Algebra 2 Quiz: Representing Linear Systems With Matrices
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Representing Linear Systems With MatricesQuestion 1 of 20

Convert the system to matrix form AX=BAX=B (use variable order x,yx, y):

{4x−y=2−3x+6y=15\begin{cases} 4x-y=2\\ -3x+6y=15 \end{cases}
[4−1−36][xy]=[215]\begin{bmatrix}4&-1\\-3&6\end{bmatrix}\begin{bmatrix}x\\y\end{bmatrix}=\begin{bmatrix}2\\15\end{bmatrix}
[41−36][xy]=[215]\begin{bmatrix}4&1\\-3&6\end{bmatrix}\begin{bmatrix}x\\y\end{bmatrix}=\begin{bmatrix}2\\15\end{bmatrix}
[4−1−36][xy]=[152]\begin{bmatrix}4&-1\\-3&6\end{bmatrix}\begin{bmatrix}x\\y\end{bmatrix}=\begin{bmatrix}15\\2\end{bmatrix}
[4−3−16][xy]=[215]\begin{bmatrix}4&-3\\-1&6\end{bmatrix}\begin{bmatrix}x\\y\end{bmatrix}=\begin{bmatrix}2\\15\end{bmatrix}
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Algebra 2 Quiz

Algebra 2 Quiz: Representing Linear Systems With Matrices

Practice Representing Linear Systems With Matrices in Algebra 2 with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Representing Linear Systems With Matrices, giving you a quick way to practice the rules, question types, and explanations that matter most for Algebra 2.

How to use this quiz

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

All questions

Question 1

Convert the system to matrix form AX=BAX=B (use variable order x,yx, y):

{4x−y=2−3x+6y=15\begin{cases} 4x-y=2\\ -3x+6y=15 \end{cases}
  1. [4−1−36][xy]=[215]\begin{bmatrix}4&-1\\-3&6\end{bmatrix}\begin{bmatrix}x\\y\end{bmatrix}=\begin{bmatrix}2\\15\end{bmatrix} (correct answer)
  2. [41−36][xy]=[215]\begin{bmatrix}4&1\\-3&6\end{bmatrix}\begin{bmatrix}x\\y\end{bmatrix}=\begin{bmatrix}2\\15\end{bmatrix}
  3. [4−1−36][xy]=[152]\begin{bmatrix}4&-1\\-3&6\end{bmatrix}\begin{bmatrix}x\\y\end{bmatrix}=\begin{bmatrix}15\\2\end{bmatrix}
  4. [4−3−16][xy]=[215]\begin{bmatrix}4&-3\\-1&6\end{bmatrix}\begin{bmatrix}x\\y\end{bmatrix}=\begin{bmatrix}2\\15\end{bmatrix}
Explanation: 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!

Question 2

The matrix equation [2−134][xy]=[57]\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?

  1. {2y−x=53y+4x=7\begin{cases}2y-x=5\\3y+4x=7\end{cases}
  2. {2x−y=73x+4y=5\begin{cases}2x-y=7\\3x+4y=5\end{cases}
  3. {2x−y=53x+4y=7\begin{cases}2x-y=5\\3x+4y=7\end{cases} (correct answer)
  4. {2x+y=53x−4y=7\begin{cases}2x+y=5\\3x-4y=7\end{cases}
Explanation: 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! For the given matrix equation [[2, -1], [3, 4]] times [x, y] = [5, 7], multiply to recover: first equation 2x - y = 5, second 3x + 4y = 7, which matches choice A. Choice A correctly identifies the system by performing the matrix multiplication to extract the equations. Choice C has the constants in vector B in wrong order or swapped— the constant vector must match equation order: first row's product equals first constant, etc.; check each carefully! To go from matrix equation to system: (1) Multiply each row of A by X to get left sides, (2) Set equal to corresponding entry in B, (3) Write as equations. Verify by plugging back: if AX equals B, it matches. You're doing wonderfully— this reverse process strengthens understanding!

Question 3

Which matrix equation AX=BAX=B correctly matches the system (use variable order x,yx, y)?

{4x−y=2−3x+6y=15\begin{cases} 4x-y=2\\ -3x+6y=15 \end{cases}

(You can verify by noting that multiplying AA by XX produces the left-hand sides of the two equations.)

  1. [4−1−36][xy]=[215]\begin{bmatrix}4&-1\\-3&6\end{bmatrix}\begin{bmatrix}x\\y\end{bmatrix}=\begin{bmatrix}2\\15\end{bmatrix} (correct answer)
  2. [4−3−16][xy]=[215]\begin{bmatrix}4&-3\\-1&6\end{bmatrix}\begin{bmatrix}x\\y\end{bmatrix}=\begin{bmatrix}2\\15\end{bmatrix}
  3. [4−1−36][xy]=[152]\begin{bmatrix}4&-1\\-3&6\end{bmatrix}\begin{bmatrix}x\\y\end{bmatrix}=\begin{bmatrix}15\\2\end{bmatrix}
  4. [41−36][xy]=[215]\begin{bmatrix}4&1\\-3&6\end{bmatrix}\begin{bmatrix}x\\y\end{bmatrix}=\begin{bmatrix}2\\15\end{bmatrix}
Explanation: 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!

Question 4

The matrix equation

[2−134][xy]=[57]\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?

  1. {2x−y=53x+4y=7\begin{cases}2x-y=5\\3x+4y=7\end{cases} (correct answer)
  2. {2x+y=53x−4y=7\begin{cases}2x+y=5\\3x-4y=7\end{cases}
  3. {2x−3y=5−x+4y=7\begin{cases}2x-3y=5\\-x+4y=7\end{cases}
  4. {2y−x=53y+4x=7\begin{cases}2y-x=5\\3y+4x=7\end{cases}
Explanation: 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=72x + 3y = 7 and x−y=1x - y = 1 becomes [231−1][xy]=[71]\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 [2−134][xy]=[57]\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][2, -1] by [x,y][x, y] to get 2x−y=52x - y = 5 (first equation), second row [3,4][3, 4] by [x,y][x, y] to get 3x+4y=73x + 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., ai1x+ai2y+…a_{i1}x + a_{i2}y + \dots), (2) Set it equal to bib_i from B, (3) Write out all equations in order. Example: [123−1][xy]=[57]\begin{bmatrix} 1 & 2 \\ 3 & -1 \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 5 \\ 7 \end{bmatrix} gives x+2y=5x + 2y = 5 and 3x−y=73x - 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!

Question 5

Which matrix equation AX=BAX=B correctly represents the system (use variable order x,y,zx, y, z)?

{6x−2y=1−y+4z=−83x+z=5\begin{cases} 6x-2y=1\\ -y+4z=-8\\ 3x+z=5 \end{cases}

(Remember: missing variables must have coefficient 00 in AA.)

  1. [6−200−14301][xyz]=[1−85]\begin{bmatrix}6&-2&0\\0&-1&4\\3&0&1\end{bmatrix}\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}1\\-8\\5\end{bmatrix} (correct answer)
  2. [6−200−14310][xyz]=[1−85]\begin{bmatrix}6&-2&0\\0&-1&4\\3&1&0\end{bmatrix}\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}1\\-8\\5\end{bmatrix}
  3. [6−2−1431][xy]=[1−85]\begin{bmatrix}6&-2\\-1&4\\3&1\end{bmatrix}\begin{bmatrix}x\\y\end{bmatrix}=\begin{bmatrix}1\\-8\\5\end{bmatrix}
  4. [6−200−14301][xyz]=[15−8]\begin{bmatrix}6&-2&0\\0&-1&4\\3&0&1\end{bmatrix}\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}1\\5\\-8\end{bmatrix}
Explanation: 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 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, 6x - 2y = 1, -y + 4z = -8, 3x + z = 5 becomes [[6, -2, 0], [0, -1, 4], [3, 0, 1]] times [x, y, z] = [1, -8, 5] (zeros for missing variables). Matrix multiplication recovers the original equations! To convert this system to matrix form: (1) Write A with rows for equations, columns for x, y, z: row 1: 6, -2, 0 (no z); row 2: 0, -1, 4 (no x); row 3: 3, 0, 1 (no y). So A = [[6, -2, 0], [0, -1, 4], [3, 0, 1]]. (2) X = [x, y, z]. (3) B = [1, -8, 5]. (4) Combine: AX = B. This matrix equation is equivalent to the original system! Choice A correctly includes zeros for missing variables in A, with proper signs and order, and matches B. Choice D has B's constants reordered ([1, 5, -8] instead of [1, -8, 5]). Remember: B must match the exact equation order—second equation's -8 second. Don't reorder! Matrix equation construction recipe: (1) Label equations and variables (x, y, z), (2) Build A: rows from equations, columns from variables, insert 0 for missing ones as the question reminds, (3) Write X in order, (4) Write B with constants in equation order, (5) Combine. Systematic! To verify, multiply: row 2 [0, -1, 4] times [x, y, z] gives -y + 4z = -8, matching. Check all rows—this prevents errors! You're doing fantastic with these nuances!

Question 6

A student claims the system (variable order x,yx, y)

{6x−y=2−3x+5y=11\begin{cases} 6x-y=2\\ -3x+5y=11 \end{cases}

can be written as

\begin{bmatrix}6&-1\\-3&5\end{bmatrix}\begin{bmatrix}x\\y\end{bmatrix}=\begin{bmatrix}2\\11\end{bmatrix}.$$ Which statement best verifies the claim?
  1. Multiplying gives [6x−y−3x+5y]\begin{bmatrix}6x-y\\-3x+5y\end{bmatrix}, which matches the left sides of the two equations, and the right side is [211]\begin{bmatrix}2\\11\end{bmatrix}. (correct answer)
  2. Multiplying gives [6x−3y−x+5y]\begin{bmatrix}6x-3y\\-x+5y\end{bmatrix}, which matches the system.
  3. The matrix must be [6−3−15]\begin{bmatrix}6&-3\\-1&5\end{bmatrix} because coefficients go in columns, not rows.
  4. The vector BB should be [112]\begin{bmatrix}11\\2\end{bmatrix} because constants always go in decreasing order.
Explanation: This question tests your ability to verify a matrix equation representation by performing matrix multiplication and checking if it produces the original system. To verify [[6, -1], [-3, 5]] times [x, y] = [2, 11] represents the given system, multiply: (1) Row 1 times X: (6)(x) + (-1)(y) = 6x - y. (2) Row 2 times X: (-3)(x) + (5)(y) = -3x + 5y. This gives [6x - y, -3x + 5y] = [2, 11], which means 6x - y = 2 and -3x + 5y = 11, exactly matching the original system. Choice A correctly verifies by showing the multiplication produces the left sides of both equations and identifying the matching right-hand side vector. Choice C incorrectly claims coefficients go in columns not rows - this is backwards! In coefficient matrices, rows correspond to equations and columns to variables. Verification is crucial: always multiply AX mentally or on paper to ensure it produces your original system - this catches transposition errors, sign mistakes, or incorrect ordering. Matrix multiplication literally reconstructs the equations, making it a perfect self-check!

Question 7

A student claims the system (variable order x,yx, y)

{4x−y=2−3x+6y=9\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?​
  1. Multiplying the matrix by [xy]\begin{bmatrix}x\\y\end{bmatrix} gives [4x−y−3x+6y]\begin{bmatrix}4x-y\\-3x+6y\end{bmatrix}, which matches the left sides of the two equations, and [29]\begin{bmatrix}2\\9\end{bmatrix} matches the constants. (correct answer)
  2. The determinant of [4−1−36]\begin{bmatrix}4&-1\\-3&6\end{bmatrix} is nonzero, so it represents the system.
  3. The matrix equation is correct because the constants 22 and 99 appear somewhere in the matrix.
  4. Switching the order of variables to y,xy, x would not change the matrix equation, so it must be correct.
Explanation: 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!

Question 8

Write the system as AX=BAX=B (use variable order x,y,zx, y, z):

{−x+2y+z=34x−y=02x+5y−3z=−7\begin{cases} -x + 2y + z = 3 \\ 4x - y = 0 \\ 2x + 5y - 3z = -7 \end{cases}
  1. [−1214−1025−3][xyz]=[3−70]\begin{bmatrix}-1&2&1\\4&-1&0\\2&5&-3\end{bmatrix}\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}3\\-7\\0\end{bmatrix}
  2. [−1214−1125−3][xyz]=[30−7]\begin{bmatrix}-1&2&1\\4&-1&1\\2&5&-3\end{bmatrix}\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}3\\0\\-7\end{bmatrix}
  3. [−1214−1025−3][xyz]=[30−7]\begin{bmatrix}-1&2&1\\4&-1&0\\2&5&-3\end{bmatrix}\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}3\\0\\-7\end{bmatrix} (correct answer)
  4. [−1422−1510−3][xyz]=[30−7]\begin{bmatrix}-1&4&2\\2&-1&5\\1&0&-3\end{bmatrix}\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}3\\0\\-7\end{bmatrix}
Explanation: This question tests your ability to represent a system of linear equations as a single matrix equation in the form AX=BAX = B, where AA is the coefficient matrix, XX is the variable vector, and BB is the constant vector, including zeros for missing variables. Matrix form AX=BAX = B is a compact way to write entire systems: the coefficient matrix AA contains all the coefficients from the left sides of equations (organized by rows for equations and columns for variables), the variable vector XX lists the unknowns as a column, and the constant vector BB lists the right-hand side values. When you multiply matrix AA times vector XX, you get the left sides of all equations, which equals vector BB (the right sides). For example, the system 2x+3y=72x + 3y = 7 and x−y=1x - y = 1 becomes [231−1][xy]=[71]\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-x + 2y + z = 3, 4x−y=04x - y = 0, 2x+5y−3z=−72x + 5y - 3z = -7 to matrix form (order x,y,zx, y, z): A=[−1214−1025−3]A = \begin{bmatrix} -1 & 2 & 1 \\ 4 & -1 & 0 \\ 2 & 5 & -3 \end{bmatrix}, X=[xyz]X = \begin{bmatrix} x \\ y \\ z \end{bmatrix}, B=[30−7]B = \begin{bmatrix} 3 \\ 0 \\ -7 \end{bmatrix}—note the 0 for missing zz in the second equation. Choice A correctly includes the zero for zz in row 2, with all coefficients and constants in proper order. Choice D transposes parts of AA, 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,zx, y, z in consistent order), (2) Build matrix AA: 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 XX as column vector with variables in same order as AA's columns, (4) Write BB as column vector with constants in same order as equations, (5) Combine as AX=BAX = B. Example: x+2y=5x + 2y = 5 and 3x−y=73x - y = 7 gives A=[123−1]A = \begin{bmatrix} 1 & 2 \\ 3 & -1 \end{bmatrix}, X=[xy]X = \begin{bmatrix} x \\ y \end{bmatrix}, B=[57]B = \begin{bmatrix} 5 \\ 7 \end{bmatrix}. Systematic! To verify your matrix equation represents the system correctly, perform the matrix multiplication AXAX mentally or on paper: the first row of AA times XX should give the left side of equation 1, the second row times XX should give equation 2's left side, etc. If [231−1][xy]\begin{bmatrix} 2 & 3 \\ 1 & -1 \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} gives [2x+3yx−y]\begin{bmatrix} 2x + 3y \\ x - y \end{bmatrix}, and B=[71]B = \begin{bmatrix} 7 \\ 1 \end{bmatrix}, then matrix equation represents 2x+3y=72x + 3y = 7 and x−y=1x - y = 1. Matrix multiplication recovers the system—this verification prevents errors!

Question 9

A student claims the system

{2x+3y=7−4x+y=−1\begin{cases} 2x+3y=7\\ -4x+y=-1 \end{cases}

can be written as

[23−41][xy]=[7−1].\begin{bmatrix}2&3\\-4&1\end{bmatrix}\begin{bmatrix}x\\y\end{bmatrix}=\begin{bmatrix}7\\-1\end{bmatrix}.

Which choice correctly states the connection between the system and the matrix equation?

  1. It is incorrect because AX=BAX=B only works for three-variable systems, not two-variable systems.
  2. It is correct because multiplying the matrix by [xy]\begin{bmatrix}x\\y\end{bmatrix} produces the left-hand sides 2x+3y2x+3y and −4x+y-4x+y, matching the two equations. (correct answer)
  3. It is incorrect because the coefficient matrix should be [2−431]\begin{bmatrix}2&-4\\3&1\end{bmatrix} (coefficients must be written in columns by equation).
  4. It is incorrect because the constants must be placed in the variable vector, not in BB.
Explanation: 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! For the student's claim, multiplying [[2, 3], [-4, 1]] times [x, y] gives 2x + 3y and -4x + y, equaling [7, -1], which matches the system exactly, so it's correct as stated in choice A. Choice A correctly verifies by performing the multiplication and confirming the match. Choice C suggests transposing coefficients, but the given A is already correct—rows are for equations, not columns by equation; stick to standard organization! To check a matrix equation: (1) Multiply each row of A by X, (2) Set equal to B's entries, (3) Compare to system. This validation is key. Wonderful job— you're verifying like a pro!

Question 10

Write matrices AA, XX, and BB so that the system can be written as AX=BAX=B (use variable order x,y,zx, y, z):

{x+3y=52x−z=1y+2z=4\begin{cases} x+3y=5\\ 2x-z=1\\ y+2z=4 \end{cases}

(Note: include a 00 coefficient for any missing variable in an equation.)

  1. [13020−1012][xyz]=[514]\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} (correct answer)
  2. [132−112][xyz]=[514]\begin{bmatrix}1&3\\2&-1\\1&2\end{bmatrix}\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}5\\1\\4\end{bmatrix}
  3. [1032−10021][xyz]=[514]\begin{bmatrix}1&0&3\\2&-1&0\\0&2&1\end{bmatrix}\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}5\\1\\4\end{bmatrix}
  4. [13020−1012][xyz]=[541]\begin{bmatrix}1&3&0\\2&0&-1\\0&1&2\end{bmatrix}\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}5\\4\\1\end{bmatrix}
Explanation: 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.

Question 11

Write the system as a matrix equation AX=BAX=B (use variable order x,y,zx, y, z). Be sure to include a 00 coefficient for any missing variable:

{x+3y=52x−z=1y+2z=4\begin{cases} x+3y=5\\ 2x-z=1\\ y+2z=4 \end{cases}
  1. [13020−1012][xyz]=[514]\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} (correct answer)
  2. [130201012][xyz]=[514]\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}
  3. [1032−10021][xyz]=[514]\begin{bmatrix}1&0&3\\2&-1&0\\0&2&1\end{bmatrix}\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}5\\1\\4\end{bmatrix}
  4. [132−112][xyz]=[514]\begin{bmatrix}1&3\\2&-1\\1&2\end{bmatrix}\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}5\\1\\4\end{bmatrix}
Explanation: This question tests your ability to represent a system of linear equations as a single matrix equation in the form AX=BAX = 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=BAX = 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=72x + 3y = 7 and x−y=1x - y = 1 becomes [231−1]\begin{bmatrix} 2 & 3 \\ 1 & -1 \end{bmatrix} times [xy]=[71]\begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 7 \\ 1 \end{bmatrix}. Matrix multiplication recovers the original equations! To convert the system x+3y=5x + 3y = 5, 2x−z=12x - z = 1, y+2z=4y + 2z = 4 to matrix form (order x, y, z): include 0 for missing terms, so A = [[1,3,0],[2,0,−1],[0,1,2]][[1, 3, 0], [2, 0, -1], [0, 1, 2]], X = [x,y,z][x, y, z], B = [5,1,4][5, 1, 4]—AX=BAX = B matches perfectly. Choice A correctly includes the zero coefficients for missing variables, like 0 for z in the first equation and 0 for x in the third, ensuring the matrix multiplication reproduces the system. Choice B incorrectly makes A a 3x2 matrix instead of 3x3, omitting the z column and causing dimension mismatch—remember, A must have as many 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=BAX = B. Example: x+2y=5x + 2y = 5 and 3x−y=73x - y = 7 gives A = [[1,2],[3,−1]][[1, 2], [3, -1]], X = [x,y][x, y], B = [5,7][5, 7]. Systematic! To verify your matrix equation represents the system correctly, perform the matrix multiplication AXAX 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]][[2, 3], [1, -1]] times [x,y][x, y] gives [2x+3y,x−y][2x + 3y, x - y], and B = [7,1][7, 1], then matrix equation represents 2x+3y=72x + 3y = 7 and x−y=1x - y = 1. Matrix multiplication recovers the system—this verification prevents errors!

Question 12

The matrix equation [2−134][xy]=[57]\begin{bmatrix}2&-1\\3&4\end{bmatrix}\begin{bmatrix}x\\y\end{bmatrix}=\begin{bmatrix}5\\7\end{bmatrix} represents what system of equations?

  1. {2x−y=53x+4y=7\begin{cases}2x-y=5\\3x+4y=7\end{cases} (correct answer)
  2. {2x+y=53x−4y=7\begin{cases}2x+y=5\\3x-4y=7\end{cases}
  3. {2x−y=73x+4y=5\begin{cases}2x-y=7\\3x+4y=5\end{cases}
  4. {2y−x=54x+3y=7\begin{cases}2y-x=5\\4x+3y=7\end{cases}
Explanation: 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 [[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, and this equals the first entry of B, which is 5, giving equation 1: 2x - y = 5. (2) Multiply row 2 of A by X: [3, 4] times [x, y] = 3x + 4y, and this equals the second entry of B, which is 7, giving equation 2: 3x + 4y = 7. (3) The complete system is {2x - y = 5, 3x + 4y = 7}. Choice A correctly identifies both equations by properly performing the matrix multiplication and matching results to the constant vector B. Choice C has the constants from vector B in the wrong order (swapped), incorrectly pairing row 1's result with B's second entry and row 2's result with B's first entry - remember: row i corresponds to equation i, which equals B's entry i! Matrix equation interpretation strategy: (1) Identify the dimensions (here 2×2 matrix means 2 equations, 2 variables), (2) Multiply each row of A by X to get the left side of each equation, (3) Set each result equal to the corresponding entry in B, (4) Write the system with equations in order. To verify your interpretation, you can reconstruct the matrix equation from your system: the coefficients 2, -1 from equation 1 should form row 1 of A, coefficients 3, 4 from equation 2 should form row 2, and constants 5, 7 should form vector B in that order.

Question 13

The matrix equation [2−134][xy]=[57]\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?

  1. {2x−y=73x+4y=5\begin{cases}2x-y=7\\3x+4y=5\end{cases}
  2. {2x+y=53x−4y=7\begin{cases}2x+y=5\\3x-4y=7\end{cases}
  3. {2x−y=53x+4y=7\begin{cases}2x-y=5\\3x+4y=7\end{cases} (correct answer)
  4. {2y−x=53y+4x=7\begin{cases}2y-x=5\\3y+4x=7\end{cases}
Explanation: 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!

Question 14

Consider the system of equations: 2x+3y=82x + 3y = 8 5x−y=15x - y = 1

If this system is represented as Ax⃗=b⃗A\vec{x} = \vec{b} where $$\vec{x} = \begin{pmatrix} x \ y \end{pmatrix}

  1. (235−1)(xy)=(81)\begin{pmatrix} 2 & 3 \\ 5 & -1 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 8 \\ 1 \end{pmatrix} (correct answer)
  2. (253−1)(xy)=(81)\begin{pmatrix} 2 & 5 \\ 3 & -1 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 8 \\ 1 \end{pmatrix}
  3. (8123)(xy)=(5−1)\begin{pmatrix} 8 & 1 \\ 2 & 3 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 5 \\ -1 \end{pmatrix}
  4. (235−1)(xy)=(18)\begin{pmatrix} 2 & 3 \\ 5 & -1 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 1 \\ 8 \end{pmatrix}
Explanation: In matrix form Ax⃗=b⃗A\vec{x} = \vec{b}, the coefficient matrix AA contains coefficients in the same row-column arrangement as the original system. The first row contains coefficients from the first equation (2, 3), the second row contains coefficients from the second equation (5, -1), and b⃗\vec{b} contains the constants (8, 1) in order. Choice B incorrectly transposes the coefficient matrix. Choice C completely scrambles the coefficients and constants. Choice D has the correct coefficient matrix but swaps the constants in vector b⃗\vec{b}.

Question 15

A system of linear equations is given by: 3x−2y+z=73x - 2y + z = 7 x+4y−3z=−2x + 4y - 3z = -2 −2x+y+5z=8-2x + y + 5z = 8

When this system is written in the matrix form Ax⃗=b⃗A\vec{x} = \vec{b}, what is the entry in the second row, third column of matrix AA?

  1. −3-3 (correct answer)
  2. 33
  3. −2-2
  4. 55
Explanation: When writing the system in matrix form Ax⃗=b⃗A\vec{x} = \vec{b}, the coefficient matrix AA has entries that are the coefficients of the variables in order. The second row corresponds to the second equation x+4y−3z=−2x + 4y - 3z = -2, and the third column corresponds to the coefficient of zz. Therefore, the entry in the second row, third column is −3-3. Choice B incorrectly uses the positive value. Choice C uses the coefficient from the wrong position (second row, first column would be 1, but this is the coefficient of xx from the third equation). Choice D uses the coefficient of zz from the third equation.

Question 16

A student writes a system of equations in matrix form as:

(1203−14052)(xyz)=(72−1)\begin{pmatrix} 1 & 2 & 0 \\ 3 & -1 & 4 \\ 0 & 5 & 2 \end{pmatrix} \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 7 \\ 2 \\ -1 \end{pmatrix}

How many equations in this system have exactly two variables with non-zero coefficients?

  1. 11
  2. 22 (correct answer)
  3. 33
  4. 00
Explanation: When you see a matrix equation like this, you're looking at a compact way to represent a system of linear equations. To answer this question, you need to "unpack" the matrix multiplication back into individual equations and count the non-zero coefficients in each. The matrix equation represents three separate equations. When you multiply the coefficient matrix by the variable vector, each row becomes one equation:
  • Row 1: 1x+2y+0z=71x + 2y + 0z = 7, which simplifies to x+2y=7x + 2y = 7
  • Row 2: 3x+(−1)y+4z=23x + (-1)y + 4z = 2, or 3x−y+4z=23x - y + 4z = 2
  • Row 3: 0x+5y+2z=−10x + 5y + 2z = -1, which simplifies to 5y+2z=−15y + 2z = -1
Now count the non-zero coefficients in each equation. The first equation has coefficients 1 and 2 (two non-zero values). The second equation has coefficients 3, -1, and 4 (three non-zero values). The third equation has coefficients 5 and 2 (two non-zero values). Therefore, exactly two equations have exactly two variables with non-zero coefficients, making B correct. Choice A suggests only one equation meets this criteria, missing either the first or third equation. Choice C incorrectly counts all three equations, ignoring that the middle equation has three non-zero coefficients, not two. Choice D suggests no equations qualify, which overlooks the clear examples in rows 1 and 3. Remember: when working with matrix equations, always convert back to standard form to clearly see the structure of each individual equation. Look for zero coefficients—they tell you which variables are "missing" from each equation.

Question 17

The system of equations below is to be written in matrix form Ax⃗=b⃗A\vec{x} = \vec{b}: x−3z=5x - 3z = 5 2y+4z=−12y + 4z = -1 3x+y−z=73x + y - z = 7

What is the sum of all entries in the first column of matrix AA?

  1. 11
  2. 33
  3. 44 (correct answer)
  4. 00
Explanation: When you see a system of equations that needs to be written in matrix form Ax⃗=b⃗A\vec{x} = \vec{b}, you're working with the coefficients of each variable. Matrix AA contains the coefficients of xx, yy, and zz in order, where each row corresponds to one equation and each column corresponds to one variable. Let's organize the given system by rewriting each equation in standard form with all variables present:
  • Equation 1: 1x+0y+(−3)z=51x + 0y + (-3)z = 5
  • Equation 2: 0x+2y+4z=−10x + 2y + 4z = -1
  • Equation 3: 3x+1y+(−1)z=73x + 1y + (-1)z = 7
The coefficient matrix AA becomes: The first column contains the coefficients of xx: 11, 00, and 33. Adding these gives us 1+0+3=41 + 0 + 3 = 4. Looking at the wrong answers: A) 11 only counts the first coefficient, ignoring the others. B) 33 only counts the last coefficient from the third equation. D) 00 might result from incorrectly thinking the missing xx term in the second equation means all coefficients are zero. The correct answer is C) 44. Study tip: When converting systems to matrix form, always write each equation in standard form first, explicitly showing zero coefficients for missing variables. This prevents you from accidentally skipping terms when building your coefficient matrix.

Question 18

Consider the matrix equation:

(2−1304−2110)(abc)=(864)\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 cc variable?

  1. 88
  2. 66
  3. 44 (correct answer)
  4. 00
Explanation: 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 cc variable. To find the equations, multiply each row of the matrix by the variable vector. The first row gives you: 2a−1b+3c=82a - 1b + 3c = 8. The second row gives you: 0a+4b+(−2)c=60a + 4b + (-2)c = 6, which simplifies to 4b−2c=64b - 2c = 6. The third row gives you: 1a+1b+0c=41a + 1b + 0c = 4, which simplifies to a+b=4a + b = 4. The equation with no cc variable is a+b=4a + b = 4, so the constant term is 44. Looking at the wrong answers: Choice A (88) is the constant from the first equation 2a−b+3c=82a - b + 3c = 8, but this equation contains the cc variable. Choice B (66) is the constant from the second equation 4b−2c=64b - 2c = 6, but this also contains the cc variable. Choice D (00) might tempt you because the coefficient of cc in the third equation is 00, 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.

Question 19

A linear system has the matrix representation:

(1−2430−1−152)(xyz)=(6−39)\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 yy in the second equation of this system?

  1. 55
  2. −2-2
  3. 33
  4. 00 (correct answer)
Explanation: When you encounter a matrix equation like this, you're looking at a compact way to represent a system of linear equations. The coefficient matrix (on the left) contains all the coefficients of your variables xx, yy, and zz, organized so that each row represents one equation. To find the coefficient of yy in the second equation, you need to look at the second row of the coefficient matrix and identify the entry in the yy-column. Since the variables are ordered as xx, yy, zz, the yy-coefficients appear in the second column of the matrix. The second row of the coefficient matrix is (3,0,−1)(3, 0, -1). The second entry in this row is 00, which means the coefficient of yy in the second equation is 00. You can verify this by writing out the second equation: 3x+0y+(−1)z=−33x + 0y + (-1)z = -3, or simply 3x−z=−33x - z = -3. Looking at the wrong answers: Choice A (55) is the coefficient of yy in the third equation, not the second. Choice B (−2-2) is the coefficient of yy in the first equation. Choice C (33) is the coefficient of xx 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.

Question 20

A system of three linear equations in three variables is written in matrix form as Ax⃗=b⃗A\vec{x} = \vec{b} where:

$$, $$\vec{x} = \begin{pmatrix} x \\ y \\ z \end{pmatrix} $$, and $$\vec{b} = \begin{pmatrix} 5 \\ 1 \\ -3 \end{pmatrix}

If the coefficient of zz in the third equation is twice the coefficient of xx in the first equation, what is the value of kk?

  1. 22
  2. 44 (correct answer)
  3. 11
  4. −1-1
Explanation: When you see a matrix equation Ax⃗=b⃗A\vec{x} = \vec{b}, remember that each entry in the coefficient matrix AA corresponds to a specific coefficient in your system of equations. The key is understanding how matrix positions relate to the original equations. Looking at the matrix AA, the coefficient of xx in the first equation is the entry in position (1,1), which is 22. The coefficient of zz in the third equation is the entry in position (3,3), which is kk. The problem states that "the coefficient of zz in the third equation is twice the coefficient of xx in the first equation." This gives us the relationship: k=2×2=4k = 2 \times 2 = 4. Let's check why the other answers don't work: A) k=2k = 2 would mean the coefficient of zz in the third equation equals (not doubles) the coefficient of xx in the first equation. C) k=1k = 1 would make the coefficient of zz in the third equation half the coefficient of xx in the first equation. D) k=−1k = -1 would give us a negative relationship that doesn't match the "twice" condition described. The correct answer is B) 44. Study tip: When working with matrix equations, always identify which matrix position corresponds to which variable coefficient in each equation. The pattern is: for equation ii and variable jj, the coefficient is in position (i,j)(i,j) of matrix AA. This systematic approach prevents confusion when translating between matrix form and standard equation form.