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).

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.

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.

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.

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 $3 \times 2$, and scalar multiplication only changes entry values (not the number of rows or columns), the product $5A$ also has dimensions $3 \times 2$. Choice B is correct because it correctly states dimensional invariance. Choice D incorrectly claims the dimensions change, suggesting a $3 \times 2$ matrix becomes $15 \times 10$ 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.

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.

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.

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.

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.

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.