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 asks for matrix addition A + B. To add matrices, we add corresponding entries: (2) + (-4) = -2 for position (1,1), (-1) + (5) = 4 for position (1,2), (0) + (1) = 1 for position (2,1), and (3) + (-2) = 1 for position (2,2). The result is the matrix [[-2, 4], [1, 1]]. Note that some students might subtract instead of add, which would give a different result.

The operation is scalar multiplication, where each entry of the matrix is multiplied by the scalar -3. For -3A, compute $-3 \times -1 = 3$ for the (1,1) entry; $-3 \times 2 = -6$ for (1,2); $-3 \times 0 = 0$ for (2,1); and $-3 \times 4 = -12$ for (2,2). This scaling applies uniformly to every element, representing a transformation that enlarges and reflects the coordinates. The result is the matrix $$\begin{bmatrix} 3 & -6 \\ 0 & -12 \end{bmatrix}$$. Choice D uses the scalar 1 instead of -3, likely from misreading the problem.

The operation is scalar multiplication by -2, multiplying every entry of the matrix by this scalar. For -2A, (1,1): -20 = 0; (1,2): -2(-3) = 6; (2,1): -22 = -4; (2,2): -21 = -2. This reflects and scales the encoding matrix uniformly. The result is the matrix [[0, 6], [-4, -2]]. Choice B flips the sign of the (1,2) entry, possibly from mishandling the negative scalar.

This question involves matrix subtraction, where we subtract corresponding entries of matrix B from matrix A. To compute A - B, we subtract each entry in position (i,j) of matrix B from the corresponding entry in position (i,j) of matrix A. For entry (1,1): 0 - (-2) = 2, for (1,2): 4 - 1 = 3, for (2,1): (-3) - 3 = -6, and for (2,2): 2 - 0 = 2. The result is the matrix [[2,3],[-6,2]].

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 requires matrix addition, where we add corresponding entries of matrices A and B. For matrices A = [[0, 3], [-5, 2]] and B = [[1, -4], [2, 1]], we compute: entry (1,1): 0 + 1 = 1, entry (1,2): 3 + (-4) = -1, entry (2,1): -5 + 2 = -3, entry (2,2): 2 + 1 = 3. The result is [[1, -1], [-3, 3]].