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

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

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

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

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

This problem involves scalar multiplication of a matrix, where each entry is multiplied by the scalar 3. For matrix A = $\begin{pmatrix} 3 & -2 \\ 0 & 1 \end{pmatrix}$, we compute $3A$ by multiplying each entry: $3(3) = 9$, $3(-2) = -6$, $3(0) = 0$, $3(1) = 3$. The result is $\begin{pmatrix} 9 & -6 \\ 0 & 3 \end{pmatrix}$. Choice C correctly shows this scalar multiplication.

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

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

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

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