{bmatrix}$, $B = \begin{bmatrix} 3 \\ 0 \\ -7 \end{bmatrix}$—note the 0 for missing $z$ in the second equation. Choice A correctly includes the zero for $z$ in row 2, with all coefficients and constants in proper order. Choice D transposes parts of $A$, like putting rows into columns, which scrambles the coefficients—stick to rows as equations and columns as variables! Matrix equation construction recipe: (1) Label equations (first, second, third, etc.) and variables ($x, y, z$ in consistent order), (2) Build matrix $A$: make rows from equations (row i = coefficients from equation i), make columns from variables (column j = coefficient of variable j in each equation), include 0 for missing variables, (3) Write $X$ as column vector with variables in same order as $A$'s columns, (4) Write $B$ as column vector with constants in same order as equations, (5) Combine as $AX = B$. Example: $x + 2y = 5$ and $3x - y = 7$ gives $A = \begin{bmatrix} 1 & 2 \\ 3 & -1 \end{bmatrix}$, $X = \begin{bmatrix} x \\ y \end{bmatrix}$, $B = \begin{bmatrix} 5 \\ 7 \end{bmatrix}$. Systematic! To verify your matrix equation represents the system correctly, perform the matrix multiplication $AX$ mentally or on paper: the first row of $A$ times $X$ should give the left side of equation 1, the second row times $X$ should give equation 2's left side, etc. If $$\begin{bmatrix} 2 & 3 \\ 1 & -1 \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix}$$ gives $$\begin{bmatrix} 2x + 3y \\ x - y \end{bmatrix}$$, and $B = \begin{bmatrix} 7 \\ 1 \end{bmatrix}$, then matrix equation represents $2x + 3y = 7$ and $x - y = 1$. Matrix multiplication recovers the system—this verification prevents errors!

This question tests your ability to represent a system of linear equations as a single matrix equation in the form $AX = B$, where $A$ is the coefficient matrix, $X$ is the variable vector, and $B$ is the constant vector. Matrix form $AX = B$ is a compact way to write entire systems: the coefficient matrix $A$ contains all the coefficients from the left sides of equations (organized by rows for equations and columns for variables), the variable vector $X$ lists the unknowns as a column, and the constant vector $B$ lists the right-hand side values. When you multiply matrix $A$ times vector $X$, you get the left sides of all equations, which equals vector $B$ (the right sides). For example, the system $2x + 3y = 7$ and $x - y = 1$ becomes $$\begin{bmatrix} 2 & 3 \\ 1 & -1 \end{bmatrix}$$ times $$\begin{bmatrix} x \\ y \end{bmatrix}$$ = $$\begin{bmatrix} 7 \\ 1 \end{bmatrix}$$. Matrix multiplication recovers the original equations! For the system $3x + 2y - z = 7$ and $-x + 4y + 5z = -2$, the matrices are $A = \begin{bmatrix} 3 & 2 & -1 \\ -1 & 4 & 5 \end{bmatrix}$ (rows for equations, columns for $x, y, z$), $X = \begin{bmatrix} x \\ y \\ z \end{bmatrix}$, $B = \begin{bmatrix} 7 \\ -2 \end{bmatrix}$ (constants in equation order), so $AX = B$ matches the system perfectly. Choice A correctly constructs the coefficient matrix $A$ with proper row-column organization, variable vector $X$ in correct order, and constant vector $B$ matching the equations. Choice B has the rows of matrix $A$ swapped, putting the second equation's coefficients in the first row, which mismatches the equation order. Remember: rows must correspond to the order of equations as given—don't rearrange them! Matrix equation construction recipe: (1) Label equations (first, second, third, etc.) and variables ($x, y, z$ in consistent order), (2) Build matrix $A$: make rows from equations (row i = coefficients from equation i), make columns from variables (column j = coefficient of variable j in each equation), include 0 for missing variables, (3) Write $X$ as column vector with variables in same order as $A$'s columns, (4) Write $B$ as column vector with constants in same order as equations, (5) Combine as $AX = B$. Example: $x + 2y = 5$ and $3x - y = 7$ gives $A = \begin{bmatrix} 1 & 2 \\ 3 & -1 \end{bmatrix}$, $X = \begin{bmatrix} x \\ y \end{bmatrix}$, $B = \begin{bmatrix} 5 \\ 7 \end{bmatrix}$. Systematic! To verify your matrix equation represents the system correctly, perform the matrix multiplication $AX$ mentally or on paper: the first row of $A$ times $X$ should give the left side of equation 1, the second row times $X$ should give equation 2's left side, etc. If $$\begin{bmatrix} 2 & 3 \\ 1 & -1 \end{bmatrix}$$ times $$\begin{bmatrix} x \\ y \end{bmatrix}$$ gives $$\begin{bmatrix} 2x + 3y \\ x - y \end{bmatrix}$$, and $B = \begin{bmatrix} 7 \\ 1 \end{bmatrix}$, then matrix equation represents $2x + 3y = 7$ and $x - y = 1$. Matrix multiplication recovers the system—this verification prevents errors!

This question tests your ability to convert a matrix equation AX = B back into the equivalent system of linear equations by performing the matrix multiplication. Matrix form AX = B is a compact way to write entire systems: the coefficient matrix A contains all the coefficients from the left sides of equations (organized by rows for equations and columns for variables), the variable vector X lists the unknowns as a column, and the constant vector B lists the right-hand side values. When you multiply matrix A times vector X, you get the left sides of all equations, which equals vector B (the right sides). For example, the system $2x + 3y = 7$ and $x - y = 1$ becomes $$ \begin{bmatrix} 2 & 3 \\ 1 & -1 \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 7 \\ 1 \end{bmatrix} $$. Matrix multiplication recovers the original equations! To convert $$ \begin{bmatrix} 2 & -1 \\ 3 & 4 \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 5 \\ 7 \end{bmatrix} $$ back to a system: multiply first row $[2, -1]$ by $[x, y]$ to get $2x - y = 5$ (first equation), second row $[3, 4]$ by $[x, y]$ to get $3x + 4y = 7$ (second equation)—this matches the original system. Choice A correctly identifies the system by properly multiplying each row of A by X to form the left sides and setting them equal to the corresponding entries in B. Choice C has the coefficients mixed up, like using -3 for y in the first equation instead of -1, likely from misreading the matrix entries. Remember: the entry in row i, column j of A is the coefficient of variable j in equation i—read carefully! To convert any AX = B to a system, (1) For each row i of A, multiply that row by X to get the left side of equation i (e.g., $a_{i1}x + a_{i2}y + \dots$), (2) Set it equal to $b_i$ from B, (3) Write out all equations in order. Example: $$ \begin{bmatrix} 1 & 2 \\ 3 & -1 \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 5 \\ 7 \end{bmatrix} $$ gives $x + 2y = 5$ and $3x - y = 7$. Straightforward! Always verify by plugging back: if you multiply your derived system's A by X, you should recover the original matrix equation. This double-check ensures accuracy—you've got this!

This question tests your ability to convert a matrix equation AX = B back into the equivalent system of linear equations by performing the matrix multiplication, including handling zeros in A. Matrix form AX = B is a compact way to write entire systems: the coefficient matrix A contains all the coefficients from the left sides of equations (organized by rows for equations and columns for variables), the variable vector X lists the unknowns as a column, and the constant vector B lists the right-hand side values. When you multiply matrix A times vector X, you get the left sides of all equations, which equals vector B (the right sides). For example, the system 2x + 3y = 7 and x - y = 1 becomes [[2, 3], [1, -1]] times [x, y] = [7, 1]. Matrix multiplication recovers the original equations! To convert [[1, 0, -2], [0, 3, 1], [4, -1, 0]] times [x, y, z] = [6, -5, 8]: row 1 gives x - 2z = 6, row 2 gives 3y + z = -5, row 3 gives 4x - y = 8—note zeros mean missing variables like no y in equation 1. Choice A correctly identifies the system by multiplying each row accurately, preserving signs and omitting terms with zero coefficients. Choice C incorrectly changes the constant in the first equation to -6, perhaps from misreading the sign—always match B's entries directly! To convert any AX = B to a system, (1) For each row i of A, multiply that row by X to get the left side of equation i (e.g., $a_{i1}$x + $a_{i2}$y + ...), (2) Set it equal to b_i from B, (3) Write out all equations in order. Example: [[1, 2], [3, -1]] [x, y] = [5, 7] gives x + 2y = 5 and 3x - y = 7. Straightforward! Always verify by plugging back: if you multiply your derived system's A by X, you should recover the original matrix equation. This double-check ensures accuracy—you've got this!

This question tests your ability to identify the coefficient matrix A in the matrix equation $AX = B$ for a system of linear equations, where A holds the coefficients organized by rows for equations and columns for variables. Matrix form $AX = B$ is a compact way to write entire systems: the coefficient matrix A contains all the coefficients from the left sides of equations (organized by rows for equations and columns for variables), the variable vector X lists the unknowns as a column, and the constant vector B lists the right-hand side values. When you multiply matrix A times vector X, you get the left sides of all equations, which equals vector B (the right sides). For example, for $-x + 4y + z = 6$, $2x - y + 3z = -1$, $5x + 0y - 2z = 7$, A is $\begin{bmatrix} -1 & 4 & 1 \\ 2 & -1 & 3 \\ 5 & 0 & -2 \end{bmatrix}$. Matrix multiplication recovers the original equations! To find A for this system: (1) Use equation 1 coefficients as row 1: -1 (x), 4 (y), 1 (z). (2) Equation 2 as row 2: 2, -1, 3. (3) Equation 3 as row 3: 5, 0, -2 (note 0 for missing y). Columns are for variables in order x, y, z. So A = $\begin{bmatrix} -1 & 4 & 1 \\ 2 & -1 & 3 \\ 5 & 0 & -2 \end{bmatrix}$. This captures all coefficients correctly! Choice A correctly constructs the coefficient matrix A with proper row-column organization and includes 0 for the missing y in the third equation. Choice B has the rows and columns transposed, which swaps the structure—rows must be equations, not variables. Remember: rows correspond to equations, columns to variables. Don't transpose! Matrix equation construction recipe: (1) Label equations and variables (x, y, z), (2) Build matrix A: row i = coefficients from equation i, column j = coefficient of variable j across equations, include 0 for missing terms, (3) Ensure A is m x n where m is number of equations, n is number of variables. Systematic! To verify A, mentally multiply a row by X: row 3 $[5, 0, -2]$ times $[x, y, z]$ should give $5x - 2z$, matching the equation (ignoring constant). If it matches all rows, A is correct—this prevents errors! You're building strong skills—great job!

This question tests your ability to represent a system of linear equations as a single matrix equation in the form AX = B, where A is the coefficient matrix, X is the variable vector, and B is the constant vector. Matrix form AX = B is a compact way to write entire systems: the coefficient matrix A contains all the coefficients from the left sides of equations (organized by rows for equations and columns for variables), the variable vector X lists the unknowns as a column, and the constant vector B lists the right-hand side values. When you multiply matrix A times vector X, you get the left sides of all equations, which equals vector B (the right sides). For example, 4x - y = 2 and -3x + 6y = 15 becomes [[4, -1], [-3, 6]] times [x, y] = [2, 15]. Matrix multiplication recovers the original equations! To convert this system to matrix form: (1) Write A with row 1: 4 (x), -1 (y); row 2: -3, 6. So A = [[4, -1], [-3, 6]]. (2) X = [x, y]. (3) B = [2, 15]. (4) Combine: AX = B. This matrix equation is equivalent to the original system! Choice A correctly constructs A with proper coefficients (including signs), X in order, and B matching the constants. Choice C has B's constants swapped ([15, 2] instead of [2, 15]). The constant vector must match equation order: first equation's 2 first, second's 15 second. Always double-check order! Matrix equation construction recipe: (1) Label equations and variables (x, y), (2) Build A: rows from equations, columns from variables, careful with signs, (3) Write X as column matching A, (4) Write B with constants in order, (5) Combine. Systematic! To verify, as the question suggests, multiply A by X: [[4, -1]] times [x, y] gives 4x - y, equals 2; second row -3x + 6y = 15. If it produces the left-hand sides correctly, it's good—this prevents errors! You're excelling—keep practicing!

This question tests your ability to represent a system of linear equations as a single matrix equation in the form AX = B, using specified variable names and order like s and ℓ. Matrix form AX = B is a compact way to write entire systems: the coefficient matrix A contains all the coefficients from the left sides of equations (organized by rows for equations and columns for variables), the variable vector X lists the unknowns as a column, and the constant vector B lists the right-hand side values. When you multiply matrix A times vector X, you get the left sides of all equations, which equals vector B (the right sides). For example, 2s + ℓ = 11 and -s + 3ℓ = 4 becomes [[2, 1], [-1, 3]] times [s, ℓ] = [11, 4]. Matrix multiplication recovers the original equations! To convert this system to matrix form: (1) Write A with row 1: 2 (s), 1 (ℓ); row 2: -1, 3. So A = [[2, 1], [-1, 3]]. (2) X = [s, ℓ] (order matches columns: s first, ℓ second). (3) B = [11, 4]. (4) Combine: AX = B. This matrix equation is equivalent to the original system! Choice A correctly uses the variable order s, ℓ in X, with A and B matching the coefficients and constants. Choice C has X as [ℓ, s] (reversed order). Remember: X's order must match A's columns—first column for s, second for ℓ. Don't swap variables! Matrix equation construction recipe: (1) Label equations and variables in given order (s, ℓ), (2) Build A: rows from equations, columns from variables in order, (3) Write X as column with variables in that order, (4) Write B with constants matching equations, (5) Combine. Systematic! To verify, multiply: first row [2, 1] times [s, ℓ] gives 2s + ℓ = 11; second [-1, 3] gives -s + 3ℓ = 4. Matches perfectly—this prevents errors! Great job applying this to real-world contexts like store sales!