If \(A = \begin{bmatrix} 1 & 2 \\ -1 & 3 \end{bmatrix}\) and \(B = \begin{bmatrix} 2 & 0 \\ 1 & -1 \end{bmatrix}\), then \((AB)'\) is equal to
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \(\begin{bmatrix} 4 & 1 \\ -2 & -3 \end{bmatrix}\)
The product AB is found first, row by column:
\( \displaystyle AB = \begin{bmatrix} 1 \times 2 + 2 \times 1 & 1 \times 0 + 2 \times (-1) \\ -1 \times 2 + 3 \times 1 & -1 \times 0 + 3 \times (-1) \end{bmatrix} \)
\( \displaystyle AB = \begin{bmatrix} 4 & -2 \\ 1 & -3 \end{bmatrix} \)
The transpose turns rows into columns:
\( \displaystyle (AB)' = \begin{bmatrix} 4 & 1 \\ -2 & -3 \end{bmatrix} \)
The same matrix comes from the reversal law \((AB)' = B'A'\).
Why the other options are incorrect
Option (2) \(\begin{bmatrix} 4 & -2 \\ 1 & -3 \end{bmatrix}\) — Incorrect. This is AB itself; it has not been transposed.
Option (3) \(\begin{bmatrix} 2 & 2 \\ 4 & -1 \end{bmatrix}\) — Incorrect. This is \(A'B'\), the product taken in the wrong order; \((AB)'\) equals \(B'A'\), not \(A'B'\).
Option (4) \(\begin{bmatrix} 4 & 1 \\ 2 & -3 \end{bmatrix}\) — Incorrect. The element in row 2, column 1 of \((AB)'\) is \(-2\), not 2.