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ACT Math · Learn by Concept

ACT Math Help: Matrices

Review real example questions for Matrices in ACT Math.

Question 1 / 10

0 of 10 answered

Matrices AAA and BBB represent two coded messages combined by addition. What is A+BA+BA+B if A=[−134−2],B=[5−3−21]?A=\begin{bmatrix}-1 & 3\\ 4 & -2\end{bmatrix},\quad B=\begin{bmatrix}5 & -3\\ -2 & 1\end{bmatrix}?A=[−14​3−2​],B=[5−2​−31​]?

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Question 1

Matrices AAA and BBB represent two coded messages combined by addition. What is A+BA+BA+B if A=[−134−2],B=[5−3−21]?A=\begin{bmatrix}-1 & 3\\ 4 & -2\end{bmatrix},\quad B=\begin{bmatrix}5 & -3\\ -2 & 1\end{bmatrix}?A=[−14​3−2​],B=[5−2​−31​]?

  1. [−50−8−3]\begin{bmatrix}-5 & 0\\ -8 & -3\end{bmatrix}[−5−8​0−3​]
  2. [−66−83]\begin{bmatrix}-6 & 6\\ -8 & 3\end{bmatrix}[−6−8​63​]
  3. [462−1]\begin{bmatrix}4 & 6\\ 2 & -1\end{bmatrix}[42​6−1​]
  4. [402−1]\begin{bmatrix}4 & 0\\ 2 & -1\end{bmatrix}[42​0−1​] (correct answer)

Explanation: This problem involves adding matrices AAA and BBB to combine coded messages. Matrix addition is performed by adding corresponding entries. For the entries: (1,1) is −1+5=4-1 + 5 = 4−1+5=4; (1,2) is 3+(−3)=03 + (-3) = 03+(−3)=0; (2,1) is 4+(−2)=24 + (-2) = 24+(−2)=2; (2,2) is −2+1=−1-2 + 1 = -1−2+1=−1. The sum A+BA + BA+B is [402−1]\begin{bmatrix} 4 & 0 \\ 2 & -1 \end{bmatrix}[42​0−1​]. Choice C incorrectly adds 3+(−3)3 + (-3)3+(−3) as 6, likely an arithmetic mistake in signs.

Question 2

Which of the following matrices is equal to 3[2−140]3 \begin{bmatrix} 2 & -1 \\ 4 & 0 \end{bmatrix}3[24​−10​]?

  1. [5273]\begin{bmatrix} 5 & 2 \\ 7 & 3 \end{bmatrix}[57​23​]
  2. [6−3120]\begin{bmatrix} 6 & -3 \\ 12 & 0 \end{bmatrix}[612​−30​] (correct answer)
  3. [6−1120]\begin{bmatrix} 6 & -1 \\ 12 & 0 \end{bmatrix}[612​−10​]
  4. [5−370]\begin{bmatrix} 5 & -3 \\ 7 & 0 \end{bmatrix}[57​−30​]

Explanation: The correct answer is B. Scalar matrix multiplication multiplies every entry in the matrix by the scalar. 3 × 2 = 6, 3 × (−1) = −3, 3 × 4 = 12, 3 × 0 = 0. This gives [6, −3; 12, 0]. A ([5, 2; 7, 3]) results from adding 3 to each entry instead of multiplying. C ([6, −1; 12, 0]) correctly multiplies the first column but leaves the −1 entry unchanged — the student multiplied 3 by 2 and 4 but forgot to multiply 3 by −1. D ([5, −3; 7, 0]) adds 3 to the first column entries but correctly multiplies the second column. Pro tip: in scalar multiplication, every single entry gets multiplied — never leave any entry untouched.

Question 3

If A=(3−214)A = \begin{pmatrix} 3 & -2 \\ 1 & 4 \end{pmatrix}A=(31​−24​), what is −A-A−A?

  1. (−3−21−4)\begin{pmatrix} -3 & -2 \\ 1 & -4 \end{pmatrix}(−31​−2−4​)
  2. (3−2−14)\begin{pmatrix} 3 & -2 \\ -1 & 4 \end{pmatrix}(3−1​−24​)
  3. (−32−1−4)\begin{pmatrix} -3 & 2 \\ -1 & -4 \end{pmatrix}(−3−1​2−4​) (correct answer)
  4. (3214)\begin{pmatrix} 3 & 2 \\ 1 & 4 \end{pmatrix}(31​24​)

Explanation: This problem involves finding the negative of a matrix, where each entry is multiplied by -1. For matrix A = (3−214)\begin{pmatrix} 3 & -2 \\ 1 & 4 \end{pmatrix}(31​−24​), we compute -A by multiplying each entry by -1: (−1)(3)=−3(-1)(3) = -3(−1)(3)=−3, (−1)(−2)=2(-1)(-2) = 2(−1)(−2)=2, (−1)(1)=−1(-1)(1) = -1(−1)(1)=−1, (−1)(4)=−4(-1)(4) = -4(−1)(4)=−4. The result is (−32−1−4)\begin{pmatrix} -3 & 2 \\ -1 & -4 \end{pmatrix}(−3−1​2−4​). Choice C correctly shows this negation.

Question 4

If A=(01−24)A = \begin{pmatrix} 0 & 1 \\ -2 & 4 \end{pmatrix}A=(0−2​14​), what is −3A-3A−3A?

  1. (0−36−12)\begin{pmatrix} 0 & -3 \\ 6 & -12 \end{pmatrix}(06​−3−12​) (correct answer)
  2. (03−612)\begin{pmatrix} 0 & 3 \\ -6 & 12 \end{pmatrix}(0−6​312​)
  3. (0−3−612)\begin{pmatrix} 0 & -3 \\ -6 & 12 \end{pmatrix}(0−6​−312​)
  4. (01−24)\begin{pmatrix} 0 & 1 \\ -2 & 4 \end{pmatrix}(0−2​14​)

Explanation: This problem involves scalar multiplication of a matrix, where each entry is multiplied by the scalar -3. For matrix A = [[0,1],[-2,4]], we compute -3A by multiplying each entry: (-3)(0) = 0, (-3)(1) = -3, (-3)(-2) = 6, (-3)(4) = -12. The result is [[0,-3],[6,-12]]. Choice A correctly shows this scalar multiplication.

Question 5

Which of the following is the product ABABAB for matrices A=(012−1)A = \begin{pmatrix} 0 & 1 \\ 2 & -1 \end{pmatrix}A=(02​1−1​) and B=(3210)B = \begin{pmatrix} 3 & 2 \\ 1 & 0 \end{pmatrix}B=(31​20​)?

  1. (3064)\begin{pmatrix} 3 & 0 \\ 6 & 4 \end{pmatrix}(36​04​)
  2. (1254)\begin{pmatrix} 1 & 2 \\ 5 & 4 \end{pmatrix}(15​24​)
  3. (1052)\begin{pmatrix} 1 & 0 \\ 5 & 2 \end{pmatrix}(15​02​)
  4. (1054)\begin{pmatrix} 1 & 0 \\ 5 & 4 \end{pmatrix}(15​04​) (correct answer)

Explanation: This problem requires matrix multiplication AB, where entry (i,j) equals row i of A dotted with column j of B. For entry (1,1): (0)(3) + (1)(1) = 0 + 1 = 1; entry (1,2): (0)(2) + (1)(0) = 0 + 0 = 0; entry (2,1): (2)(3) + (-1)(1) = 6 - 1 = 5; entry (2,2): (2)(2) + (-1)(0) = 4 - 0 = 4. The result is [[1,0],[5,4]]. Choice D correctly shows this matrix multiplication.

Question 6

If A=(3−201)A = \begin{pmatrix} 3 & -2 \\ 0 & 1 \end{pmatrix}A=(30​−21​), what is 3A3A3A?

  1. (3−603)\begin{pmatrix} 3 & -6 \\ 0 & 3 \end{pmatrix}(30​−63​)
  2. (6−402)\begin{pmatrix} 6 & -4 \\ 0 & 2 \end{pmatrix}(60​−42​)
  3. (9−603)\begin{pmatrix} 9 & -6 \\ 0 & 3 \end{pmatrix}(90​−63​) (correct answer)
  4. (9−201)\begin{pmatrix} 9 & -2 \\ 0 & 1 \end{pmatrix}(90​−21​)

Explanation: This problem involves scalar multiplication of a matrix, where each entry is multiplied by the scalar 3. For matrix A = (3−201)\begin{pmatrix} 3 & -2 \\ 0 & 1 \end{pmatrix}(30​−21​), we compute 3A3A3A by multiplying each entry: 3(3)=93(3) = 93(3)=9, 3(−2)=−63(-2) = -63(−2)=−6, 3(0)=03(0) = 03(0)=0, 3(1)=33(1) = 33(1)=3. The result is (9−603)\begin{pmatrix} 9 & -6 \\ 0 & 3 \end{pmatrix}(90​−63​). Choice C correctly shows this scalar multiplication.

Question 7

What is the determinant of matrix (2−134)\begin{pmatrix} 2 & -1 \\ 3 & 4 \end{pmatrix}(23​−14​)?

  1. 11 (correct answer)
  2. -11
  3. 10
  4. -10

Explanation: This problem asks for the determinant of a 2×2 matrix. For matrix [[a,b],[c,d]], the determinant is ad - bc. Here we have [[2,-1],[3,4]], so the determinant is (2)(4) - (-1)(3) = 8 - (-3) = 8 + 3 = 11. The answer is 11.

Question 8

What is the determinant of matrix (3122)\begin{pmatrix} 3 & 1 \\ 2 & 2 \end{pmatrix}(32​12​)?

  1. 3
  2. 5
  3. 4 (correct answer)
  4. 6

Explanation: This problem asks for the determinant of a 2×2 matrix. For matrix [[a,b],[c,d]], the determinant is ad - bc. Here we have [[3,1],[2,2]], so the determinant is (3)(2) - (1)(2) = 6 - 2 = 4. The answer is 4.

Question 9

What is the determinant of matrix (4213)\begin{pmatrix} 4 & 2 \\ 1 & 3 \end{pmatrix}(41​23​)?

  1. 4
  2. 8
  3. 10 (correct answer)
  4. 14

Explanation: This problem asks for the determinant of a 2×2 matrix. For matrix [[a,b],[c,d]], the determinant is ad - bc. Here we have [[4,2],[1,3]], so the determinant is (4)(3) - (2)(1) = 12 - 2 = 10. The answer is 10.

Question 10

What is the determinant of the matrix [−3241]?\begin{bmatrix}-3 & 2\\ 4 & 1\end{bmatrix}?[−34​21​]?

  1. 111111
  2. 555
  3. −11-11−11 (correct answer)
  4. −5-5−5

Explanation: The operation is finding the determinant of a 2x2 matrix using the formula ad−bcad - bcad−bc. For the matrix [−3241]\begin{bmatrix} -3 & 2 \\ 4 & 1 \end{bmatrix}[−34​21​], it's (−3)(1)−(2)(4)=−3−8=−11(-3)(1) - (2)(4) = -3 - 8 = -11(−3)(1)−(2)(4)=−3−8=−11. This computation is essential for determining if the matrix is invertible. The result is -11. Choice D switches to bc−adbc - adbc−ad, which reverses the sign incorrectly.