Scalar Multiplication of Matrices - Precalculus

1. Given the scalar $k=\tfrac{1}{2}$ and matrix $M=\begin{bmatrix}4 & 0 & -2\\6 & -4 & 2\end{bmatrix}$ (a $2\times3$ matrix), what is $kM$?
   Answer: $\begin{bmatrix}2 & 0 & -1\\3 & -2 & 1\end{bmatrix}$ 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 = 1/2 and matrix M = [[4, 0, -2], [6, -4, 2]], we multiply each entry by 1/2: row 1 becomes 2, 0, -1; row 2 becomes 3, -2, 1, giving the result [[2, 0, -1], [3, -2, 1]]. Choice A is correct because it multiplies every entry by the scalar correctly. Choice D omits multiplying the entries, leaving them unchanged when all entries must be affected. 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 the scalar is a fraction, ensure accurate division for each entry.
2. Given $A=\begin{bmatrix}1 & 0\\-2 & 3\end{bmatrix}$ and $B=\begin{bmatrix}2 & -1\\4 & 0\end{bmatrix}$, calculate $2A-3B$.​
   Answer: $\begin{bmatrix}-4 & 3\\-16 & 6\end{bmatrix}$ 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.
3. For the matrix $A=\begin{bmatrix}2 & -1\\0 & 3\end{bmatrix}$ (a $2\times 2$ matrix), what is $-2A$?​​
   Answer: $\begin{bmatrix}-4 & 2\\0 & -6\end{bmatrix}$ 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).
4. For $A=\begin{bmatrix}-3 & 2 & 0\\1 & -4 & 5\end{bmatrix}$, which matrix represents $0\cdot A$?​
   Answer: $\begin{bmatrix}0 & 0 & 0\\0 & 0 & 0\end{bmatrix}$ 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.
5. For the $2\times 2$ matrix $A=\begin{bmatrix}3 & -2\\ 1 & 0\end{bmatrix},$ which matrix represents $0A$?
   Answer: $\begin{bmatrix}0 & 0\\ 0 & 0\end{bmatrix}$ 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.
6. Let $A=\begin{bmatrix}2 & -5\\ 1 & 0\end{bmatrix}.$ What is the effect of multiplying $A$ by $-1$ (i.e., what is $-A$)?
   Answer: $\begin{bmatrix}-2 & 5\\ -1 & 0\end{bmatrix}$ 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 matrix A = [[2, -5], [1, 0]], we get -A = [[-1·2, -1·(-5)], [-1·1, -1·0]] = [[-2, 5], [-1, 0]]. Choice A is correct because it multiplies every entry by the scalar correctly, with -1 reversing all signs: positive 2 becomes -2, negative -5 becomes positive 5, positive 1 becomes -1, and 0 remains 0. Choice C has correct magnitudes but wrong signs, failing to account for negative scalar reversing signs on the negative entry -5, which should become positive 5. 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). The operation -A is particularly useful as it represents the additive inverse of matrix A.
7. Multiply the scalar $k=-1$ by the matrix $M=\begin{bmatrix}2 & 0 & -5\end{bmatrix}$ (a $1\times3$ matrix). What is $kM$?​
   Answer: $\begin{bmatrix}-2 & 0 & 5\end{bmatrix}$ 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 M = [[2, 0, -5]], we get [[-2, 0, 5]]. Choice A is correct because it multiplies every entry by the scalar correctly. Choice B 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). 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.
8. Let $A$ be the $2 \times 3$ matrix $A=\begin{bmatrix}2 & -1 & 0\\ 3 & 4 & -2\end{bmatrix}.$ What is $-2A$?
   Answer: $\begin{bmatrix}-4 & 2 & 0\\ -6 & -8 & 4\end{bmatrix}$ 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 $(kA)_{ij} = k \cdot a_{ij}$ for every entry at row i, column j. For scalar k = -2 and matrix A = $\begin{bmatrix} 2 & -1 & 0 \\ 3 & 4 & -2 \end{bmatrix}$, we multiply each entry by -2: the entry in row 1, column 1 becomes $-2 \cdot 2 = -4$, row 1 column 2 becomes $-2 \cdot (-1) = 2$, and similarly for others, giving us the result matrix $\begin{bmatrix} -4 & 2 & 0 \\ -6 & -8 & 4 \end{bmatrix}$. Choice B is correct because it multiplies every entry by the scalar correctly. Choice A only multiplies some entries (such as the first row) instead of multiplying every single entry in the matrix by the scalar. 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.
9. Two matrices $A = \begin{bmatrix} 2 & -1 \\ 4 & 3 \end{bmatrix}$ and $B = \begin{bmatrix} -6 & 3 \\ -12 & -9 \end{bmatrix}$ are related by scalar multiplication. If matrix $C = \begin{bmatrix} 1 & 5 \\ -2 & 0 \end{bmatrix}$ undergoes the same scalar transformation that relates $A$ to $B$, what is the trace of the resulting matrix?
   Answer: $-3$ 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 $A$ and $B$, examine corresponding elements. Looking at the first element: $2k = -6$, so $k = -3$. You can verify this works for all elements: $-3 \times 2 = -6$, $-3 \times (-1) = 3$, $-3 \times 4 = -12$, and $-3 \times 3 = -9$. Indeed, $B = -3A$. Now apply this same scalar multiplication to matrix $C$: $-3C = -3\begin{bmatrix} 1 & 5 \\ -2 & 0 \end{bmatrix} = \begin{bmatrix} -3 & -15 \\ 6 & 0 \end{bmatrix}$ The trace of a matrix is the sum of its diagonal elements. For the resulting matrix, the trace is $-3 + 0 = -3$. Looking at the wrong answers: Choice A ($3$) likely comes from finding the positive trace of the original matrix $C$ ($1 + 0 = 1$) or making a sign error. Choice C ($-18$) might result from multiplying the scalar by the sum of all elements in $C$ rather than just the diagonal elements. Choice D ($-6$) 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.
10. Let $A$ and $B$ be $2\times 2$ matrices $A=\begin{bmatrix}1 & 0\\ -2 & 3\end{bmatrix},\quad B=\begin{bmatrix}4 & -1\\ 2 & 0\end{bmatrix}.$ Calculate $2A+B$.
    Answer: $\begin{bmatrix}6 & -1\\ -2 & 6\end{bmatrix}$ 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.
11. A research team collects data in matrix $D = \begin{bmatrix} 0.6 & -0.3 \\ 0.9 & 0.4 \end{bmatrix}$. 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?
    Answer: $176$ 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 $D$ by 100: $100D = \begin{bmatrix} 60 & -30 \\ 90 & 40 \end{bmatrix}$ Next, apply the confidence factor of 0.8: $0.8 \times 100D = \begin{bmatrix} 48 & -24 \\ 72 & 32 \end{bmatrix}$ Now find the absolute values: $|48| = 48$, $|-24| = 24$, $|72| = 72$, $|32| = 32$ Sum these: $48 + 24 + 72 + 32 = 176$ Answer C (176) is correct. Answer A (140) likely comes from forgetting the confidence factor—if you only converted to percentages, you'd get $60 + 30 + 90 + 40 = 220$, but this doesn't match exactly, suggesting a calculation error along the way. Answer B (1.76) represents the sum before converting to percentages: $|0.6| + |-0.3| + |0.9| + |0.4| = 2.2$, then incorrectly applying 0.8 as $2.2 \times 0.8 = 1.76$. 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.
12. Given the scalar $k=\tfrac{1}{2}$ and the $2\times 2$ matrix $M=\begin{bmatrix}4 & -2\\ 0 & 6\end{bmatrix},$ what is $kM$?
    Answer: $\begin{bmatrix}2 & -1\\ 0 & 3\end{bmatrix}$ 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 $(kA)_{ij} = k \cdot a_{ij}$ for every entry at row i, column j. For scalar $k = \frac{1}{2}$ and matrix $ M = \begin{bmatrix} 4 & -2 \\ 0 & 6 \end{bmatrix} $, we multiply each entry by $\frac{1}{2}$: the entry in row 1, column 1 becomes $\frac{1}{2} \cdot 4 = 2$, row 1 column 2 becomes $\frac{1}{2} \cdot (-2) = -1$, and similarly for others, giving us the result matrix $\begin{bmatrix} 2 & -1 \\ 0 & 3 \end{bmatrix}$. Choice A is correct because it multiplies every entry by the scalar correctly. Choice C makes an arithmetic error in computing the scalar, using 2 (the reciprocal) instead of $\frac{1}{2}$, calculating 2 times each entry instead of half. 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).
13. Given $A=\begin{bmatrix}1 & -2\\ 3 & 4\end{bmatrix} \quad\text{and}\quad B=\begin{bmatrix}0 & 5\\ -1 & 2\end{bmatrix},$ which property is demonstrated by the equation $3(A+B)=3A+3B$?
    Answer: Distributive property over matrix addition 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 C is correct because it properly identifies the distributive property over matrix addition being demonstrated by the equation 3(A+B) = 3A + 3B. Choice A refers to the associative property with scalars, which would be something like (km)A = k(mA), not the property shown in this equation. 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. This distributive property is fundamental to linear algebra and allows us to factor out common scalars from matrix expressions.
14. Let $A=\begin{bmatrix}1 & 0\\-2 & 3\end{bmatrix}$ and $B=\begin{bmatrix}4 & -1\\0 & 2\end{bmatrix}$. Which property is demonstrated by the statement $3(A+B)=3A+3B$?
    Answer: Distributive property over matrix addition 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: k(aᵢⱼ + bᵢⱼ) = kaᵢⱼ + kbᵢⱼ for any entry position. Choice B is correct because it correctly identifies the distributive property over matrix addition. Choice D confuses scalar multiplication with matrix multiplication, attempting to apply commutative property which doesn't always hold for matrix multiplication. 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. The distributive property is key for simplifying expressions involving scalars and matrix sums.
15. Matrix $R = \begin{bmatrix} x & 2x \\ -x & 3x \end{bmatrix}$ represents a scaled version of matrix $S = \begin{bmatrix} 1 & 2 \\ -1 & 3 \end{bmatrix}$. If the determinant of $R$ equals 20, what is the value of $x^2$?
    Answer: $4$ 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 $R$ is matrix $S$ scaled by factor $x$. To find $x$, you'll need to calculate the determinant of $R$. For a 2×2 matrix $\begin{bmatrix} a & b \\ c & d \end{bmatrix}$, the determinant is $ad - bc$. 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.
16. Calculate $2A+3B$ where $A=\begin{bmatrix}1 & 0\\-2 & 4\end{bmatrix},\quad B=\begin{bmatrix}-1 & 2\\3 & 0\end{bmatrix}.$​​
    Answer: $\begin{bmatrix}-1 & 6\\5 & 8\end{bmatrix}$ 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·[[1, 0], [-2, 4]] = [[2, 0], [-4, 8]] and 3B = 3·[[-1, 2], [3, 0]] = [[-3, 6], [9, 0]], then add corresponding entries to get [[2+(-3), 0+6], [-4+9, 8+0]] = [[-1, 6], [5, 8]]. Choice A is correct because it properly applies the distributive property with specific values, correctly computing both scalar multiplications and the subsequent addition. Choice B makes an arithmetic error in computing the entry at row 1, column 2, calculating 4 instead of 6. 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. 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.
17. Let $A=\begin{bmatrix}2 & -4 & 6\end{bmatrix}$ (a $1\times3$ matrix). What is $0A$?
    Answer: $\begin{bmatrix}0 & 0 & 0\end{bmatrix}$ 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.
18. If $kA=B$ where $A=\begin{bmatrix}1 & -2\\3 & 0\end{bmatrix}$ and $B=\begin{bmatrix}-3 & 6\\-9 & 0\end{bmatrix}$, what is the real scalar $k$?
    Answer: $k=-3$ 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 can solve for any entry, e.g., k·1 = -3 implies k = -3, and verify for others: -3·(-2)=6, -3·3=-9, -3·0=0, matching B. Choice C is correct because it gives the scalar that scales A to B correctly. Choice A has the wrong sign, failing to account for the sign changes in the entries. 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. You can find the scalar by dividing corresponding non-zero entries of B by those of A.
19. For $A=\begin{bmatrix}0 & -3 & 2\\5 & 1 & -1\end{bmatrix}$ (a $2\times3$ matrix), what are the dimensions of $-4A$?
    Answer: $2\times3$ 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.
20. Let $A$ be the $2\times 3$ matrix $A=\begin{bmatrix}-1 & 2 & 0\\ 3 & -4 & 5\end{bmatrix}.$ What is the effect of multiplying $A$ by $-1$ (i.e., what is $-A$)?
    Answer: $\begin{bmatrix}1 & -2 & 0\\ -3 & 4 & -5\end{bmatrix}$ 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 $A = \begin{bmatrix} -1 & 2 & 0 \\ 3 & -4 & 5 \end{bmatrix}$, $-A = \begin{bmatrix} 1 & -2 & 0 \\ -3 & 4 & -5 \end{bmatrix}$. 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.

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