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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.

Question 1 / 20

0 of 20 answered

Matrix R=[x2x−x3x]R = \begin{bmatrix} x & 2x \\ -x & 3x \end{bmatrix}R=[x−x​2x3x​] represents a scaled version of matrix S=[12−13]S = \begin{bmatrix} 1 & 2 \\ -1 & 3 \end{bmatrix}S=[1−1​23​]. If the determinant of RRR equals 20, what is the value of x2x^2x2?

Select an answer to continue

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.

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

Matrix R=[x2x−x3x]R = \begin{bmatrix} x & 2x \\ -x & 3x \end{bmatrix}R=[x−x​2x3x​] represents a scaled version of matrix S=[12−13]S = \begin{bmatrix} 1 & 2 \\ -1 & 3 \end{bmatrix}S=[1−1​23​]. If the determinant of RRR equals 20, what is the value of x2x^2x2?

  1. 555
  2. 202020
  3. 444 (correct answer)
  4. 252525

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 RRR is matrix SSS scaled by factor xxx. To find xxx, you'll need to calculate the determinant of RRR. For a 2×2 matrix [abcd]\begin{bmatrix} a & b \\ c & d \end{bmatrix}[ac​bd​], the determinant is ad−bcad - bcad−bc. For matrix R=[x2x−x3x]R = \begin{bmatrix} x & 2x \\ -x & 3x \end{bmatrix}R=[x−x​2x3x​]: det⁡(R)=(x)(3x)−(2x)(−x)=3x2+2x2=5x2\det(R) = (x)(3x) - (2x)(-x) = 3x^2 + 2x^2 = 5x^2det(R)=(x)(3x)−(2x)(−x)=3x2+2x2=5x2 Since the determinant equals 20: 5x2=205x^2 = 205x2=20 x2=4x^2 = 4x2=4 Let's examine the wrong answers. Choice A (555) is the coefficient from our determinant calculation - you might pick this if you confused the coefficient with the final answer. Choice B (202020) is simply the given determinant value, which you'd choose if you mistakenly thought x2x^2x2 equaled the determinant directly. Choice D (252525) could result from calculation errors, perhaps incorrectly computing the determinant as x2x^2x2 instead of 5x25x^25x2. The correct answer is C (444). Study tip: Remember that when a matrix is scaled by factor kkk, its determinant is scaled by k2k^2k2 (for 2×2 matrices). You can verify this: det⁡(S)=1(3)−2(−1)=5\det(S) = 1(3) - 2(-1) = 5det(S)=1(3)−2(−1)=5, and det⁡(R)=x2⋅det⁡(S)=5x2\det(R) = x^2 \cdot \det(S) = 5x^2det(R)=x2⋅det(S)=5x2, confirming our approach.

Question 2

Let A=[2−103]A=\begin{bmatrix}2 & -1\\0 & 3\end{bmatrix}A=[20​−13​] (a 2×22\times22×2 matrix). What is −2A-2A−2A?

  1. [−420−6]\begin{bmatrix}-4 & 2\\0 & -6\end{bmatrix}[−40​2−6​] (correct answer)
  2. [0−3−21]\begin{bmatrix}0 & -3\\-2 & 1\end{bmatrix}[0−2​−31​]
  3. [0−3−25]\begin{bmatrix}0 & -3\\-2 & 5\end{bmatrix}[0−2​−35​]
  4. [−4−3−2−6]\begin{bmatrix}-4 & -3\\-2 & -6\end{bmatrix}[−4−2​−3−6​]

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 3

Given A=[10−23]A=\begin{bmatrix}1 & 0\\-2 & 3\end{bmatrix}A=[1−2​03​] and B=[2−140]B=\begin{bmatrix}2 & -1\\4 & 0\end{bmatrix}B=[24​−10​], calculate 2A−3B2A-3B2A−3B.

  1. [−43−166]\begin{bmatrix}-4 & 3\\-16 & 6\end{bmatrix}[−4−16​36​] (correct answer)
  2. [−4−3−166]\begin{bmatrix}-4 & -3\\-16 & 6\end{bmatrix}[−4−16​−36​]
  3. [8−316−6]\begin{bmatrix}8 & -3\\16 & -6\end{bmatrix}[816​−3−6​]
  4. [−11−83]\begin{bmatrix}-1 & 1\\-8 & 3\end{bmatrix}[−1−8​13​]

Explanation: This question tests understanding of scalar multiplication of matrices. Scalar multiplication distributes over matrix addition: k(A+B)=kA+kBk(A + B) = kA + kBk(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−3B2A - 3B2A−3B, we first perform scalar multiplication: 2A=[20−46]2A = \begin{bmatrix} 2 & 0 \\-4 & 6 \end{bmatrix}2A=[2−4​06​] and 3B=[6−3120]3B = \begin{bmatrix} 6 & -3 \\12 & 0 \end{bmatrix}3B=[612​−30​], then subtract corresponding entries to get [−43−166]\begin{bmatrix} -4 & 3 \\-16 & 6 \end{bmatrix}[−4−16​36​]. 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+3B2A + 3B2A+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 4

For A=[−3201−45]A=\begin{bmatrix}-3 & 2 & 0\\1 & -4 & 5\end{bmatrix}A=[−31​2−4​05​], which matrix represents 0⋅A0\cdot A0⋅A?

  1. [−3201−45]\begin{bmatrix}-3 & 2 & 0\\1 & -4 & 5\end{bmatrix}[−31​2−4​05​]
  2. [000000]\begin{bmatrix}0 & 0 & 0\\0 & 0 & 0\end{bmatrix}[00​00​00​] (correct answer)
  3. [320145]\begin{bmatrix}3 & 2 & 0\\1 & 4 & 5\end{bmatrix}[31​24​05​]
  4. [−320]\begin{bmatrix}-3 & 2 & 0\end{bmatrix}[−3​2​0​]

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 5

Let AAA and BBB be 2×22\times 22×2 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}.A=[1−2​03​],B=[42​−10​]. Calculate 2A+B2A+B2A+B.

  1. [7−1−26]\begin{bmatrix}7 & -1\\ -2 & 6\end{bmatrix}[7−2​−16​]
  2. [6−206]\begin{bmatrix}6 & -2\\ 0 & 6\end{bmatrix}[60​−26​]
  3. [6−1−26]\begin{bmatrix}6 & -1\\ -2 & 6\end{bmatrix}[6−2​−16​] (correct answer)
  4. [5−1−23]\begin{bmatrix}5 & -1\\ -2 & 3\end{bmatrix}[5−2​−13​]

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 6

A research team collects data in matrix D=[0.6−0.30.90.4]D = \begin{bmatrix} 0.6 & -0.3 \\ 0.9 & 0.4 \end{bmatrix}D=[0.60.9​−0.30.4​]. To convert the data to percentages and then apply a confidence factor of 0.8, they perform scalar multiplication. What is the sum of the absolute values of all entries in the final matrix?

  1. 140140140
  2. 1.761.761.76
  3. 176176176 (correct answer)
  4. 17.617.617.6

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 DDD by 100: 100D=[60−309040]100D = \begin{bmatrix} 60 & -30 \\ 90 & 40 \end{bmatrix}100D=[6090​−3040​] Next, apply the confidence factor of 0.8: 0.8×100D=[48−247232]0.8 \times 100D = \begin{bmatrix} 48 & -24 \\ 72 & 32 \end{bmatrix}0.8×100D=[4872​−2432​] Now find the absolute values: ∣48∣=48|48| = 48∣48∣=48, ∣−24∣=24|-24| = 24∣−24∣=24, ∣72∣=72|72| = 72∣72∣=72, ∣32∣=32|32| = 32∣32∣=32 Sum these: 48+24+72+32=17648 + 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 = 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∣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 = 1.762.2×0.8=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 7

Given M=[40−21−35]M=\begin{bmatrix}4 & 0 & -2\\1 & -3 & 5\end{bmatrix}M=[41​0−3​−25​] (a 2×32\times32×3 matrix), what is 12M\tfrac{1}{2}M21​M?​

  1. [20−1tfrac12−3252]\begin{bmatrix}2 & 0 & -1\\tfrac{1}{2} & -\tfrac{3}{2} & \tfrac{5}{2}\end{bmatrix}[2tfrac12​0−23​​−125​​] (correct answer)
  2. [80−42−610]\begin{bmatrix}8 & 0 & -4\\2 & -6 & 10\end{bmatrix}[82​0−6​−410​]
  3. [9212−32tfrac32−52112]\begin{bmatrix}\tfrac{9}{2} & \tfrac{1}{2} & -\tfrac{3}{2}\\tfrac{3}{2} & -\tfrac{5}{2} & \tfrac{11}{2}\end{bmatrix}[29​tfrac32​21​−25​​−23​211​​]
  4. [20−2tfrac12−352]\begin{bmatrix}2 & 0 & -2\\tfrac{1}{2} & -3 & \tfrac{5}{2}\end{bmatrix}[2tfrac12​0−3​−225​​]

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 = 1/2 and matrix M = [[4, 0, -2], [1, -3, 5]], we multiply each entry by 1/2: the entry in row 1, column 1 becomes (1/2)·4 = 2, row 1 column 2: (1/2)·0=0, row 1 column 3: (1/2)·(-2)=-1, row 2 column 1: (1/2)·1=1/2, row 2 column 2: (1/2)·(-3)=-3/2, row 2 column 3: (1/2)·5=5/2, giving us the result matrix [[2, 0, -1], [1/2, -3/2, 5/2]]. Choice A is correct because it multiplies every entry by the scalar correctly. Choice B makes an arithmetic error in computing multiple entries, calculating whole numbers like 8 and -6 instead of halves like 2 and -3/2. 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 8

Let AAA be the 2×32\times 32×3 matrix A=[2−1034−2].A=\begin{bmatrix}2 & -1 & 0\\ 3 & 4 & -2\end{bmatrix}.A=[23​−14​0−2​]. What is 3A3A3A?

  1. [6−30912−6]\begin{bmatrix}6 & -3 & 0\\ 9 & 12 & -6\end{bmatrix}[69​−312​0−6​] (correct answer)
  2. [523671]\begin{bmatrix}5 & 2 & 3\\ 6 & 7 & 1\end{bmatrix}[56​27​31​]
  3. [6−1034−2]\begin{bmatrix}6 & -1 & 0\\ 3 & 4 & -2\end{bmatrix}[63​−14​0−2​]
  4. [6−30912−2]\begin{bmatrix}6 & -3 & 0\\ 9 & 12 & -2\end{bmatrix}[69​−312​0−2​]

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 9

Let A=[2−310]A=\begin{bmatrix}2 & -3\\1 & 0\end{bmatrix}A=[21​−30​] and B=[−46−20]B=\begin{bmatrix}-4 & 6\\-2 & 0\end{bmatrix}B=[−4−2​60​]. If kA=BkA=BkA=B, what is kkk?​​

  1. k=−2k=-2k=−2 (correct answer)
  2. k=2k=2k=2
  3. k=−1k=-1k=−1
  4. k=12k=\tfrac{1}{2}k=21​

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. To find k such that kA = B, we need k·[[2, -3], [1, 0]] = [[-4, 6], [-2, 0]], which means k·2 = -4, k·(-3) = 6, k·1 = -2, and k·0 = 0. From the first equation, k = -4/2 = -2; from the second equation, k = 6/(-3) = -2; from the third equation, k = -2/1 = -2; and the fourth equation is satisfied for any k since 0 times any number equals 0. Choice A is correct because k = -2 satisfies all four equations, making every entry of kA equal to the corresponding entry of B. Choice B would give kA = [[4, -6], [2, 0]], which has wrong signs for all non-zero entries. 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. When solving for a scalar k in an equation kA = B, check that the same value of k works for all corresponding entries.

Question 10

What is the effect of multiplying a matrix AAA by −1-1−1 (that is, computing −1⋅A-1\cdot A−1⋅A)?​​

  1. Each entry of AAA is increased by 1.
  2. The dimensions of AAA double.
  3. Each entry of AAA changes sign (is multiplied by −1-1−1). (correct answer)
  4. Only the diagonal entries of AAA change sign.

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 = -1 has the specific effect of negating all entries: for any matrix A, the product (-1)A has entries (-1)·aᵢⱼ = -aᵢⱼ for every position (i,j), meaning each positive entry becomes negative and each negative entry becomes positive. Choice C is correct because it correctly states that each entry of A changes sign when multiplied by -1. Choice D 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). 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 11

For A=[0−3251−1]A=\begin{bmatrix}0 & -3 & 2\\5 & 1 & -1\end{bmatrix}A=[05​−31​2−1​] (a 2×32\times32×3 matrix), what are the dimensions of −4A-4A−4A?

  1. 2×32\times32×3 (correct answer)
  2. 3×23\times23×2
  3. 8×128\times128×12
  4. 6×16\times16×1

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 12

Let A=[2−46]A=\begin{bmatrix}2 & -4 & 6\end{bmatrix}A=[2​−4​6​] (a 1×31\times31×3 matrix). What is 0A0A0A?

  1. [2−46]\begin{bmatrix}2 & -4 & 6\end{bmatrix}[2​−4​6​]
  2. [000]\begin{bmatrix}0 & 0 & 0\end{bmatrix}[0​0​0​] (correct answer)
  3. [0−46]\begin{bmatrix}0 & -4 & 6\end{bmatrix}[0​−4​6​]
  4. [0]\begin{bmatrix}0\end{bmatrix}[0​]

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 = [[2, -4, 6]], 0A = [[0, 0, 0]]. Choice B is correct because it correctly gives the zero matrix. Choice A omits multiplying by zero, leaving the original matrix unchanged. 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 13

For the 2×22\times 22×2 matrix A=[1−430]A=\begin{bmatrix}1 & -4\\3 & 0\end{bmatrix}A=[13​−40​], which matrix represents 2A2A2A?​​

  1. [2−830]\begin{bmatrix}2 & -8\\3 & 0\end{bmatrix}[23​−80​]
  2. [3−252]\begin{bmatrix}3 & -2\\5 & 2\end{bmatrix}[35​−22​]
  3. [1−2tfrac320]\begin{bmatrix}1 & -2\\tfrac{3}{2} & 0\end{bmatrix}[1tfrac32​−20​]
  4. [2−860]\begin{bmatrix}2 & -8\\6 & 0\end{bmatrix}[26​−80​] (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 14

Let AAA be the 2×32\times 32×3 matrix A=[−1203−45].A=\begin{bmatrix}-1 & 2 & 0\\ 3 & -4 & 5\end{bmatrix}.A=[−13​2−4​05​]. What is the effect of multiplying AAA by −1-1−1 (i.e., what is −A-A−A)?

  1. [1−20−34−5]\begin{bmatrix}1 & -2 & 0\\ -3 & 4 & -5\end{bmatrix}[1−3​−24​0−5​] (correct answer)
  2. [−2104−36]\begin{bmatrix}-2 & 1 & 0\\ 4 & -3 & 6\end{bmatrix}[−24​1−3​06​]
  3. [0314−56]\begin{bmatrix}0 & 3 & 1\\ 4 & -5 & 6\end{bmatrix}[04​3−5​16​]
  4. [120345]\begin{bmatrix}1 & 2 & 0\\ 3 & 4 & 5\end{bmatrix}[13​24​05​]

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=−1k = -1k=−1 has the specific effect of negating all entries: for A=[−1203−45]A = \begin{bmatrix} -1 & 2 & 0 \\ 3 & -4 & 5 \end{bmatrix}A=[−13​2−4​05​], −A=[1−20−34−5]-A = \begin{bmatrix} 1 & -2 & 0 \\ -3 & 4 & -5 \end{bmatrix}−A=[1−3​−24​0−5​]. Choice A is correct because it multiplies every entry by the scalar correctly. Choice D has correct magnitudes but wrong signs, failing to account for negative scalar reversing signs. 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

Using the matrices A=[2−103],B=[14−20],A=\begin{bmatrix}2 & -1\\0 & 3\end{bmatrix},\quad B=\begin{bmatrix}1 & 4\\-2 & 0\end{bmatrix},A=[20​−13​],B=[1−2​40​], which property is demonstrated by the equality 3(A+B)=3A+3B3(A+B)=3A+3B3(A+B)=3A+3B?​​

  1. Associative property with scalars
  2. Distributive property over matrix addition (correct answer)
  3. Identity property of scalar multiplication
  4. Zero property of scalar multiplication

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: for any entry position (i,j), we have k(aᵢⱼ + bᵢⱼ) = kaᵢⱼ + kbᵢⱼ. Choice B is correct because it correctly identifies that the equality 3(A+B) = 3A + 3B demonstrates the distributive property of scalar multiplication over matrix addition. Choice A refers to associativity with scalars, which would be something like (kc)A = k(cA), not the property shown in this example. 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 distributive property is fundamental to scalar multiplication and allows us to factor out common scalars or distribute them as needed in matrix expressions.

Question 16

A manufacturing company uses matrix P=[40−2135]P = \begin{bmatrix} 4 & 0 & -2 \\ 1 & 3 & 5 \end{bmatrix}P=[41​03​−25​] to represent profit margins (in thousands of dollars) for different product lines. Due to economic conditions, all profits and losses will be reduced to 40% of their current values. If the company needs the sum of all entries in the resulting matrix to be positive, what constraint must be satisfied?

  1. The constraint requires additional analysis since 0.4(11)=4.40.4(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 > 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 > 00.4(13)=5.2>0
  4. The constraint is already satisfied since 0.4(11)=4.4>00.4(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 PPP: 4+0+(−2)+1+3+5=114 + 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 = 4.40.4×11=4.4, which is positive. Looking at the wrong answers: Choice A suggests the calculation 0.4(11)=4.40.4(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 > 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 17

Two matrices A=[2−143]A = \begin{bmatrix} 2 & -1 \\ 4 & 3 \end{bmatrix}A=[24​−13​] and B=[−63−12−9]B = \begin{bmatrix} -6 & 3 \\ -12 & -9 \end{bmatrix}B=[−6−12​3−9​] are related by scalar multiplication. If matrix C=[15−20]C = \begin{bmatrix} 1 & 5 \\ -2 & 0 \end{bmatrix}C=[1−2​50​] undergoes the same scalar transformation that relates AAA to BBB, what is the trace of the resulting matrix?

  1. 333
  2. −3-3−3 (correct answer)
  3. −18-18−18
  4. −6-6−6

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 AAA and BBB, examine corresponding elements. Looking at the first element: 2k=−62k = -62k=−6, so k=−3k = -3k=−3. You can verify this works for all elements: −3×2=−6-3 \times 2 = -6−3×2=−6, −3×(−1)=3-3 \times (-1) = 3−3×(−1)=3, −3×4=−12-3 \times 4 = -12−3×4=−12, and −3×3=−9-3 \times 3 = -9−3×3=−9. Indeed, B=−3AB = -3AB=−3A. Now apply this same scalar multiplication to matrix CCC: −3C=−3[15−20]=[−3−1560]-3C = -3\begin{bmatrix} 1 & 5 \\ -2 & 0 \end{bmatrix} = \begin{bmatrix} -3 & -15 \\ 6 & 0 \end{bmatrix}−3C=−3[1−2​50​]=[−36​−150​] The trace of a matrix is the sum of its diagonal elements. For the resulting matrix, the trace is −3+0=−3-3 + 0 = -3−3+0=−3. Looking at the wrong answers: Choice A (333) likely comes from finding the positive trace of the original matrix CCC (1+0=11 + 0 = 11+0=1) or making a sign error. Choice C (−18-18−18) might result from multiplying the scalar by the sum of all elements in CCC rather than just the diagonal elements. Choice D (−6-6−6) 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 18

Let A=[1−234]A=\begin{bmatrix}1 & -2\\3 & 4\end{bmatrix}A=[13​−24​] and B=[05−12]B=\begin{bmatrix}0 & 5\\-1 & 2\end{bmatrix}B=[0−1​52​]. What is 3(A+B)3(A+B)3(A+B)?

  1. [39618]\begin{bmatrix}3 & 9\\6 & 18\end{bmatrix}[36​918​] (correct answer)
  2. [1326]\begin{bmatrix}1 & 3\\2 & 6\end{bmatrix}[12​36​]
  3. [3366]\begin{bmatrix}3 & 3\\6 & 6\end{bmatrix}[36​36​]
  4. [315618]\begin{bmatrix}3 & 15\\6 & 18\end{bmatrix}[36​1518​]

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 19

A matrix P=[302−141]P=\begin{bmatrix}3 & 0 & 2\\-1 & 4 & 1\end{bmatrix}P=[3−1​04​21​] represents prices (in dollars). If all prices increase by 50%, the new price matrix is 1.5P1.5P1.5P. What is 1.5P1.5P1.5P?​​

  1. [4.503−1.561.5]\begin{bmatrix}4.5 & 0 & 3\\-1.5 & 6 & 1.5\end{bmatrix}[4.5−1.5​06​31.5​] (correct answer)
  2. [1.501−0.520.5]\begin{bmatrix}1.5 & 0 & 1\\-0.5 & 2 & 0.5\end{bmatrix}[1.5−0.5​02​10.5​]
  3. [302−141]\begin{bmatrix}3 & 0 & 2\\-1 & 4 & 1\end{bmatrix}[3−1​04​21​]
  4. [4.51.530.562.5]\begin{bmatrix}4.5 & 1.5 & 3\\0.5 & 6 & 2.5\end{bmatrix}[4.50.5​1.56​32.5​]

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 = 1.5 and matrix P = [[3, 0, 2], [-1, 4, 1]], we multiply each entry by 1.5: the entry in row 1, column 1 becomes 1.5·3 = 4.5, the entry in row 1, column 2 becomes 1.5·0 = 0, the entry in row 1, column 3 becomes 1.5·2 = 3, the entry in row 2, column 1 becomes 1.5·(-1) = -1.5, the entry in row 2, column 2 becomes 1.5·4 = 6, and the entry in row 2, column 3 becomes 1.5·1 = 1.5, giving us the result matrix [[4.5, 0, 3], [-1.5, 6, 1.5]]. Choice A is correct because it multiplies every entry by the scalar 1.5 correctly, representing a 50% price increase. Choice D makes arithmetic errors in computing several entries, including changing the zero in position (1,2) to 1.5 when it should remain 0. 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 20

For the 3×23\times 23×2 matrix A=[1−3024−1],A=\begin{bmatrix}1 & -3\\ 0 & 2\\ 4 & -1\end{bmatrix},A=​104​−32−1​​, what are the dimensions of 5A5A5A?

  1. 2×32\times 32×3
  2. 3×23\times 23×2 (correct answer)
  3. 5×55\times 55×5
  4. 15×1015\times 1015×10

Explanation: This question tests understanding of scalar multiplication of matrices. When a matrix is multiplied by a scalar kkk, each entry in the matrix is multiplied by kkk, and the dimensions of the matrix remain unchanged—only the entry values change. Since A has dimensions 3×23 \times 23×2, and scalar multiplication only changes entry values (not the number of rows or columns), the product 5A5A5A also has dimensions 3×23 \times 23×2. Choice B is correct because it correctly states dimensional invariance. Choice D incorrectly claims the dimensions change, suggesting a 3×23 \times 23×2 matrix becomes 15×1015 \times 1015×10 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.