Representing Linear Systems with Matrices - Algebra 2

Convert the system to $AX=B$ (use variable order $x, y, z$ ): $\begin{cases} 5x-2y+z=0\\ -x+3y+4z=10\\ 2x+y-3z=-6 \end{cases}$
Answer: $\begin{bmatrix}5&-2&1\\-1&3&4\\2&1&-3\end{bmatrix}\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}0\\10\\-6\end{bmatrix}$ 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. To convert the system 5x - 2y + z = 0, -x + 3y + 4z = 10, and 2x + y - 3z = -6 to matrix form: (1) Build coefficient matrix A row by row: row 1 is [5, -2, 1] from equation 1, row 2 is [-1, 3, 4] from equation 2 (note the -x gives coefficient -1), row 3 is [2, 1, -3] from equation 3. So A = [[5, -2, 1], [-1, 3, 4], [2, 1, -3]]. (2) Variable vector X = [x, y, z] as column. (3) Constant vector B = [0, 10, -6] as column. Choice A correctly assembles all components with proper signs and order. Choice C has an error in the second row of A: it shows [1, 3, 4] instead of [-1, 3, 4], missing the negative sign from -x in the second equation - remember that -x means the coefficient is -1, not 1! Critical detail for negative leading coefficients: When an equation begins with a negative variable like -x + 3y + 4z = 10, the coefficient of x is -1, not 1. Don't lose the negative sign when building matrix A! Similarly, -3z has coefficient -3, and -y would have coefficient -1. To verify your matrix equation, multiply A times X: row 1 [5, -2, 1] times [x, y, z] gives 5x - 2y + z = 0; row 2 [-1, 3, 4] times [x, y, z] gives -x + 3y + 4z = 10; row 3 [2, 1, -3] times [x, y, z] gives 2x + y - 3z = -6 - perfect match with our original system!
A linear system has the matrix representation: $\begin{pmatrix} 1 & -2 & 4 \\ 3 & 0 & -1 \\ -1 & 5 & 2 \end{pmatrix} \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 6 \\ -3 \\ 9 \end{pmatrix}$ What is the coefficient of $y$ in the second equation of this system?
Answer: $0$ When you encounter a matrix equation like this, you're looking at a compact way to represent a system of linear equations. The coefficient matrix (on the left) contains all the coefficients of your variables $x$ , $y$ , and $z$ , organized so that each row represents one equation. To find the coefficient of $y$ in the second equation, you need to look at the second row of the coefficient matrix and identify the entry in the $y$ -column. Since the variables are ordered as $x$ , $y$ , $z$ , the $y$ -coefficients appear in the second column of the matrix. The second row of the coefficient matrix is $(3, 0, -1)$ . The second entry in this row is $0$ , which means the coefficient of $y$ in the second equation is $0$ . You can verify this by writing out the second equation: $3x + 0y + (-1)z = -3$ , or simply $3x - z = -3$ . Looking at the wrong answers: Choice A ( $5$ ) is the coefficient of $y$ in the third equation, not the second. Choice B ( $-2$ ) is the coefficient of $y$ in the first equation. Choice C ( $3$ ) is the coefficient of $x$ in the second equation, showing confusion between which variable and which position in the row. Study tip: Remember that in a coefficient matrix, rows represent equations and columns represent variables. Always double-check which row (equation number) and which column (variable) the question is asking about to avoid mixing up positions.
A student claims the system $\begin{cases} 3x-y=6\\ 2x+5y=-4 \end{cases}$ is represented by $\begin{bmatrix}3&-1\\2&5\end{bmatrix}\begin{bmatrix}x\\y\end{bmatrix}=\begin{bmatrix}6\\-4\end{bmatrix}.$ Which statement best connects the matrix equation to the system?
Answer: The first row of $A$ gives coefficients of $x$ and $y$ in the first equation, and the second row gives coefficients in the second equation; multiplying $A$ by $X$ produces the left-hand sides of the system. This question tests your understanding of how matrix equations connect to systems of linear equations through the fundamental relationship of matrix multiplication. When we write AX = B for a system, the coefficient matrix A organizes all variable coefficients systematically: rows correspond to equations and columns correspond to variables, so multiplying A by X reconstructs the left-hand sides of all equations simultaneously. For the system 3x - y = 6 and 2x + 5y = -4, the matrix equation [[3, -1], [2, 5]] times [x, y] = [6, -4] represents it because: (1) First row [3, -1] times [x, y] gives 3x + (-1)y = 3x - y, matching equation 1's left side, (2) Second row [2, 5] times [x, y] gives 2x + 5y, matching equation 2's left side, (3) These results equal [6, -4], the right-hand sides. This is exactly what matrix multiplication does - it performs all equation computations at once! Choice A correctly explains this relationship: rows give equation coefficients, and multiplication produces left-hand sides. Choice B incorrectly claims columns correspond to equations and rows to variables - this is backwards and would make matrix multiplication meaningless for systems. The power of matrix representation: When you multiply A times X, you're essentially computing all left-hand sides of your system simultaneously - row 1 of A contains all information about equation 1, row 2 about equation 2, etc. This organization allows us to use matrix operations (like finding inverses) to solve entire systems at once rather than equation by equation. To understand why this works, trace through the multiplication: [[3, -1], [2, 5]] times [x, y] means compute [3x + (-1)y, 2x + 5y] = [3x - y, 2x + 5y], and setting this equal to [6, -4] gives us back our original system - the matrix equation is a perfect encoding of the linear system!
Write the system with missing variables as a matrix equation $AX=B$ (use variable order $x, y, z$ ): $\begin{cases} x+3y=5\\ 2x-z=1\\ y+2z=4 \end{cases}$
Answer: $\begin{bmatrix}1&3&0\\2&0&-1\\0&1&2\end{bmatrix}\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}5\\1\\4\end{bmatrix}$ This question tests your ability to represent a system of linear equations with missing variables as a single matrix equation in the form AX = B, including zeros for absent terms. 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, x + 3y = 5, 2x - z = 1, y + 2z = 4 becomes [[1, 3, 0], [2, 0, -1], [0, 1, 2]] times [x, y, z] = [5, 1, 4] (zeros for missing terms). Matrix multiplication recovers the original equations! To convert this system to matrix form: (1) Write coefficient matrix A with rows for equations and columns for x, y, z: row 1: 1, 3, 0 (no z); row 2: 2, 0, -1 (no y); row 3: 0, 1, 2 (no x). So A = [[1, 3, 0], [2, 0, -1], [0, 1, 2]]. (2) X = [x, y, z]. (3) B = [5, 1, 4]. (4) Combine as AX = B. This matrix equation is equivalent to the original system! Choice A correctly includes zeros for missing variables in A, with proper organization, and matches X and B. Choice B is missing the z column in A (it's 3x2, but should be 3x3 for three variables). Remember: A must have a column for every variable, even if some entries are zero. Include those zeros! Matrix equation construction recipe: (1) Label equations and all variables (x, y, z), (2) Build A: rows from equations, columns from variables, put 0 where a variable is missing, (3) Write X as column with all variables in order, (4) Write B with constants in equation order, (5) Combine. Systematic! To verify, multiply AX: row 2 [2, 0, -1] times [x, y, z] gives 2x - z = 1, matching. Do this for all rows to confirm—this prevents errors! Excellent progress—you've got this!
Write the system as a matrix equation $AX=B$ (variable order $x, y, z$ ). Remember to use a coefficient of $0$ for any missing variable. $\begin{cases} x+3y=5\\ 2x-z=1\\ y+2z=4 \end{cases}$
Answer: $\begin{bmatrix}1&3&0\\2&0&-1\\0&1&2\end{bmatrix}\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}5\\1\\4\end{bmatrix}$ This question tests your ability to represent a system of linear equations as a single matrix equation in the form AX = B, paying special attention to missing variables that require coefficient 0. Matrix form AX = B requires the coefficient matrix A to have consistent dimensions: if working with variables x, y, z, then every row must have three entries, using 0 for any missing variable. To convert the system x + 3y = 5, 2x - z = 1, and y + 2z = 4 to matrix form: (1) Identify missing variables: equation 1 lacks z (coefficient 0), equation 2 lacks y (coefficient 0), equation 3 lacks x (coefficient 0). (2) Write complete coefficient matrix A = [[1, 3, 0], [2, 0, -1], [0, 1, 2]] where each row has all three coefficients in x, y, z order. (3) Write X = [x, y, z] and B = [5, 1, 4]. (4) The matrix equation is [[1, 3, 0], [2, 0, -1], [0, 1, 2]] times [x, y, z] = [5, 1, 4]. Choice A correctly includes all necessary zeros for missing variables and maintains proper coefficient positions. Choice C has an error in row 2 - it shows [2, 0, 1] instead of [2, 0, -1], missing the negative sign on the z-coefficient in equation 2x - z = 1. Critical strategy for missing variables: Before building matrix A, rewrite each equation showing all variables explicitly: x + 3y + 0z = 5, 2x + 0y + (-1)z = 1, 0x + 1y + 2z = 4. This makes the coefficient matrix obvious! To verify, multiply each row by [x, y, z]: row 1 gives x + 3y + 0z = x + 3y, row 2 gives 2x + 0y - z = 2x - z, row 3 gives 0x + y + 2z = y + 2z, perfectly matching the original equations!
The matrix equation $\begin{bmatrix}2&-1\\3&4\end{bmatrix}\begin{bmatrix}x\\y\end{bmatrix}=\begin{bmatrix}5\\7\end{bmatrix}$ represents what system of linear equations?
Answer: $\begin{cases}2x-y=5\\3x+4y=7\end{cases}$ This question tests your ability to interpret a matrix equation AX = B and recover the original system of linear equations it represents. Matrix form AX = B is a compact way to write entire systems: when you multiply the coefficient matrix A times the variable vector X, each row of A produces one equation's left side, which equals the corresponding entry in constant vector B. To recover the system from [[2, -1], [3, 4]] times [x, y] = [5, 7]: (1) Multiply row 1 of A by X: [2, -1] times [x, y] = 2x + (-1)y = 2x - y. (2) This equals the first entry of B, which is 5, giving equation 1: 2x - y = 5. (3) Multiply row 2 of A by X: [3, 4] times [x, y] = 3x + 4y. (4) This equals the second entry of B, which is 7, giving equation 2: 3x + 4y = 7. (5) The complete system is 2x - y = 5 and 3x + 4y = 7. Choice A correctly identifies both equations by properly performing the matrix multiplication and matching with the constant vector. Choice D has the constants reversed - it shows 2x - y = 7 and 3x + 4y = 5, mixing up which constant goes with which equation. Remember that row i of matrix A generates equation i! Matrix multiplication recovery process: For each row i in matrix A, multiply that row by variable vector X to get the left side of equation i, then set it equal to entry i in vector B. This systematic approach prevents errors! To verify your answer, you can work backwards: take the system of equations and convert it back to matrix form - you should get the original matrix equation!
Which choice correctly identifies $A$ , $X$ , and $B$ for the system (use variable order $x, y$ )? $\begin{cases} 4x-y=10\\ -3x+6y=-12 \end{cases}$
Answer: $A=\begin{bmatrix}4&-1\\-3&6\end{bmatrix},\;X=\begin{bmatrix}x\\y\end{bmatrix},\;B=\begin{bmatrix}10\\-12\end{bmatrix}$ This question tests your ability to identify the three components A, X, and B of a matrix equation AX = B that represents a given system of linear equations. The coefficient matrix A contains all coefficients from the left sides of equations, the variable vector X lists the unknowns as a column, and the constant vector B lists the right-hand side values - all three must be correctly identified and properly formatted. To find A, X, and B for the system 4x - y = 10 and -3x + 6y = -12: (1) Coefficient matrix A uses equation coefficients as rows: row 1 is [4, -1] from equation 1 (4 for x, -1 for -y), row 2 is [-3, 6] from equation 2, giving A = [[4, -1], [-3, 6]]. (2) Variable vector X lists variables as a column in the specified order: X = [x, y] (written vertically). (3) Constant vector B lists equation right sides as a column: B = [10, -12] (10 from equation 1, -12 from equation 2). Choice A correctly identifies all three components with proper formatting and signs. Choice D incorrectly writes X as a row vector [x, y] instead of a column vector - in matrix equations AX = B, the variable vector X must be a column so that matrix multiplication AX is defined (you can't multiply a matrix by a row vector on the right)! Matrix equation component checklist: (1) A is an m×n matrix where m = number of equations and n = number of variables, (2) X is an n×1 column vector listing variables, (3) B is an m×1 column vector listing constants, (4) The product AX must have the same dimensions as B (both m×1). To verify your components work together, check that matrix multiplication is defined: A is 2×2, X is 2×1, so AX is 2×1, which matches B's dimensions of 2×1 ✓.
Represent the system as $AX=B$ (use variable order $x, y, z$ ): $\begin{cases} 2x+y-z=5\\ x-3y+2z=1\\ 4x+y+z=9 \end{cases}$
Answer: $\begin{bmatrix}2&1&-1\\1&-3&2\\4&1&1\end{bmatrix}\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}5\\1\\9\end{bmatrix}$ 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. To convert the system 2x + y - z = 5, x - 3y + 2z = 1, and 4x + y + z = 9 to matrix form: (1) Write coefficient matrix A by using equation 1 coefficients as row 1, equation 2 coefficients as row 2, and equation 3 coefficients as row 3, with columns for each variable in order: A = [[2, 1, -1], [1, -3, 2], [4, 1, 1]] (first column is x-coefficients: 2, 1, 4; second column is y-coefficients: 1, -3, 1; third column is z-coefficients: -1, 2, 1). (2) Write variable vector X = [x, y, z] as column (order matches columns of A). (3) Write constant vector B = [5, 1, 9] as column (matching equation order). (4) Combine: [[2, 1, -1], [1, -3, 2], [4, 1, 1]] times [x, y, z] = [5, 1, 9]. 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 C has the rows and columns of matrix A transposed (swapped), putting coefficients in wrong positions - the third column should contain z-coefficients (-1, 2, 1), not the coefficients of x from different equations. 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), (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. To verify your matrix equation represents the system correctly, perform the matrix multiplication AX mentally: the first row of A times X gives 2x + y - z, which equals 5 (first entry of B), confirming the first equation is correctly represented.
For the system (variable order $x, y, z$ ): $\begin{cases} -2x+4y+z=6\\ 3x-y+0z=-5\\ x+2y-3z=7 \end{cases}$ What is the coefficient matrix $A$ in $AX=B$ ?
Answer: $A=\begin{bmatrix}-2&4&1\\3&-1&0\\1&2&-3\end{bmatrix}$ This question tests your ability to identify the coefficient matrix A from a system of linear equations when writing it in matrix form AX = B. The coefficient matrix A contains all the coefficients from the left sides of the equations, organized with rows corresponding to equations and columns corresponding to variables in the specified order (x, y, z). To extract the coefficient matrix from -2x + 4y + z = 6, 3x - y + 0z = -5, and x + 2y - 3z = 7: (1) Row 1 comes from equation 1: coefficients are -2 (for x), 4 (for y), 1 (for z). (2) Row 2 comes from equation 2: coefficients are 3 (for x), -1 (for y), 0 (for z) - note that 0z is explicitly written to show z's coefficient is 0. (3) Row 3 comes from equation 3: coefficients are 1 (for x), 2 (for y), -3 (for z). Therefore A = [[-2, 4, 1], [3, -1, 0], [1, 2, -3]]. Choice A correctly identifies all coefficients including the 0 in the second equation's z-term. Choice C incorrectly transposes the matrix structure, mixing up which coefficients belong to which variables - the first column should contain x-coefficients (-2, 3, 1), not a mix of different variable coefficients. When building coefficient matrices, work systematically: go equation by equation (each becomes a row), and within each equation, extract coefficients in the variable order specified (x, y, z). Include 0 for any missing variable - this maintains the proper matrix dimensions and ensures correct multiplication!
Write the system of linear equations as a single matrix equation $AX=B$ (use variable order $x, y$ ): $\begin{cases} 3x+2y=8\\ x-5y=-3 \end{cases}$
Answer: $\begin{bmatrix}3&2\\1&-5\end{bmatrix}\begin{bmatrix}x\\y\end{bmatrix}=\begin{bmatrix}8\\-3\end{bmatrix}$ 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!
A student claims the system (variable order $x, y$ ) $\begin{cases} 4x-y=2\\ -3x+6y=9 \end{cases}$ can be written as $\begin{bmatrix}4&-1\\-3&6\end{bmatrix}\begin{bmatrix}x\\y\end{bmatrix}=\begin{bmatrix}2\\9\end{bmatrix}.$ Which statement best verifies the connection between the system and the matrix equation?
Answer: Multiplying the matrix by $\begin{bmatrix}x\\y\end{bmatrix}$ gives $\begin{bmatrix}4x-y\\-3x+6y\end{bmatrix}$ , which matches the left sides of the two equations, and $\begin{bmatrix}2\\9\end{bmatrix}$ matches the constants. This question tests your ability to verify that a matrix equation correctly represents a given system of linear equations by checking the matrix multiplication. The key to verification is performing the matrix multiplication AX and confirming it produces the left sides of the original equations, which then equal the constant vector B. To verify that [[4, -1], [-3, 6]] times [x, y] = [2, 9] represents the system 4x - y = 2 and -3x + 6y = 9: (1) Multiply row 1 by variable vector: [4, -1] times [x, y] = 4x + (-1)y = 4x - y. (2) This should equal the first constant, 2, giving 4x - y = 2 ✓. (3) Multiply row 2 by variable vector: [-3, 6] times [x, y] = -3x + 6y. (4) This should equal the second constant, 9, giving -3x + 6y = 9 ✓. (5) Both equations match the original system perfectly! Choice A correctly identifies this verification process - multiplying the matrix by the variable vector recovers the left sides of equations. Choice B discusses the determinant, which is irrelevant for verifying the matrix equation represents the system - a nonzero determinant only tells us the system has a unique solution, not whether the representation is correct. Verification strategy: Always multiply AX to check your work! Each row of A times X should give exactly the left side of the corresponding equation. If even one coefficient is wrong, the multiplication will reveal the error immediately. This multiplication check is the gold standard for verifying matrix representations of linear systems - it directly shows that the matrix equation and the original system are mathematically equivalent!
Consider the matrix equation: $\begin{pmatrix} 2 & -1 & 3 \\ 0 & 4 & -2 \\ 1 & 1 & 0 \end{pmatrix} \begin{pmatrix} a \\ b \\ c \end{pmatrix} = \begin{pmatrix} 8 \\ 6 \\ 4 \end{pmatrix}$ If this represents a system of linear equations, what is the constant term in the equation that has no $c$ variable?
Answer: $4$ When you see a matrix equation like this, you're looking at a compact way to represent a system of linear equations. The matrix multiplication gives you three separate equations, and you need to identify which one has no $c$ variable. To find the equations, multiply each row of the matrix by the variable vector. The first row gives you: $2a - 1b + 3c = 8$ . The second row gives you: $0a + 4b + (-2)c = 6$ , which simplifies to $4b - 2c = 6$ . The third row gives you: $1a + 1b + 0c = 4$ , which simplifies to $a + b = 4$ . The equation with no $c$ variable is $a + b = 4$ , so the constant term is $4$ . Looking at the wrong answers: Choice A ( $8$ ) is the constant from the first equation $2a - b + 3c = 8$ , but this equation contains the $c$ variable. Choice B ( $6$ ) is the constant from the second equation $4b - 2c = 6$ , but this also contains the $c$ variable. Choice D ( $0$ ) might tempt you because the coefficient of $c$ in the third equation is $0$ , but the question asks for the constant term, not the coefficient. The key strategy here is to systematically write out each equation from the matrix multiplication, then identify which equation is missing the specified variable. Don't get distracted by coefficients—focus on which equation actually lacks the variable in question, then read its constant term.
Write matrices $A$ , $X$ , and $B$ so that the system can be written as $AX=B$ (use variable order $x, y, z$ ): $\begin{cases} x+3y=5\\ 2x-z=1\\ y+2z=4 \end{cases}$ (Note: include a $0$ coefficient for any missing variable in an equation.)
Answer: $\begin{bmatrix}1&3&0\\2&0&-1\\0&1&2\end{bmatrix}\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}5\\1\\4\end{bmatrix}$ 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 - note the special requirement to include 0 coefficients for missing variables. Matrix form AX = B requires complete coefficient information: if an equation is missing a variable (like equation 1 missing z, or equation 3 missing x), you must include a 0 coefficient in the appropriate position of matrix A. To convert the system x + 3y = 5, 2x - z = 1, and y + 2z = 4 to matrix form: (1) Write coefficient matrix A including zeros for missing variables: equation 1 has coefficients [1, 3, 0] (0 for missing z), equation 2 has [2, 0, -1] (0 for missing y), equation 3 has [0, 1, 2] (0 for missing x), giving A = [[1, 3, 0], [2, 0, -1], [0, 1, 2]]. (2) Write variable vector X = [x, y, z] as column. (3) Write constant vector B = [5, 1, 4] as column (matching equation order). (4) Combine: [[1, 3, 0], [2, 0, -1], [0, 1, 2]] times [x, y, z] = [5, 1, 4]. Choice A correctly constructs the coefficient matrix A with zeros in proper positions for missing variables, variable vector X in correct order, and constant vector B matching the equations. Choice B omits the z-column entirely, creating a 3×2 matrix that cannot multiply with a 3×1 variable vector - matrix dimensions must match: for three variables, you need three columns even if some entries are zero! Complete matrix construction recipe for systems with missing variables: (1) Identify all variables in the system (here: x, y, z), (2) For each equation, write coefficients for ALL variables in order, using 0 for any variable not appearing in that equation, (3) Build matrix A row by row, (4) Include all variables in vector X, (5) Match constants to equation order in vector B. To verify, multiply row 1 of A times X: 1·x + 3·y + 0·z = x + 3y = 5 ✓, confirming equation 1 is correctly represented with its missing z term handled by the 0 coefficient.
A student claims the system (variable order $x, y$ ) $\begin{cases} 2x-3y=4 \\ -x+5y=1 \end{cases}$ can be written as $\begin{bmatrix} 2 & -3 \\ -1 & 5 \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 4 \\ 1 \end{bmatrix}.$ Which choice correctly explains why this matrix equation matches the system?
Answer: Each row of $A$ lists the coefficients of $x$ and $y$ in one equation, and multiplying $A \begin{bmatrix} x \\ y \end{bmatrix}$ produces the left-hand sides $2x-3y$ and $-x+5y$ , which equal the entries of $B$ . This question tests your understanding of why a given matrix equation AX = B matches a system of linear equations, focusing on the role of 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! The student's matrix $\begin{bmatrix} 2 & -3 \\ -1 & 5 \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 4 \\ 1 \end{bmatrix}$ matches because row 1 gives $2x - 3y = 4$ , row 2 gives $-x + 5y = 1$ , equaling B's entries. Choice A correctly explains that each row of A provides the coefficients for one equation, and multiplication yields the left sides matching the system. Choice C incorrectly states that columns of A correspond to equations and rows to variables, which reverses the standard organization—remember, rows are equations, columns are variables! To understand why AX = B matches a system, (1) Compute the matrix product AX: each entry in the result is the left side of an equation, (2) Set it equal to B's entries for the right sides. Example: $\begin{bmatrix} 1 & 2 \\ 3 & -1 \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 5 \\ 7 \end{bmatrix}$ gives $[x + 2y, 3x - y] = [5, 7]$ , so $x + 2y = 5$ and $3x - y = 7$ . Clear! Always verify by multiplying: if your explanation holds, the multiplication should reproduce the system exactly. This builds confidence—you're doing great!
Convert the system to matrix form $AX=B$ (use variable order $x, y$ ): $\begin{cases} 4x-y=2\\ -3x+6y=15 \end{cases}$
Answer: $\begin{bmatrix}4&-1\\-3&6\end{bmatrix}\begin{bmatrix}x\\y\end{bmatrix}=\begin{bmatrix}2\\15\end{bmatrix}$ 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 4x - y = 2 and -3x + 6y = 15 to matrix form: A = [[4, -1], [-3, 6]], X = [x, y], B = [2, 15], as in choice A. Choice A correctly constructs A with proper coefficients, X in order, and B matching constants. Choice C has the constants in vector B in wrong order—the constant vector must match equation order: first equation's right side in first position, second in second; check each equation carefully when building B! Matrix equation construction recipe: (1) Label equations and variables, (2) Build A with rows from equations and columns from variables, (3) Write X as column, (4) Write B with constants in order, (5) Combine. Verify by multiplying AX. You're mastering this—stay consistent!
Write the system of linear equations as a single matrix equation $AX=B$ (use variable order $x, y$ ): $\begin{cases} 3x+2y=8\\ x-5y=-3 \end{cases}$
Answer: $\begin{bmatrix}3&2\\1&-5\end{bmatrix}\begin{bmatrix}x\\y\end{bmatrix}=\begin{bmatrix}8\\-3\end{bmatrix}$ 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. 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]. 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 - the second row should be [1, -5] not [2, -5], and the first column should contain x-coefficients [3, 1] not [3, 2]. 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), (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. To verify your matrix equation represents the system correctly, perform the matrix multiplication AX mentally: the first row [3, 2] times [x, y] gives 3x + 2y = 8, and the second row [1, -5] times [x, y] gives x - 5y = -3, matching our original system perfectly!
The matrix equation $\begin{bmatrix}2&-1\\3&4\end{bmatrix}\begin{bmatrix}x\\y\end{bmatrix}=\begin{bmatrix}5\\7\end{bmatrix}$ represents which system of linear equations?
Answer: $\begin{cases}2x-y=5\\3x+4y=7\end{cases}$ 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 to recover the left-hand sides. 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, [[2, -1], [3, 4]] times [x, y] = [5, 7] becomes 2x - y = 5 and 3x + 4y = 7. Matrix multiplication recovers the original equations! To find the system from this matrix equation: (1) Multiply first row of A [2, -1] by X [x, y] to get left side of equation 1: 2x - y. Set equal to B's first entry: 2x - y = 5. (2) Second row [3, 4] times X: 3x + 4y = 7. So the system is 2x - y = 5 and 3x + 4y = 7. This matches the matrix form perfectly! Choice A correctly represents the system obtained from performing the matrix multiplication AX and setting it equal to B component-wise. Choice D has the constants swapped (5 and 7 reversed). Remember: the order in B corresponds to the row order in A—first row gives first equation and first constant. Don't mix them up! To convert matrix equations to systems: (1) For each row i of A, compute the dot product with X to get the left side of equation i, (2) Set it equal to B's i-th entry, (3) Write out all equations. Systematic! Verify by plugging back: if you build A from your system, it should match the given A. For 2x - y = 5 and 3x + 4y = 7, A is indeed [[2, -1], [3, 4]], B [5, 7]. This round-trip check prevents errors! You're doing wonderfully—keep it up!
A student claims the system below can be written as $\begin{bmatrix}1&-2&3\\0&4&-1\end{bmatrix}\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}7\\5\end{bmatrix}.$ Which system does this matrix equation represent (use variable order $x, y, z$ )?
Answer: $\begin{cases}x-2y+3z=7\\4y-z=5\end{cases}$ This question tests your ability to interpret a matrix equation AX = B and recover the original system of linear equations it represents. When you multiply a coefficient matrix A by a variable vector X, each row of A produces one equation: row i of A times X gives the left side of equation i, which equals the corresponding entry in vector B. To find the system represented by [[1, -2, 3], [0, 4, -1]] times [x, y, z] = [7, 5]: (1) Multiply row 1 of A by X: [1, -2, 3] times [x, y, z] = 1·x + (-2)·y + 3·z = x - 2y + 3z, and this equals the first entry of B, which is 7, giving equation 1: x - 2y + 3z = 7. (2) Multiply row 2 of A by X: [0, 4, -1] times [x, y, z] = 0·x + 4·y + (-1)·z = 4y - z, and this equals the second entry of B, which is 5, giving equation 2: 4y - z = 5. (3) The complete system is {x - 2y + 3z = 7, 4y - z = 5}. Choice A correctly identifies both equations by properly performing the matrix multiplication, including the 0 coefficient for x in the second equation. Choice D incorrectly interprets the second equation as having 4x instead of 4y - the 4 appears in position (2,2) of the matrix, which corresponds to the y-coefficient in equation 2, not the x-coefficient! Matrix equation interpretation strategy: (1) Remember that row i gives equation i, and column j corresponds to variable j, (2) Multiply each row by the variable vector systematically, (3) Include terms with 0 coefficients (like 0·x) to avoid confusion, (4) Set each result equal to the corresponding B entry. To verify your interpretation, reconstruct the matrix: equation 1's coefficients [1, -2, 3] should be row 1, equation 2's coefficients [0, 4, -1] should be row 2, matching the original matrix exactly.
The matrix equation $\begin{bmatrix}1&0&-2\\0&3&1\\4&-1&0\end{bmatrix}\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}6\\-5\\8\end{bmatrix}$ represents which system of equations?
Answer: $\begin{cases}x-2z=6\\3y+z=-5\\4x-y=8\end{cases}$ 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!
Convert the system to matrix form $AX=B$ using variable order $x, y$ : $\begin{cases} -2x+7y=9\\ 4x+y=-6 \end{cases}$
Answer: $\begin{bmatrix}-2&7\\4&1\end{bmatrix}\begin{bmatrix}x\\y\end{bmatrix}=\begin{bmatrix}9\\-6\end{bmatrix}$ 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. To convert the system -2x + 7y = 9 and 4x + y = -6 to matrix form: (1) Write coefficient matrix A using equation 1 coefficients as row 1 and equation 2 coefficients as row 2: A = [[-2, 7], [4, 1]] (first column is x-coefficients: -2, 4; second column is y-coefficients: 7, 1). (2) Write variable vector X = [x, y] as column. (3) Write constant vector B = [9, -6] as column (equation 1's constant 9, equation 2's constant -6). Choice A correctly constructs all components with proper signs and ordering. Choice C incorrectly writes the variable vector X as a row [x, y] instead of a column - vectors in matrix equations must be columns for multiplication to work properly! Matrix construction is systematic: rows from equations, columns from variables, and always write vectors as columns. To verify: [[-2, 7], [4, 1]] times [x, y] gives [-2x + 7y, 4x + y] = [9, -6], confirming the original system!

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