Precalculus Quiz: Representing Linear Systems With Matrices
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Representing Linear Systems With MatricesQuestion 1 of 20

Which system of linear equations corresponds to the matrix equation:

\begin{pmatrix} 0 & 1 & -3 \\ 2 & -1 & 0 \\ 1 & 2 & 1 \end{pmatrix} \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 7 \\ -2 \\ 5 \end{pmatrix} $$?
−3x+y=7,2x−y=−2,x+2y+z=5-3x + y = 7, \quad 2x - y = -2, \quad x + 2y + z = 5
x+y−3z=7,2x−y+z=−2,x+2y=5x + y - 3z = 7, \quad 2x - y + z = -2, \quad x + 2y = 5
y−3z=7,2x−y=−2,x+2y+z=5y - 3z = 7, \quad 2x - y = -2, \quad x + 2y + z = 5
x−3z=7,2x−y=−2,x+2y+z=5x - 3z = 7, \quad 2x - y = -2, \quad x + 2y + z = 5
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Precalculus Quiz

Precalculus Quiz: Representing Linear Systems With Matrices

Practice Representing Linear Systems With Matrices in Precalculus 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 Precalculus.

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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.

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Question 1

Which system of linear equations corresponds to the matrix equation:

\begin{pmatrix} 0 & 1 & -3 \\ 2 & -1 & 0 \\ 1 & 2 & 1 \end{pmatrix} \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 7 \\ -2 \\ 5 \end{pmatrix} $$?
  1. −3x+y=7,2x−y=−2,x+2y+z=5-3x + y = 7, \quad 2x - y = -2, \quad x + 2y + z = 5
  2. x+y−3z=7,2x−y+z=−2,x+2y=5x + y - 3z = 7, \quad 2x - y + z = -2, \quad x + 2y = 5
  3. y−3z=7,2x−y=−2,x+2y+z=5y - 3z = 7, \quad 2x - y = -2, \quad x + 2y + z = 5 (correct answer)
  4. x−3z=7,2x−y=−2,x+2y+z=5x - 3z = 7, \quad 2x - y = -2, \quad x + 2y + z = 5
Explanation: When you encounter a matrix equation like this, you're seeing a compact way to represent a system of linear equations. The key is understanding how matrix multiplication works to translate between the two forms. To convert the matrix equation to a system, you multiply each row of the coefficient matrix by the variable vector. The first row (01−3)\begin{pmatrix} 0 & 1 & -3 \end{pmatrix} multiplied by (xyz)\begin{pmatrix} x \\ y \\ z \end{pmatrix} gives 0x+1y+(−3)z=y−3z0x + 1y + (-3)z = y - 3z, which equals 7. The second row (2−10)\begin{pmatrix} 2 & -1 & 0 \end{pmatrix} gives 2x−y+0z=2x−y2x - y + 0z = 2x - y, which equals -2. The third row (121)\begin{pmatrix} 1 & 2 & 1 \end{pmatrix} gives x+2y+zx + 2y + z, which equals 5. This produces the system: y−3z=7y - 3z = 7, 2x−y=−22x - y = -2, x+2y+z=5x + 2y + z = 5, which is exactly option C. Option A incorrectly writes the first equation as −3x+y=7-3x + y = 7, mixing up the coefficients. Option B writes the first equation as x+y−3z=7x + y - 3z = 7, adding an xx term that shouldn't exist since the first entry in row 1 is 0. It also incorrectly includes a zz term in the second equation and omits zz from the third. Option D writes the first equation as x−3z=7x - 3z = 7, again adding an xx term and omitting the yy term. Remember: each row of the coefficient matrix corresponds to one equation, and the position of each number tells you which variable it multiplies. Zero coefficients mean that variable doesn't appear in that equation.

Question 2

Consider the matrix equation:

(abcdefghi)(stu)=(pqr)\begin{pmatrix} a & b & c \\ d & e & f \\ g & h & i \end{pmatrix} \begin{pmatrix} s \\ t \\ u \end{pmatrix} = \begin{pmatrix} p \\ q \\ r \end{pmatrix}

If the corresponding system has the equation 2s−3t+7u=−52s - 3t + 7u = -5 as its second equation, what are the values of dd, ee, ff, and qq?

  1. d=−5,e=2,f=−3,q=7d = -5, e = 2, f = -3, q = 7
  2. d=2,e=−3,f=7,q=−5d = 2, e = -3, f = 7, q = -5 (correct answer)
  3. d=2,e=7,f=−3,q=−5d = 2, e = 7, f = -3, q = -5
  4. d=−3,e=2,f=7,q=−5d = -3, e = 2, f = 7, q = -5
Explanation: When you see a matrix equation like this, remember that matrix multiplication creates a system of linear equations. Each row of the coefficient matrix multiplies the variable vector to produce one equation in the system. To find the correct values, you need to understand how matrix multiplication works. The second row of the coefficient matrix (def)\begin{pmatrix} d & e & f \end{pmatrix} multiplies the variable vector (stu)\begin{pmatrix} s \\ t \\ u \end{pmatrix} to give the second component of the result vector. This multiplication yields: d⋅s+e⋅t+f⋅u=qd \cdot s + e \cdot t + f \cdot u = q Since you're told the second equation is 2s−3t+7u=−52s - 3t + 7u = -5, you can directly match coefficients:
  • The coefficient of ss is 22, so d=2d = 2
  • The coefficient of tt is −3-3, so e=−3e = -3
  • The coefficient of uu is 77, so f=7f = 7
  • The constant term is −5-5, so q=−5q = -5
This gives us d=2,e=−3,f=7,q=−5d = 2, e = -3, f = 7, q = -5, which is answer choice B. Looking at the wrong answers: Choice A incorrectly places −5-5 as the value of dd instead of qq. Choice C swaps the values of ee and ff, confusing the coefficients of tt and uu. Choice D mixes up multiple coefficients, placing −3-3 in the dd position. Study tip: Always remember that in matrix-to-system conversions, coefficients stay in their exact positions—first matrix entry becomes the first variable's coefficient, second becomes the second variable's coefficient, and so on. The result vector gives you the constants on the right side of each equation.

Question 3

For the system of equations

{−x+2y+4z=35x−y+z=−72x+3y−2z=1\begin{cases} -x+2y+4z=3\\ 5x-y+z=-7\\ 2x+3y-2z=1 \end{cases}

what is the entry a23a_{23} of the coefficient matrix AA (row 2, column 3) when the variable order is x,y,zx,y,z?​

  1. −1-1
  2. 44
  3. 11 (correct answer)
  4. 33
Explanation: This question tests understanding of how to represent a system of linear equations as a single matrix equation Ax = b. To build the coefficient matrix A from a system of equations, create a matrix where entry aᵢⱼ is the coefficient of variable j in equation i: rows represent equations, columns represent variables, and if a variable doesn't appear in an equation, its coefficient is 0. From equation 5x - y + z = -7, the coefficients are 5 for x, -1 for y, and 1 for z, so the corresponding row of the coefficient matrix is [5 -1 1], and thus a_{23} (row 2, column 3) is 1. Choice C is correct because it accurately extracts the coefficient of z in the second equation. Choice A has the wrong sign, using -1 when the actual coefficient is 1 for z in equation 2. Key to converting systems to matrix form: create the coefficient matrix A by listing coefficients in order (row i = equation i, column j = variable j), create variable vector x listing unknowns in the same order as columns, and create constant vector b listing the right-hand sides in the same order as equations. To verify your matrix equation is correct, expand Ax by multiplying: each row of A dotted with x should give the left side of the corresponding equation, and these should equal the entries of b (the right sides).

Question 4

Consider the system: 3x+2y=53x + 2y = 5 −x+4y=3-x + 4y = 3

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

  1. (32−14)(xy)=(53)\begin{pmatrix} 3 & 2 \\ -1 & 4 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 5 \\ 3 \end{pmatrix} (correct answer)
  2. (3−124)(xy)=(53)\begin{pmatrix} 3 & -1 \\ 2 & 4 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 5 \\ 3 \end{pmatrix}
  3. (5332−14)(xy)=(00)\begin{pmatrix} 5 & 3 \\ 3 & 2 \\ -1 & 4 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \end{pmatrix}
  4. (32−14)(xy)=(35)\begin{pmatrix} 3 & 2 \\ -1 & 4 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 3 \\ 5 \end{pmatrix}
Explanation: The coefficient matrix AA is formed by placing coefficients row by row: first row contains coefficients of 3x+2y=53x + 2y = 5, giving [3,2][3, 2], and second row contains coefficients of −x+4y=3-x + 4y = 3, giving [−1,4][-1, 4]. The vector b⃗\vec{b} contains the constants [5,3][5, 3]. Choice B incorrectly transposes the coefficient matrix. Choice C has wrong dimensions and incorrect constant vector. Choice D swaps the entries in the constant vector.

Question 5

A linear system has the matrix representation:

(1−2305−1402)(pqr)=(8−312)\begin{pmatrix} 1 & -2 & 3 \\ 0 & 5 & -1 \\ 4 & 0 & 2 \end{pmatrix} \begin{pmatrix} p \\ q \\ r \end{pmatrix} = \begin{pmatrix} 8 \\ -3 \\ 12 \end{pmatrix}

What is the second equation in this system when written in standard form?

  1. 0p+5q−1r=−30p + 5q - 1r = -3
  2. p−2q+3r=8p - 2q + 3r = 8
  3. 4p+2r=124p + 2r = 12
  4. 5q−r=−35q - r = -3 (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 matrix multiplication Ax⃗=b⃗A\vec{x} = \vec{b} expands into individual equations where each row of the coefficient matrix corresponds to one equation in the system. To find the second equation, focus on the second row of the coefficient matrix and the second element of the result vector. The second row is (05−1)\begin{pmatrix} 0 & 5 & -1 \end{pmatrix}, which means the coefficients are 0 for pp, 5 for qq, and -1 for rr. The second element of the result vector is -3. This gives us: 0p+5q+(−1)r=−30p + 5q + (-1)r = -3, which simplifies to 5q−r=−35q - r = -3. Choice D correctly represents this equation in its simplest form by omitting the zero coefficient term. Choice A writes out the full form 0p+5q−1r=−30p + 5q - 1r = -3, which is mathematically correct but unnecessarily includes the zero term. Standard form typically omits terms with zero coefficients. Choice B gives you the first equation (first row): p−2q+3r=8p - 2q + 3r = 8. This comes from the first row $$ \begin{pmatrix} 1 & -2 & 3 \end{pmatrix} Choice C represents the third equation (third row): $$4p + 0q + 2r = 12$$, simplified to $$4p + 2r = 12$$ by dropping the zero term. **Study tip:** Always match the row number to the equation number. The second row always gives you the second equation, and so on. When writing in standard form, drop any terms with zero coefficients for cleaner presentation.

Question 6

Two different linear systems are represented by: System 1: Ax⃗=b⃗A\vec{x} = \vec{b} where $$A = \begin{pmatrix} 1 & 2 \ 3 & 4 \end{pmatrix}

System 2: $$C\vec{x} = \vec{d}$$ where $$C = \begin{pmatrix} 1 & 3 \\ 2 & 4 \end{pmatrix}

What is the relationship between these coefficient matrices?

  1. CC is the inverse of AA, meaning C=A−1C = A^{-1}
  2. CC is the transpose of AA, meaning C=ATC = A^T (correct answer)
  3. CC is the negative of AA, meaning C=−AC = -A
  4. CC is scalar multiple of AA, meaning C=kAC = kA for some constant kk
Explanation: When you encounter questions comparing matrices, you need to systematically check the fundamental matrix relationships: transpose, inverse, scalar multiplication, and addition/subtraction. Let's examine what happens when we transpose matrix AA. The transpose operation ATA^T flips a matrix across its main diagonal, so rows become columns and columns become rows. For matrix A=(1234)A = \begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix} , the transpose is AT=(1324)A^T = \begin{pmatrix} 1 & 3 \\ 2 & 4 \end{pmatrix} . This matches matrix CC exactly, confirming that C=ATC = A^T. Now let's check why the other options fail. Option A claims C=A−1C = A^{-1}. To find A−1A^{-1}, we calculate 1det⁡(A)(4−2−31)=1−2(4−2−31)=(−211.5−0.5)\frac{1}{\det(A)} \begin{pmatrix} 4 & -2 \\ -3 & 1 \end{pmatrix} = \frac{1}{-2} \begin{pmatrix} 4 & -2 \\ -3 & 1 \end{pmatrix} = \begin{pmatrix} -2 & 1 \\ 1.5 & -0.5 \end{pmatrix} , which clearly doesn't equal CC. Option C suggests C=−AC = -A, but −A=(−1−2−3−4)-A = \begin{pmatrix} -1 & -2 \\ -3 & -4 \end{pmatrix} , which has all negative entries unlike CC. Option D proposes C=kAC = kA for some scalar kk. If this were true, then 11=32=23=44\frac{1}{1} = \frac{3}{2} = \frac{2}{3} = \frac{4}{4}, which gives us 1=1.5=0.67=11 = 1.5 = 0.67 = 1 – clearly impossible. Study tip: When comparing matrices, always check transpose first since it's the quickest calculation. Remember that (i,j)(i,j) entry of ATA^T equals the (j,i)(j,i) entry of AA.

Question 7

For the system of equations (variable order [x  y  z]T[x\;y\;z]^T)

{x−3y+2z=04x+z=5−2y+3z=−6\begin{cases} x-3y+2z=0\\ 4x+z=5\\ -2y+3z=-6 \end{cases}

what is the coefficient matrix AA in Ax=bAx=b?

  1. [1−324010−23]\begin{bmatrix}1&-3&2\\4&0&1\\0&-2&3\end{bmatrix} (correct answer)
  2. [1−324100−23]\begin{bmatrix}1&-3&2\\4&1&0\\0&-2&3\end{bmatrix}
  3. [140−30−2213]\begin{bmatrix}1&4&0\\-3&0&-2\\2&1&3\end{bmatrix}
  4. [1−32040150−23−6]\begin{bmatrix}1&-3&2&0\\4&0&1&5\\0&-2&3&-6\end{bmatrix}
Explanation: This question tests understanding of how to represent a system of linear equations as a single matrix equation Ax = b. To build the coefficient matrix A from a system of equations, create a matrix where entry aᵢⱼ is the coefficient of variable j in equation i: rows represent equations, columns represent variables, and if a variable doesn't appear in an equation, its coefficient is 0. For the system x - 3y + 2z = 0, 4x + z = 5, and -2y + 3z = -6, we must be careful about missing variables. In equation 2 (4x + z = 5), variable y doesn't appear, which means its coefficient is 0, so the row is [4 0 1] if variables are ordered x, y, z. Similarly, in equation 3 (-2y + 3z = -6), variable x doesn't appear, giving row [0 -2 3]. Choice A is correct because it properly includes zero coefficients for missing variables: [[1 -3 2], [4 0 1], [0 -2 3]]. Choice B makes an error by putting 1 instead of 0 for the coefficient of y in equation 2, incorrectly suggesting the equation is 4x + y + 0z = 5. Remember to include 0 coefficients when a variable doesn't appear in an equation: for example, if equation 2 is '4x + z = 5' in a system with variables x, y, z, the coefficient row is [4 0 1], with the 0 for the missing y.

Question 8

For the system of equations

{−2x+y−3z=64x−y+z=−2x+2y+0z=5\begin{cases} -2x + y - 3z = 6 \\ 4x - y + z = -2 \\ x + 2y + 0z = 5 \end{cases}

what is entry a23a_{23} of the coefficient matrix AA (row 2, column 3) when the variable order is (x,y,z)(x,y,z)?

  1. −3-3
  2. 00
  3. 11 (correct answer)
  4. 22
Explanation: This question tests understanding of how to represent a system of linear equations as a single matrix equation Ax=bAx = b. To build the coefficient matrix AA from a system of equations, create a matrix where entry aija_{ij} is the coefficient of variable j in equation i: rows represent equations, columns represent variables, and if a variable doesn't appear in an equation, its coefficient is 0. From equation 4x−y+z=−24x - y + z = -2, the coefficients are 4 for x, -1 for y, and 1 for z, so the corresponding row of the coefficient matrix is [4,−1,1][4, -1, 1], and a23a_{23} is the entry in row 2, column 3, which is 1 for z. Choice C is correct because it properly extracts the coefficient for z in the second equation as 1. Choice A has the wrong sign on the coefficient, using -3 for a different variable when the actual coefficient for z in equation 2 is +1. Key to converting systems to matrix form: create the coefficient matrix AA by listing coefficients in order (row i = equation i, column j = variable j), create variable vector x listing unknowns in the same order as columns, and create constant vector b listing the right-hand sides in the same order as equations. Remember to include 0 coefficients when a variable doesn't appear in an equation: for example, if equation 2 is x+3z=5x + 3z = 5 in a system with variables x, y, z, the coefficient row is [1,0,3][1, 0, 3], with the 0 for the missing y.

Question 9

A system of linear equations is given by: 2x−3y+z=72x - 3y + z = 7 x+4y−2z=−1x + 4y - 2z = -1 −3x+y+5z=8-3x + 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. −2-2 (correct answer)
  2. −1-1
  3. 44
  4. 11
Explanation: In matrix form Ax⃗=b⃗A\vec{x} = \vec{b}, matrix AA contains the coefficients of the variables. The second row corresponds to the second equation x+4y−2z=−1x + 4y - 2z = -1, and the third column corresponds to the coefficient of zz. Therefore, the entry in the second row, third column is −2-2. Choice B is the constant term from the second equation. Choice C is the coefficient of yy in the second equation. Choice D is the coefficient of xx in the second equation.

Question 10

Verify that the matrix equation is equivalent to the system by selecting the matching matrix equation. System:

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

Which matrix equation Ax=bAx=b matches this system (using variable order x,yx,y)?

  1. (21−15)(xy)=(3−4)\begin{pmatrix} 2 & 1 \\ -1 & 5 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 3 \\ -4 \end{pmatrix} (correct answer)
  2. (2−115)(xy)=(3−4)\begin{pmatrix} 2 & -1 \\ 1 & 5 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 3 \\ -4 \end{pmatrix}
  3. (21−15)(xy)=(−43)\begin{pmatrix} 2 & 1 \\ -1 & 5 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} -4 \\ 3 \end{pmatrix}
  4. (2115)(xy)=(3−4)\begin{pmatrix} 2 & 1 \\ 1 & 5 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 3 \\ -4 \end{pmatrix}
Explanation: This question tests understanding of how to represent a system of linear equations as a single matrix equation Ax=bAx = b. A system of linear equations can be written compactly as Ax=bAx = b, where AA is the coefficient matrix (containing all the coefficients of the variables), xx is the column vector of variables (the unknowns), and bb is the column vector of constants (the right-hand side values from each equation). For the system 2x+y=32x + y = 3, −x+5y=−4-x + 5y = -4, we extract the coefficients: equation 1 has coefficients 2 for x and 1 for y, giving first row [2 1][2\ 1], and equation 2 has coefficients -1 for x and 5 for y, giving second row [−1 5][-1\ 5]. The coefficient matrix is A=(21−15)A = \begin{pmatrix} 2 & 1 \\ -1 & 5 \end{pmatrix}, the variable vector is x=(xy)x = \begin{pmatrix} x \\ y \end{pmatrix}, and the constant vector is b=(3−4)b = \begin{pmatrix} 3 \\ -4 \end{pmatrix}. Choice A is correct because it properly extracts coefficients into matrix A and correctly identifies variable and constant vectors. Choice B has the wrong sign on the coefficient of x in equation 2, using 1 when the equation has -1, and swaps signs in the rows. The matrix form Ax=bAx = b is just a compact notation for the system—it doesn't solve the system, but it sets up for solution methods like using the inverse (if A is invertible) or row reduction.

Question 11

Verify the matrix equation is equivalent to the system by choosing the matching matrix form. System:

{x+y=32x−y=1\begin{cases} x+y=3\\ 2x-y=1 \end{cases}

Which matrix equation correctly represents it as Ax=bAx=b?

  1. [112−1][xy]=[31]\begin{bmatrix}1&1\\2&-1\end{bmatrix}\begin{bmatrix}x\\y\end{bmatrix}=\begin{bmatrix}3\\1\end{bmatrix} (correct answer)
  2. [121−1][xy]=[31]\begin{bmatrix}1&2\\1&-1\end{bmatrix}\begin{bmatrix}x\\y\end{bmatrix}=\begin{bmatrix}3\\1\end{bmatrix}
  3. [1121][xy]=[31]\begin{bmatrix}1&1\\2&1\end{bmatrix}\begin{bmatrix}x\\y\end{bmatrix}=\begin{bmatrix}3\\1\end{bmatrix}
  4. [112−1][xy]=[13]\begin{bmatrix}1&1\\2&-1\end{bmatrix}\begin{bmatrix}x\\y\end{bmatrix}=\begin{bmatrix}1\\3\end{bmatrix}
Explanation: This question tests understanding of how to represent a system of linear equations as a single matrix equation Ax = b. A system of linear equations can be written compactly as Ax = b, where A is the coefficient matrix (containing all the coefficients of the variables), x is the column vector of variables (the unknowns), and b is the column vector of constants (the right-hand side values from each equation). For the system x + y = 3 and 2x - y = 1, we extract the coefficients: equation 1 has coefficients 1 and 1 for variables x and y, giving first row [1 1], and equation 2 has coefficients 2 and -1, giving second row [2 -1]. The coefficient matrix is A = [[1 1], [2 -1]], the variable vector is x = [x; y], and the constant vector is b = [3; 1]. Choice A is correct because it properly extracts coefficients into matrix A = [[1 1], [2 -1]] and correctly identifies the constant vector b = [3; 1]. Choice C has the wrong sign on the coefficient of y in the second equation, using 1 when the equation has -y (coefficient -1). To verify your matrix equation is correct, expand Ax by multiplying: each row of A dotted with x should give the left side of the corresponding equation, and these should equal the entries of b (the right sides).

Question 12

For the system of equations

{−x+2y+4z=35x−y+z=−72x+3y−2z=1\begin{cases} -x+2y+4z=3\\ 5x-y+z=-7\\ 2x+3y-2z=1 \end{cases}

what is the entry a23a_{23} of the coefficient matrix AA (row 2, column 3) when the variable order is x,y,zx,y,z?

  1. −1-1
  2. 44
  3. 11 (correct answer)
  4. 33
Explanation: This question tests understanding of how to represent a system of linear equations as a single matrix equation Ax=bAx = b. To build the coefficient matrix AA from a system of equations, create a matrix where entry aija_{ij} is the coefficient of variable jj in equation ii: rows represent equations, columns represent variables, and if a variable doesn't appear in an equation, its coefficient is 0. From equation 5x−y+z=−75x - y + z = -7, the coefficients are 5 for xx, -1 for yy, and 1 for zz, so the corresponding row of the coefficient matrix is [5 -1 1], and thus a23a_{23} (row 2, column 3) is 1. Choice C is correct because it accurately extracts the coefficient of zz in the second equation. Choice A has the wrong sign, using -1 when the actual coefficient is 1 for zz in equation 2. Key to converting systems to matrix form: create the coefficient matrix AA by listing coefficients in order (row ii = equation ii, column jj = variable jj), create variable vector xx listing unknowns in the same order as columns, and create constant vector bb listing the right-hand sides in the same order as equations. To verify your matrix equation is correct, expand AxAx by multiplying: each row of AA dotted with xx should give the left side of the corresponding equation, and these should equal the entries of bb (the right sides).

Question 13

A homogeneous linear system is represented in matrix form as:

(2−1312−1312)(xyz)=(000)\begin{pmatrix} 2 & -1 & 3 \\ 1 & 2 & -1 \\ 3 & 1 & 2 \end{pmatrix} \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix}

Which statement about this system is necessarily true?

  1. The system always has the solution (x,y,z)=(0,0,0)(x,y,z) = (0,0,0) regardless of the coefficient matrix values. (correct answer)
  2. The system has a unique non-trivial solution since the coefficient matrix is 3×33 \times 3 and square.
  3. The system has infinitely many solutions because the right-hand side vector contains only zeros throughout.
  4. The system has no solution unless the coefficient matrix has a determinant equal to zero.
Explanation: A homogeneous system Ax⃗=0⃗A\vec{x} = \vec{0} always has the trivial solution x⃗=0⃗\vec{x} = \vec{0}, regardless of the coefficient matrix AA. This is because A⋅0⃗=0⃗A \cdot \vec{0} = \vec{0} for any matrix AA. Choice B is incorrect because whether additional non-trivial solutions exist depends on the determinant of AA. Choice C is incorrect because infinitely many solutions only occur when det⁡(A)=0\det(A) = 0. Choice D is incorrect because homogeneous systems always have at least the trivial solution.

Question 14

For the system of equations (note the missing zz term in the second equation)

{2x−y+3z=1−x+4y=65x+z=−2\begin{cases} 2x-y+3z=1\\ -x+4y=6\\ 5x+z=-2 \end{cases}

what is the coefficient matrix AA in Ax=bAx=b (using variable order x,y,zx,y,z)?​

  1. (2−13−140501)\begin{pmatrix}2&-1&3\\-1&4&0\\5&0&1\end{pmatrix} (correct answer)
  2. (2−13−141500)\begin{pmatrix}2&-1&3\\-1&4&1\\5&0&0\end{pmatrix}
  3. (2−13−146501)\begin{pmatrix}2&-1&3\\-1&4&6\\5&0&1\end{pmatrix}
  4. (2−134−10501)\begin{pmatrix}2&-1&3\\4&-1&0\\5&0&1\end{pmatrix}
Explanation: This question tests understanding of how to represent a system of linear equations as a single matrix equation Ax = b. To build the coefficient matrix A from a system of equations, create a matrix where entry aᵢⱼ is the coefficient of variable j in equation i: rows represent equations, columns represent variables, and if a variable doesn't appear in an equation, its coefficient is 0. In the equation -x + 4y = 6, variable z doesn't appear, which means its coefficient is 0, so the row is [-1 4 0] if variables are ordered x, y, z; similarly, in 5x + z = -2, y is missing so coefficient 0, giving [5 0 1], and first equation is [2 -1 3]. Choice A is correct because it properly extracts coefficients into matrix A, including zeros for missing variables. Choice B omits the zero coefficient for the missing variable z in equation 2, shifting all subsequent entries incorrectly; when a variable doesn't appear in an equation, we must include a 0 in that position. Remember to include 0 coefficients when a variable doesn't appear in an equation: for example, if equation 2 is 'x + 3z = 5' in a system with variables x, y, z, the coefficient row is [1 0 3], with the 0 for the missing y. Common error: don't put the constants (right-hand side values) into the coefficient matrix A—they belong in the separate vector b, while A contains only the coefficients of the variables.

Question 15

A student writes the system x+2y−z=3x + 2y - z = 3 3x−y+2z=13x - y + 2z = 1 in matrix form and claims the result is:

(123−1)(xy)=(31)\begin{pmatrix} 1 & 2 \\ 3 & -1 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 3 \\ 1 \end{pmatrix}

What error did the student make?

  1. The student wrote the coefficient matrix in column form rather than row form as required.
  2. The student incorrectly placed the constant terms in the coefficient matrix instead of the vector.
  3. The student omitted the variable zz and its coefficients from the matrix representation completely. (correct answer)
  4. The student reversed the order of the equations when constructing the coefficient matrix arrangement.
Explanation: When converting a system of linear equations to matrix form, you must account for every variable that appears in the system. The matrix equation Ax=bA\mathbf{x} = \mathbf{b} requires the coefficient matrix AA to include all variables, the variable vector x\mathbf{x} to list all unknowns, and the constant vector b\mathbf{b} to contain the right-hand side values. The correct matrix representation should be: (12−13−12)(xyz)=(31)\begin{pmatrix} 1 & 2 & -1 \\ 3 & -1 & 2 \end{pmatrix} \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 3 \\ 1 \end{pmatrix} The student's error is clear: they completely ignored the variable zz and its coefficients (−1-1 and 22), creating a 2×22 \times 2 coefficient matrix instead of the required 2×32 \times 3 matrix. This makes answer choice C correct. Looking at the other options: A is wrong because the coefficient matrix is correctly arranged in row form—each row represents one equation. B is incorrect because the constant terms (3 and 1) are properly placed in the constant vector, not mixed into the coefficient matrix. D is false because the equations appear in the correct order—the first row corresponds to the first equation, and the second row to the second equation. Study tip: When writing systems in matrix form, always count your variables first. Your coefficient matrix must have as many columns as you have variables, and your variable vector must include every unknown that appears in the system.

Question 16

Express the system as Ax=bAx=b and identify the coefficient matrix AA:

{2x+0y=−4−x+3y=1\begin{cases} 2x + 0y = -4 \\ - x + 3y = 1 \end{cases}
  1. [20−13]\begin{bmatrix}2 & 0\\ -1 & 3\end{bmatrix} (correct answer)
  2. [2−1]\begin{bmatrix}2\\ -1\end{bmatrix}
  3. [2−103]\begin{bmatrix}2 & -1\\ 0 & 3\end{bmatrix}
  4. [2013]\begin{bmatrix}2 & 0\\ 1 & 3\end{bmatrix}
Explanation: This question tests understanding of how to represent a system of linear equations as a single matrix equation Ax=bAx = b. To build the coefficient matrix AA from a system of equations, create a matrix where entry aija_{ij} is the coefficient of variable jj in equation ii: rows represent equations, columns represent variables, and if a variable doesn't appear in an equation, its coefficient is 0. For the system 2x+0y=−42x + 0y = -4 and −x+3y=1-x + 3y = 1, we extract the coefficients: equation 1 has coefficients 2 for x and 0 for y, giving first row [2 0][2\ 0], and equation 2 has coefficients -1 for x and 3 for y, giving second row [−1 3][-1\ 3]. Choice A is correct because it properly extracts coefficients into matrix AA, including the zero for the missing y in the first equation. Choice D has the wrong sign on the coefficient of x in equation 2, using +1 when the equation has -x. Remember to include 0 coefficients when a variable doesn't appear in an equation: for example, if equation 2 is 'x+3z=5x + 3z = 5' in a system with variables x, y, z, the coefficient row is [1 0 3][1\ 0\ 3], with the 0 for the missing y. To verify your matrix equation is correct, expand AxAx by multiplying: each row of AA dotted with x should give the left side of the corresponding equation, and these should equal the entries of b (the right sides).

Question 17

Given the system of equations

{2x−y+z=1−x+3y−2z=−44x+0y+5z=6\begin{cases} 2x - y + z = 1 \\ - x + 3y - 2z = -4 \\ 4x + 0y + 5z = 6 \end{cases}

write the system as a matrix equation Ax=bAx=b (using variable order (x,y,z)(x,y,z)).

  1. [2−11−13−2405][xyz]=[1−46]\begin{bmatrix}2 & -1 & 1\\ -1 & 3 & -2\\ 4 & 0 & 5\end{bmatrix}\begin{bmatrix}x\\ y\\ z\end{bmatrix}=\begin{bmatrix}1\\ -4\\ 6\end{bmatrix} (correct answer)
  2. [2−11−132405][xyz]=[1−46]\begin{bmatrix}2 & -1 & 1\\ -1 & 3 & 2\\ 4 & 0 & 5\end{bmatrix}\begin{bmatrix}x\\ y\\ z\end{bmatrix}=\begin{bmatrix}1\\ -4\\ 6\end{bmatrix}
  3. [2−11−13−2450][xyz]=[1−46]\begin{bmatrix}2 & -1 & 1\\ -1 & 3 & -2\\ 4 & 5 & 0\end{bmatrix}\begin{bmatrix}x\\ y\\ z\end{bmatrix}=\begin{bmatrix}1\\ -4\\ 6\end{bmatrix}
  4. [2−11−13−2405][xyz]=[146]\begin{bmatrix}2 & -1 & 1\\ -1 & 3 & -2\\ 4 & 0 & 5\end{bmatrix}\begin{bmatrix}x\\ y\\ z\end{bmatrix}=\begin{bmatrix}1\\ 4\\ 6\end{bmatrix}
Explanation: This question tests understanding of how to represent a system of linear equations as a single matrix equation Ax = b. A system of linear equations can be written compactly as Ax = b, where A is the coefficient matrix (containing all the coefficients of the variables), x is the column vector of variables (the unknowns), and b is the column vector of constants (the right-hand side values from each equation). For the system 2x - y + z = 1, -x + 3y - 2z = -4, and 4x + 0y + 5z = 6, we extract the coefficients: equation 1 has [2 -1 1], equation 2 has [-1 3 -2], equation 3 has [4 0 5]; the coefficient matrix is A = [[2 -1 1]; [-1 3 -2]; [4 0 5]], x = [x; y; z], and b = [1; -4; 6]. Choice A is correct because it properly extracts coefficients into matrix A, including the zero for y in the third equation. Choice B has the wrong sign on the coefficient of z in equation 2, using +2 when the equation has -2z. Remember to include 0 coefficients when a variable doesn't appear in an equation: for example, if equation 2 is 'x + 3z = 5' in a system with variables x, y, z, the coefficient row is [1 0 3], with the 0 for the missing y. To verify your matrix equation is correct, expand Ax by multiplying: each row of A dotted with x should give the left side of the corresponding equation, and these should equal the entries of b (the right sides).

Question 18

Verify the matrix form by choosing the matrix equation that is equivalent to the system

{x+5y=−4−2x+y=3\begin{cases} x + 5y = -4\\ -2x + y = 3 \end{cases}
  1. [15−21][xy]=[−43]\begin{bmatrix}1 & 5\\ -2 & 1\end{bmatrix}\begin{bmatrix}x\\ y\end{bmatrix}=\begin{bmatrix}-4\\ 3\end{bmatrix} (correct answer)
  2. [1−251][xy]=[−43]\begin{bmatrix}1 & -2\\ 5 & 1\end{bmatrix}\begin{bmatrix}x\\ y\end{bmatrix}=\begin{bmatrix}-4\\ 3\end{bmatrix}
  3. [15−21][xy]=[43]\begin{bmatrix}1 & 5\\ -2 & 1\end{bmatrix}\begin{bmatrix}x\\ y\end{bmatrix}=\begin{bmatrix}4\\ 3\end{bmatrix}
  4. [15−4−213][xy]=[00]\begin{bmatrix}1 & 5 & -4\\ -2 & 1 & 3\end{bmatrix}\begin{bmatrix}x\\ y\end{bmatrix}=\begin{bmatrix}0\\ 0\end{bmatrix}
Explanation: This question tests understanding of how to represent a system of linear equations as a single matrix equation Ax=bAx = b. A system of linear equations can be written compactly as Ax=bAx = b, where AA is the coefficient matrix (containing all the coefficients of the variables), xx is the column vector of variables (the unknowns), and bb is the column vector of constants (the right-hand side values from each equation). For the system x+5y=−4x + 5y = -4, −2x+y=3-2x + y = 3, we extract the coefficients: equation 1 has 1 for x and 5 for y, giving first row [1 5][1\ 5], and equation 2 has -2 for x and 1 for y, giving second row [−2 1][-2\ 1]; the coefficient matrix is A=[15−21]A = \begin{bmatrix} 1 & 5 \\ -2 & 1 \end{bmatrix}, the variable vector is x=[xy]x = \begin{bmatrix} x \\ y \end{bmatrix}, and the constant vector is b=[−43]b = \begin{bmatrix} -4 \\ 3 \end{bmatrix}. Choice A is correct because it properly extracts coefficients into matrix A and correctly identifies variable and constant vectors. Choice B transposes the coefficient matrix, listing coefficients by column instead of by row, which reverses the structure (variables and equations get mixed up). Key to converting systems to matrix form: create the coefficient matrix A by listing coefficients in order (row i = equation i, column j = variable j), create variable vector x listing unknowns in the same order as columns, and create constant vector b listing the right-hand sides in the same order as equations. To verify your matrix equation is correct, expand Ax by multiplying: each row of A dotted with x should give the left side of the corresponding equation, and these should equal the entries of b (the right sides).

Question 19

Given the matrix equation

\begin{bmatrix}0 & 3 & -1\\ 2 & -1 & 0\\ -4 & 0 & 5\end{bmatrix}\begin{bmatrix}x\\ y\\ z\end{bmatrix}=\begin{bmatrix}2\\ -3\\ 1\end{bmatrix},$$ which system of equations does it represent?
  1. {3y−z=22x−y=−3−4x+5z=1\begin{cases} 3y-z=2\\ 2x-y=-3\\ -4x+5z=1 \end{cases} (correct answer)
  2. {3x−z=22x−y=−3−4x+5z=1\begin{cases} 3x-z=2\\ 2x-y=-3\\ -4x+5z=1 \end{cases}
  3. {3y−z=22x−y=3−4x+5z=1\begin{cases} 3y-z=2\\ 2x-y=3\\ -4x+5z=1 \end{cases}
  4. {3y−z=22x−y=−3−4x+5y=1\begin{cases} 3y-z=2\\ 2x-y=-3\\ -4x+5y=1 \end{cases}
Explanation: This question tests understanding of how to represent a system of linear equations as a single matrix equation Ax = b. The matrix equation Ax = b is equivalent to the system of equations: when you multiply the coefficient matrix A by the variable vector x using matrix-vector multiplication, each entry of the result equals the corresponding entry of the constant vector b, giving you the original equations. The matrix equation [[0 3 -1]; [2 -1 0]; [-4 0 5]] [x; y; z] = [2; -3; 1] expands by performing the matrix-vector multiplication: the first entry of Ax is 0x + 3y -1z = 2, giving equation 3y - z = 2; the second entry gives 2x -1y + 0z = -3, or 2x - y = -3; the third gives -4x + 0y + 5z = 1, or -4x + 5z = 1. Choice A is correct because it accurately represents the system as Ax = b with proper structure, including zeros for missing variables. Choice C has the wrong sign on the constant in equation 2, using +3 instead of -3. Remember to include 0 coefficients when a variable doesn't appear in an equation: for example, if the first row is [0 3 -1], it means no x term, so coefficient 0 for x. To verify your matrix equation is correct, expand Ax by multiplying: each row of A dotted with x should give the left side of the corresponding equation, and these should equal the entries of b (the right sides).

Question 20

For the system of equations

{−x+2y=65x−y=−3\begin{cases} -x + 2y = 6\\ 5x - y = -3 \end{cases}

write the system as a matrix equation Ax=bAx=b.

  1. [−125−1][xy]=[6−3]\begin{bmatrix}-1 & 2\\ 5 & -1\end{bmatrix}\begin{bmatrix}x\\ y\end{bmatrix}=\begin{bmatrix}6\\ -3\end{bmatrix} (correct answer)
  2. [−152−1][xy]=[6−3]\begin{bmatrix}-1 & 5\\ 2 & -1\end{bmatrix}\begin{bmatrix}x\\ y\end{bmatrix}=\begin{bmatrix}6\\ -3\end{bmatrix}
  3. [125−1][xy]=[6−3]\begin{bmatrix}1 & 2\\ 5 & -1\end{bmatrix}\begin{bmatrix}x\\ y\end{bmatrix}=\begin{bmatrix}6\\ -3\end{bmatrix}
  4. [−125−1][xy]=[−6−3]\begin{bmatrix}-1 & 2\\ 5 & -1\end{bmatrix}\begin{bmatrix}x\\ y\end{bmatrix}=\begin{bmatrix}-6\\ -3\end{bmatrix}
Explanation: This question tests understanding of how to represent a system of linear equations as a single matrix equation Ax=bAx = b. A system of linear equations can be written compactly as Ax=bAx = b, where A is the coefficient matrix (containing all the coefficients of the variables), x is the column vector of variables (the unknowns), and b is the column vector of constants (the right-hand side values from each equation). For the system −x+2y=6-x + 2y = 6, 5x−y=−35x - y = -3, we extract the coefficients: equation 1 has -1 for x and 2 for y, giving first row [−1 2][-1 \ 2], and equation 2 has 5 for x and -1 for y, giving second row [5 −1][5 \ -1]; the coefficient matrix is A=[−125−1]A = \begin{bmatrix} -1 & 2 \\ 5 & -1 \end{bmatrix}, the variable vector is x=[xy]x = \begin{bmatrix} x \\ y \end{bmatrix}, and the constant vector is b=[6−3]b = \begin{bmatrix} 6 \\ -3 \end{bmatrix}. Choice A is correct because it properly extracts coefficients into matrix A and correctly identifies variable and constant vectors. Choice C has the wrong sign on the coefficient of x in equation 1, using 1 instead of -1. Key to converting systems to matrix form: create the coefficient matrix A by listing coefficients in order (row i = equation i, column j = variable j), create variable vector x listing unknowns in the same order as columns, and create constant vector b listing the right-hand sides in the same order as equations. Remember to include the correct signs when extracting coefficients; for example, -y means coefficient -1 for y.