Precalculus Quiz: Scalar Multiplication Of Matrices
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Scalar Multiplication Of MatricesQuestion 1 of 20

Let AA be the 2×32\times 3 matrix A=[2−1034−2].A=\begin{bmatrix}2 & -1 & 0\\ 3 & 4 & -2\end{bmatrix}. What is 3A3A?

[6−30912−6]\begin{bmatrix}6 & -3 & 0\\ 9 & 12 & -6\end{bmatrix}
[523671]\begin{bmatrix}5 & 2 & 3\\ 6 & 7 & 1\end{bmatrix}
[6−1034−2]\begin{bmatrix}6 & -1 & 0\\ 3 & 4 & -2\end{bmatrix}
[6−30912−2]\begin{bmatrix}6 & -3 & 0\\ 9 & 12 & -2\end{bmatrix}
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Precalculus Quiz

Precalculus Quiz: Scalar Multiplication Of Matrices

Practice Scalar Multiplication Of 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 Scalar Multiplication Of 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

Let AA be the 2×32\times 3 matrix A=[2−1034−2].A=\begin{bmatrix}2 & -1 & 0\\ 3 & 4 & -2\end{bmatrix}. What is 3A3A?

  1. [6−30912−6]\begin{bmatrix}6 & -3 & 0\\ 9 & 12 & -6\end{bmatrix} (correct answer)
  2. [523671]\begin{bmatrix}5 & 2 & 3\\ 6 & 7 & 1\end{bmatrix}
  3. [6−1034−2]\begin{bmatrix}6 & -1 & 0\\ 3 & 4 & -2\end{bmatrix}
  4. [6−30912−2]\begin{bmatrix}6 & -3 & 0\\ 9 & 12 & -2\end{bmatrix}
Explanation: This question tests understanding of scalar multiplication of matrices. Scalar multiplication of a matrix means multiplying every entry of the matrix by the scalar: if k is a scalar and A is a matrix, then (kA)ᵢⱼ = k·aᵢⱼ for every entry at row i, column j. For scalar k = 3 and matrix A = [[2, -1, 0], [3, 4, -2]], we multiply each entry by 3: the entry in row 1, column 1 becomes 3·2 = 6, the entry in row 1, column 2 becomes 3·(-1) = -3, the entry in row 1, column 3 becomes 3·0 = 0, the entry in row 2, column 1 becomes 3·3 = 9, the entry in row 2, column 2 becomes 3·4 = 12, and the entry in row 2, column 3 becomes 3·(-2) = -6, giving us the result matrix [[6, -3, 0], [9, 12, -6]]. Choice A is correct because it multiplies every entry by the scalar correctly: 3·2 = 6, 3·(-1) = -3, 3·0 = 0, 3·3 = 9, 3·4 = 12, and 3·(-2) = -6. Choice D makes an arithmetic error in computing the bottom-right entry, calculating -2 instead of -6. Key to scalar multiplication: remember that every single entry of the matrix gets multiplied by the scalar—there are no exceptions, and the dimensions stay exactly the same.

Question 2

Let A=[40−26].A=\begin{bmatrix}4 & 0\\ -2 & 6\end{bmatrix}. What is 12A\tfrac{1}{2}A?

  1. [20−13]\begin{bmatrix}2 & 0\\ -1 & 3\end{bmatrix} (correct answer)
  2. [20−23]\begin{bmatrix}2 & 0\\ -2 & 3\end{bmatrix}
  3. [180−14112]\begin{bmatrix}\tfrac{1}{8} & 0\\ -\tfrac{1}{4} & \tfrac{1}{12}\end{bmatrix}
  4. [4.50.5−1.56.5]\begin{bmatrix}4.5 & 0.5\\ -1.5 & 6.5\end{bmatrix}
Explanation: This question tests understanding of scalar multiplication of matrices. Scalar multiplication of a matrix means multiplying every entry of the matrix by the scalar: if k is a scalar and A is a matrix, then (kA)ᵢⱼ = k·aᵢⱼ for every entry at row i, column j. For scalar k = ½ and matrix A = [[4, 0], [-2, 6]], we multiply each entry by ½: the entry in row 1, column 1 becomes ½·4 = 2, the entry in row 1, column 2 becomes ½·0 = 0, the entry in row 2, column 1 becomes ½·(-2) = -1, and the entry in row 2, column 2 becomes ½·6 = 3, giving us the result matrix [[2, 0], [-1, 3]]. Choice A is correct because it multiplies every entry by the scalar correctly: ½·4 = 2, ½·0 = 0, ½·(-2) = -1, and ½·6 = 3. Choice C treats scalar multiplication as exponentiation, raising entries to the power k instead of multiplying them by k, resulting in fractional values that don't match the correct computation. Key to scalar multiplication: remember that every single entry of the matrix gets multiplied by the scalar—there are no exceptions, and the dimensions stay exactly the same. When multiplying by a fraction like ½, you're dividing each entry by 2.

Question 3

For the matrix A=[2−103]A=\begin{bmatrix}2 & -1\\0 & 3\end{bmatrix} (a 2×22\times 2 matrix), what is −2A-2A?​​

  1. [−420−6]\begin{bmatrix}-4 & 2\\0 & -6\end{bmatrix} (correct answer)
  2. [0−3−21]\begin{bmatrix}0 & -3\\-2 & 1\end{bmatrix}
  3. [2−3−23]\begin{bmatrix}2 & -3\\-2 & 3\end{bmatrix}
  4. [4−206]\begin{bmatrix}4 & -2\\0 & 6\end{bmatrix}
Explanation: This question tests understanding of scalar multiplication of matrices. Scalar multiplication of a matrix means multiplying every entry of the matrix by the scalar: if k is a scalar and A is a matrix, then (kA)ᵢⱼ = k·aᵢⱼ for every entry at row i, column j. For scalar k = -2 and matrix A = [[2, -1], [0, 3]], we multiply each entry by -2: the entry in row 1, column 1 becomes (-2)·2 = -4, the entry in row 1, column 2 becomes (-2)·(-1) = 2, the entry in row 2, column 1 becomes (-2)·0 = 0, and the entry in row 2, column 2 becomes (-2)·3 = -6, giving us the result matrix [[-4, 2], [0, -6]]. Choice A is correct because it multiplies every entry by the scalar -2 correctly, accounting for the sign changes. Choice D has correct magnitudes but wrong signs, failing to account for the negative scalar reversing signs. Key to scalar multiplication: remember that every single entry of the matrix gets multiplied by the scalar—there are no exceptions, and the dimensions stay exactly the same. Special scalars to remember: multiplying by 1 leaves the matrix unchanged, multiplying by 0 gives the zero matrix (all entries become 0), and multiplying by -1 negates every entry (flips all signs).

Question 4

Let A=[2−103]A=\begin{bmatrix}2 & -1\\0 & 3\end{bmatrix} (a 2×22\times2 matrix). What is −2A-2A?

  1. [−420−6]\begin{bmatrix}-4 & 2\\0 & -6\end{bmatrix} (correct answer)
  2. [0−3−21]\begin{bmatrix}0 & -3\\-2 & 1\end{bmatrix}
  3. [0−3−25]\begin{bmatrix}0 & -3\\-2 & 5\end{bmatrix}
  4. [−4−3−2−6]\begin{bmatrix}-4 & -3\\-2 & -6\end{bmatrix}
Explanation: This question tests understanding of scalar multiplication of matrices. Scalar multiplication of a matrix means multiplying every entry of the matrix by the scalar: if k is a scalar and A is a matrix, then (kA)ᵢⱼ = k·aᵢⱼ for every entry at row i, column j. For scalar k = -2 and matrix A = [[2, -1], [0, 3]], we multiply each entry by -2: the entry in row 1, column 1 becomes -2·2 = -4, row 1 column 2 -2·(-1)=2, row 2 column 1 -2·0=0, row 2 column 2 -2·3=-6, giving us the result matrix [[-4, 2], [0, -6]]. Choice A is correct because it multiplies every entry by the scalar correctly. Choice B makes an arithmetic error in computing specific entries, calculating wrong values like 0 for top left instead of -4. Key to scalar multiplication: remember that every single entry of the matrix gets multiplied by the scalar—there are no exceptions, and the dimensions stay exactly the same. The effect of a negative scalar is to scale and reverse the signs of all entries.

Question 5

For the 2×22\times 2 matrix A=[3−210],A=\begin{bmatrix}3 & -2\\ 1 & 0\end{bmatrix}, which matrix represents 0A0A?

  1. [3−210]\begin{bmatrix}3 & -2\\ 1 & 0\end{bmatrix}
  2. [0000]\begin{bmatrix}0 & 0\\ 0 & 0\end{bmatrix} (correct answer)
  3. [0−210]\begin{bmatrix}0 & -2\\ 1 & 0\end{bmatrix}
  4. [3000]\begin{bmatrix}3 & 0\\ 0 & 0\end{bmatrix}
Explanation: This question tests understanding of scalar multiplication of matrices. The effect of scalar multiplication depends on the value of the scalar: positive scalars scale all entries proportionally, negative scalars also reverse signs, and special values like 0 or 1 have unique effects (0 gives the zero matrix, 1 leaves the matrix unchanged). Multiplying by k = 0 has the specific effect of producing the zero matrix: for A = [[3, -2], [1, 0]], 0A = [[0, 0], [0, 0]]. Choice B is correct because it multiplies every entry by the scalar correctly. Choice A omits multiplying one or more entries, leaving them unchanged when all entries must be affected. Special scalars to remember: multiplying by 1 leaves the matrix unchanged, multiplying by 0 gives the zero matrix (all entries become 0), and multiplying by -1 negates every entry (flips all signs). Key to scalar multiplication: remember that every single entry of the matrix gets multiplied by the scalar—there are no exceptions, and the dimensions stay exactly the same.

Question 6

Let A=[1−2034−1]A=\begin{bmatrix}1 & -2 & 0\\3 & 4 & -1\end{bmatrix} (a 2×32\times3 matrix). What are the dimensions of 5A5A?

  1. 2×32\times3 (correct answer)
  2. 5×55\times5
  3. 10×1510\times15
  4. 3×23\times2
Explanation: This question tests understanding of scalar multiplication of matrices. When a matrix is multiplied by a scalar k, each entry in the matrix is multiplied by k, and the dimensions of the matrix remain unchanged—only the entry values change. Since A has dimensions 2×3, and scalar multiplication only changes entry values (not the number of rows or columns), the product 5A also has dimensions 2×3. Choice A is correct because it correctly states dimensional invariance. Choice C incorrectly claims the dimensions change, suggesting a 2×3 matrix becomes 10×15 or something similar, when scalar multiplication never changes dimensions. Key to scalar multiplication: remember that every single entry of the matrix gets multiplied by the scalar—there are no exceptions, and the dimensions stay exactly the same. To check your work, verify that your result matrix has the same dimensions as the original.

Question 7

For the 2×22\times 2 matrix A=[1−430]A=\begin{bmatrix}1 & -4\\3 & 0\end{bmatrix}, which matrix represents 2A2A?​​

  1. [2−830]\begin{bmatrix}2 & -8\\3 & 0\end{bmatrix}
  2. [3−252]\begin{bmatrix}3 & -2\\5 & 2\end{bmatrix}
  3. [1−2tfrac320]\begin{bmatrix}1 & -2\\tfrac{3}{2} & 0\end{bmatrix}
  4. [2−860]\begin{bmatrix}2 & -8\\6 & 0\end{bmatrix} (correct answer)
Explanation: This question tests understanding of scalar multiplication of matrices. Scalar multiplication of a matrix means multiplying every entry of the matrix by the scalar: if k is a scalar and A is a matrix, then (kA)ᵢⱼ = k·aᵢⱼ for every entry at row i, column j. For scalar k = 2 and matrix A = [[1, -4], [3, 0]], we multiply each entry by 2: the entry in row 1, column 1 becomes 2·1 = 2, the entry in row 1, column 2 becomes 2·(-4) = -8, the entry in row 2, column 1 becomes 2·3 = 6, and the entry in row 2, column 2 becomes 2·0 = 0, giving us the result matrix [[2, -8], [6, 0]]. Choice D is correct because it multiplies every entry by the scalar 2 correctly. Choice A makes an arithmetic error in computing the entry at row 2, column 1, calculating 3 instead of 6. Key to scalar multiplication: remember that every single entry of the matrix gets multiplied by the scalar—there are no exceptions, and the dimensions stay exactly the same. To check your work, verify that every entry in your result is exactly the scalar times the corresponding entry in the original matrix, and that your result matrix has the same dimensions as the original.

Question 8

Let A=[1−234]A=\begin{bmatrix}1 & -2\\3 & 4\end{bmatrix} and B=[05−12]B=\begin{bmatrix}0 & 5\\-1 & 2\end{bmatrix}. What is 3(A+B)3(A+B)?

  1. [39618]\begin{bmatrix}3 & 9\\6 & 18\end{bmatrix} (correct answer)
  2. [1326]\begin{bmatrix}1 & 3\\2 & 6\end{bmatrix}
  3. [3366]\begin{bmatrix}3 & 3\\6 & 6\end{bmatrix}
  4. [315618]\begin{bmatrix}3 & 15\\6 & 18\end{bmatrix}
Explanation: This question tests understanding of scalar multiplication of matrices. Scalar multiplication distributes over matrix addition: k(A + B) = kA + kB, meaning you can multiply the scalar by each matrix separately and then add the results, or add the matrices first and then multiply by the scalar. The distributive property k(A + B) = kA + kB holds because scalar multiplication affects each entry independently: showing that k(aᵢⱼ + bᵢⱼ) = kaᵢⱼ + kbᵢⱼ for any entry position. Choice A is correct because it properly applies the distributive property with specific values. Choice C applies the distributive property incorrectly, claiming k(A + B) ≠ kA + kB, which contradicts the fundamental property of scalar multiplication. When combining scalar multiplication with matrix addition (like 2A + 3B), always perform the scalar multiplications first (multiply every entry of A by 2, every entry of B by 3), then add the resulting matrices entry-by-entry. Key to scalar multiplication: remember that every single entry of the matrix gets multiplied by the scalar—there are no exceptions, and the dimensions stay exactly the same.

Question 9

Given A=[10−23]A=\begin{bmatrix}1 & 0\\-2 & 3\end{bmatrix} and B=[2−140]B=\begin{bmatrix}2 & -1\\4 & 0\end{bmatrix}, calculate 2A−3B2A-3B.​

  1. [−43−166]\begin{bmatrix}-4 & 3\\-16 & 6\end{bmatrix} (correct answer)
  2. [−4−3−166]\begin{bmatrix}-4 & -3\\-16 & 6\end{bmatrix}
  3. [8−316−6]\begin{bmatrix}8 & -3\\16 & -6\end{bmatrix}
  4. [−11−83]\begin{bmatrix}-1 & 1\\-8 & 3\end{bmatrix}
Explanation: This question tests understanding of scalar multiplication of matrices. Scalar multiplication distributes over matrix addition: k(A + B) = kA + kB, meaning you can multiply the scalar by each matrix separately and then add the results, or add the matrices first and then multiply by the scalar. To compute 2A - 3B, we first perform scalar multiplication: 2A = [[2, 0], [-4, 6]] and 3B = [[6, -3], [12, 0]], then subtract corresponding entries to get [[-4, 3], [-16, 6]]. Choice A is correct because it properly applies the distributive property with specific values. Choice C has correct magnitudes but wrong signs, failing to account for negative scalar reversing signs in the subtraction. When combining scalar multiplication with matrix addition (like 2A + 3B), always perform the scalar multiplications first (multiply every entry of A by 2, every entry of B by 3), then add the resulting matrices entry-by-entry. Key to scalar multiplication: remember that every single entry of the matrix gets multiplied by the scalar—there are no exceptions, and the dimensions stay exactly the same.

Question 10

For A=[0−3251−1]A=\begin{bmatrix}0 & -3 & 2\\5 & 1 & -1\end{bmatrix} (a 2×32\times3 matrix), what are the dimensions of −4A-4A?

  1. 2×32\times3 (correct answer)
  2. 3×23\times2
  3. 8×128\times12
  4. 6×16\times1
Explanation: This question tests understanding of scalar multiplication of matrices. When a matrix is multiplied by a scalar k, each entry in the matrix is multiplied by k, and the dimensions of the matrix remain unchanged—only the entry values change. Since A has dimensions 2×3, and scalar multiplication only changes entry values (not the number of rows or columns), the product -4A also has dimensions 2×3. Choice A is correct because it correctly states dimensional invariance. Choice C incorrectly claims the dimensions change, suggesting a 2×3 matrix becomes 8×12 or something similar, when scalar multiplication never changes dimensions. Key to scalar multiplication: remember that every single entry of the matrix gets multiplied by the scalar—there are no exceptions, and the dimensions stay exactly the same. To check your work, verify that every entry in your result is exactly the scalar times the corresponding entry in the original matrix, and that your result matrix has the same dimensions as the original.

Question 11

Let A=[1−234]A=\begin{bmatrix}1 & -2\\3 & 4\end{bmatrix} and B=[05−12]B=\begin{bmatrix}0 & 5\\-1 & 2\end{bmatrix}. What is 3(A+B)3(A+B)?​

  1. [39618]\begin{bmatrix}3 & 9\\6 & 18\end{bmatrix} (correct answer)
  2. [1326]\begin{bmatrix}1 & 3\\2 & 6\end{bmatrix}
  3. [3366]\begin{bmatrix}3 & 3\\6 & 6\end{bmatrix}
  4. [315618]\begin{bmatrix}3 & 15\\6 & 18\end{bmatrix}
Explanation: This question tests understanding of scalar multiplication of matrices. Scalar multiplication distributes over matrix addition: k(A + B) = kA + kB, meaning you can multiply the scalar by each matrix separately and then add the results, or add the matrices first and then multiply by the scalar. The distributive property k(A + B) = kA + kB holds because scalar multiplication affects each entry independently: showing that k(aᵢⱼ + bᵢⱼ) = kaᵢⱼ + kbᵢⱼ for any entry position. Choice A is correct because it properly applies the distributive property with specific values. Choice C applies the distributive property incorrectly, claiming k(A + B) ≠ kA + kB, which contradicts the fundamental property of scalar multiplication. When combining scalar multiplication with matrix addition (like 2A + 3B), always perform the scalar multiplications first (multiply every entry of A by 2, every entry of B by 3), then add the resulting matrices entry-by-entry. Key to scalar multiplication: remember that every single entry of the matrix gets multiplied by the scalar—there are no exceptions, and the dimensions stay exactly the same.

Question 12

Given the scalar k=−2k=-2 and the 2×22\times 2 matrix M=[1−304],M=\begin{bmatrix}1 & -3\\ 0 & 4\end{bmatrix}, what is kMkM?

  1. [−130−4]\begin{bmatrix}-1 & 3\\ 0 & -4\end{bmatrix}
  2. [−260−8]\begin{bmatrix}-2 & 6\\ 0 & -8\end{bmatrix} (correct answer)
  3. [−1−60−4]\begin{bmatrix}-1 & -6\\ 0 & -4\end{bmatrix}
  4. [3−5−22]\begin{bmatrix}3 & -5\\ -2 & 2\end{bmatrix}
Explanation: This question tests understanding of scalar multiplication of matrices. When a matrix is multiplied by a scalar k, each entry in the matrix is multiplied by k, and the dimensions of the matrix remain unchanged—only the entry values change. For scalar k = -2 and matrix M = [[1, -3], [0, 4]], we multiply each entry by -2: the entry in row 1, column 1 becomes (-2)·1 = -2, the entry in row 1, column 2 becomes (-2)·(-3) = 6, the entry in row 2, column 1 becomes (-2)·0 = 0, and the entry in row 2, column 2 becomes (-2)·4 = -8, giving us the result matrix [[-2, 6], [0, -8]]. Choice B is correct because it multiplies every entry by the scalar correctly, accounting for the negative scalar reversing signs where appropriate. Choice A only multiplies some entries (such as the diagonal) instead of multiplying every single entry in the matrix by the scalar. Special scalars to remember: multiplying by 1 leaves the matrix unchanged, multiplying by 0 gives the zero matrix (all entries become 0), and multiplying by -1 negates every entry (flips all signs). When multiplying by a negative scalar like -2, remember that positive entries become negative and negative entries become positive.

Question 13

Matrix R=[x2x−x3x]R = \begin{bmatrix} x & 2x \\ -x & 3x \end{bmatrix} represents a scaled version of matrix S=[12−13]S = \begin{bmatrix} 1 & 2 \\ -1 & 3 \end{bmatrix} . If the determinant of RR equals 20, what is the value of x2x^2?

  1. 55
  2. 2020
  3. 44 (correct answer)
  4. 2525
Explanation: When you see a matrix described as a "scaled version" of another matrix, you're dealing with scalar multiplication - each element of the original matrix is multiplied by the same constant. Here, matrix RR is matrix SS scaled by factor xx. To find xx, you'll need to calculate the determinant of RR. For a 2×2 matrix [abcd]\begin{bmatrix} a & b \\ c & d \end{bmatrix}, the determinant is ad−bcad - bc. For matrix $$R = \begin{bmatrix} x & 2x \ -x & 3x \end{bmatrix} $$\det(R) = (x)(3x) - (2x)(-x) = 3x^2 + 2x^2 = 5x^2$$ Since the determinant equals 20: $$5x^2 = 20$$ $$x^2 = 4$$ Let's examine the wrong answers. Choice A ($$5$$) is the coefficient from our determinant calculation - you might pick this if you confused the coefficient with the final answer. Choice B ($$20$$) is simply the given determinant value, which you'd choose if you mistakenly thought $$x^2$$ equaled the determinant directly. Choice D ($$25$$) could result from calculation errors, perhaps incorrectly computing the determinant as $$x^2$$ instead of $$5x^2$$. The correct answer is C ($$4$$). **Study tip:** Remember that when a matrix is scaled by factor $$k$$, its determinant is scaled by $$k^2$$ (for 2×2 matrices). You can verify this: $$\det(S) = 1(3) - 2(-1) = 5$$, and $$\det(R) = x^2 \cdot \det(S) = 5x^2$$, confirming our approach.

Question 14

For A=[−3201−45]A=\begin{bmatrix}-3 & 2 & 0\\1 & -4 & 5\end{bmatrix}, which matrix represents 0⋅A0\cdot A?

  1. [−3201−45]\begin{bmatrix}-3 & 2 & 0\\1 & -4 & 5\end{bmatrix}
  2. [000000]\begin{bmatrix}0 & 0 & 0\\0 & 0 & 0\end{bmatrix} (correct answer)
  3. [320145]\begin{bmatrix}3 & 2 & 0\\1 & 4 & 5\end{bmatrix}
  4. [−320]\begin{bmatrix}-3 & 2 & 0\end{bmatrix}
Explanation: This question tests understanding of scalar multiplication of matrices. The effect of scalar multiplication depends on the value of the scalar: positive scalars scale all entries proportionally, negative scalars also reverse signs, and special values like 0 or 1 have unique effects (0 gives the zero matrix, 1 leaves the matrix unchanged). Multiplying by k = 0 has the specific effect of producing the zero matrix: for A = [[-3, 2, 0], [1, -4, 5]], we get [[0, 0, 0], [0, 0, 0]]. Choice B is correct because it multiplies every entry by the scalar correctly. Choice A omits multiplying one or more entries, leaving them unchanged when all entries must be affected. Special scalars to remember: multiplying by 1 leaves the matrix unchanged, multiplying by 0 gives the zero matrix (all entries become 0), and multiplying by -1 negates every entry (flips all signs). To check your work, verify that every entry in your result is exactly the scalar times the corresponding entry in the original matrix, and that your result matrix has the same dimensions as the original.

Question 15

Let AA and BB be 2×22\times 2 matrices A=[10−23],B=[4−120].A=\begin{bmatrix}1 & 0\\ -2 & 3\end{bmatrix},\quad B=\begin{bmatrix}4 & -1\\ 2 & 0\end{bmatrix}. Calculate 2A+B2A+B.

  1. [7−1−26]\begin{bmatrix}7 & -1\\ -2 & 6\end{bmatrix}
  2. [6−206]\begin{bmatrix}6 & -2\\ 0 & 6\end{bmatrix}
  3. [6−1−26]\begin{bmatrix}6 & -1\\ -2 & 6\end{bmatrix} (correct answer)
  4. [5−1−23]\begin{bmatrix}5 & -1\\ -2 & 3\end{bmatrix}
Explanation: This question tests understanding of scalar multiplication of matrices. Scalar multiplication distributes over matrix addition: k(A + B) = kA + kB, meaning you can multiply the scalar by each matrix separately and then add the results, or add the matrices first and then multiply by the scalar. To compute 2A + B, we first perform scalar multiplication: 2A = [[2, 0], [-4, 6]] and B = [[4, -1], [2, 0]], then add corresponding entries to get [[6, -1], [-2, 6]]. Choice C is correct because it properly applies the distributive property with specific values. Choice D makes an arithmetic error in computing specific entry, calculating -2 + 2 = -2 instead of 0 for row 2 column 1. When combining scalar multiplication with matrix addition (like 2A + 3B), always perform the scalar multiplications first (multiply every entry of A by 2, every entry of B by 3), then add the resulting matrices entry-by-entry. Don't confuse scalar multiplication (multiply each entry by a number) with matrix multiplication (row-column dot products)—they are completely different operations with different rules and different results.

Question 16

A research team collects data in matrix $$D = \begin{bmatrix} 0.6 & -0.3 \ 0.9 & 0.4 \end{bmatrix}

  1. 140140
  2. 1.761.76
  3. 176176 (correct answer)
  4. 17.617.6
Explanation: When you encounter matrix problems involving real-world data transformations, break down each operation step by step. This question tests scalar multiplication and absolute value calculations. First, convert the data to percentages by multiplying matrix DD by 100: Next, apply the confidence factor of 0.8: Now find the absolute values: ∣48∣=48|48| = 48, ∣−24∣=24|-24| = 24, ∣72∣=72|72| = 72, ∣32∣=32|32| = 32 Sum these: 48+24+72+32=17648 + 24 + 72 + 32 = 176 Answer C (176) is correct. Answer A (140) likely comes from forgetting the confidence factor—if you only converted to percentages, you'd get 60+30+90+40=22060 + 30 + 90 + 40 = 220, but this doesn't match exactly, suggesting a calculation error along the way. Answer B (1.76) represents the sum before converting to percentages: ∣0.6∣+∣−0.3∣+∣0.9∣+∣0.4∣=2.2|0.6| + |-0.3| + |0.9| + |0.4| = 2.2, then incorrectly applying 0.8 as 2.2×0.8=1.762.2 \times 0.8 = 1.76. Answer D (17.6) occurs if you correctly find 176 but then divide by 10, possibly misunderstanding the percentage conversion step. Strategy tip: In multi-step matrix problems, perform operations in the correct sequence and double-check each transformation. Write out intermediate matrices to avoid losing track of decimal places and signs.

Question 17

A manufacturing company uses matrix $$P = \begin{bmatrix} 4 & 0 & -2 \ 1 & 3 & 5 \end{bmatrix}

  1. The constraint requires additional analysis since 0.4(11)=4.40.4(11) = 4.4 but matrix structure matters
  2. The constraint cannot be satisfied since 0.4(9)=3.6>00.4(9) = 3.6 > 0 but losses exceed gains
  3. The constraint is already satisfied since 0.4(13)=5.2>00.4(13) = 5.2 > 0
  4. The constraint is already satisfied since 0.4(11)=4.4>00.4(11) = 4.4 > 0 (correct answer)
Explanation: When you encounter matrix scaling problems, the key insight is that scalar multiplication affects every entry uniformly, and the sum of all entries scales by the same factor. First, let's find the sum of all entries in matrix PP: 4+0+(−2)+1+3+5=114 + 0 + (-2) + 1 + 3 + 5 = 11. When all values are reduced to 40% of their current values, we multiply the entire matrix by 0.4. Since scalar multiplication distributes over addition, the sum of entries in the new matrix will be 0.4×11=4.40.4 \times 11 = 4.4, which is positive. Looking at the wrong answers: Choice A suggests the calculation 0.4(11)=4.40.4(11) = 4.4 is correct but claims "matrix structure matters" - this is incorrect because when finding the sum of all entries, individual positions don't affect the total. Choice B contains a calculation error, using 9 instead of 11 as the sum, and makes an unsupported claim about "losses exceeding gains." Choice C also has a calculation error, using 13 instead of 11 as the original sum. Choice D correctly identifies that 0.4(11)=4.4>00.4(11) = 4.4 > 0, meaning the constraint is already satisfied. Study tip: When dealing with scalar multiplication of matrices, remember that it affects all entries equally. If you need the sum of entries, you can either multiply each entry individually then add, or find the original sum first and then multiply - both give the same result. Always double-check your arithmetic when adding matrix entries.

Question 18

Two matrices A=[2−143]A = \begin{bmatrix} 2 & -1 \\ 4 & 3 \end{bmatrix} and B=[−63−12−9]B = \begin{bmatrix} -6 & 3 \\ -12 & -9 \end{bmatrix} are related by scalar multiplication. If matrix C=[15−20]C = \begin{bmatrix} 1 & 5 \\ -2 & 0 \end{bmatrix} undergoes the same scalar transformation that relates AA to BB, what is the trace of the resulting matrix?

  1. 33
  2. −3-3 (correct answer)
  3. −18-18
  4. −6-6
Explanation: When you encounter problems about scalar multiplication of matrices, you need to identify the scalar that transforms one matrix into another, then apply that same scalar to a new matrix. To find the scalar relating matrices AA and BB, examine corresponding elements. Looking at the first element: 2k=−62k = -6, so k=−3k = -3. You can verify this works for all elements: −3×2=−6-3 \times 2 = -6, −3×(−1)=3-3 \times (-1) = 3, −3×4=−12-3 \times 4 = -12, and −3×3=−9-3 \times 3 = -9. Indeed, B=−3AB = -3A. Now apply this same scalar multiplication to matrix CC: The trace of a matrix is the sum of its diagonal elements. For the resulting matrix, the trace is −3+0=−3-3 + 0 = -3. Looking at the wrong answers: Choice A (33) likely comes from finding the positive trace of the original matrix CC (1+0=11 + 0 = 1) or making a sign error. Choice C (−18-18) might result from multiplying the scalar by the sum of all elements in CC rather than just the diagonal elements. Choice D (−6-6) could come from doubling the correct trace or confusing the scalar value with the final answer. Remember: scalar multiplication affects every element of a matrix by the same factor, and the trace is always the sum of diagonal elements only. When working with matrix transformations, verify your scalar with multiple elements before proceeding.

Question 19

A financial model uses matrix $$F = \begin{bmatrix} 250 & -180 \ 320 & -95 \end{bmatrix}

  1. 1.21.12=1514\frac{1.2}{1.12} = \frac{15}{14}
  2. 1.121.2=1415\frac{1.12}{1.2} = \frac{14}{15} (correct answer)
  3. 1.12−1.2=−0.081.12 - 1.2 = -0.08
  4. 1.121.2=0.9333\frac{1.12}{1.2} = 0.9333
Explanation: When you encounter matrix scaling problems involving corrections, think about what operations were intended versus what actually happened, then determine how to bridge the gap. The intended operation was to multiply matrix FF by 1.121.12 (a 12% increase). Instead, the matrix was multiplied by 1.21.2. To find the correction factor, you need to determine what additional scalar will transform the incorrect result into the correct one. Let's call the correction factor kk. The relationship is: k×(1.2×F)=1.12×Fk \times (1.2 \times F) = 1.12 \times F Dividing both sides by FF: k×1.2=1.12k \times 1.2 = 1.12 Solving for kk: k=1.121.2=1415k = \frac{1.12}{1.2} = \frac{14}{15} This means multiplying the incorrect result by 1415\frac{14}{15} will give you the intended outcome. Answer choice A reverses the fraction, giving 1.21.12\frac{1.2}{1.12}, which would amplify the error rather than correct it. Choice C uses subtraction (1.12−1.21.12 - 1.2), but you need a multiplicative correction factor, not an additive one. Choice D gives the decimal equivalent 0.93330.9333, but the exact fraction form in choice B is more precise and mathematically complete. The correct answer is B: 1.121.2=1415\frac{1.12}{1.2} = \frac{14}{15}. Strategy tip: In correction problems, always set up the equation "correction factor × incorrect result = intended result" and solve for the correction factor. Watch out for reversed fractions—they're common trap answers.

Question 20

Matrix $$A = \begin{bmatrix} -3 & 6 \ 2 & -8 \end{bmatrix}

  1. Multiply by −1.5-1.5; entry is −3-3 (correct answer)
  2. Multiply by −0.5-0.5; entry is −1-1
  3. Multiply by 1.51.5; entry is 33
  4. Multiply by −1.5-1.5; entry is 33
Explanation: To reverse all changes, we multiply by -1. Then to apply a 50% buffer to the reversed changes, we multiply by 1.5. Combined: multiply by (-1)(1.5) = -1.5. The entry at position (2,1) is 2, so after scalar multiplication: 2(-1.5) = -3.