If \(A = \begin{bmatrix}1&2\\0&-1\end{bmatrix}\) and \(B = \begin{bmatrix}3&1\\2&4\end{bmatrix}\), then \((AB)'\) is equal to
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \(\begin{bmatrix}7&-2\\9&-4\end{bmatrix}\)
First the product, row by column:
\( \displaystyle AB = \begin{bmatrix}3 + 4&1 + 8\\0 - 2&0 - 4\end{bmatrix} = \begin{bmatrix}7&9\\-2&-4\end{bmatrix} \)
The transpose writes the rows as columns:
\( \displaystyle (AB)' = \begin{bmatrix}7&-2\\9&-4\end{bmatrix} \)
This agrees with the reversal rule \((AB)' = B'A'\).
Why the other options are incorrect
Option (2) \(\begin{bmatrix}7&9\\-2&-4\end{bmatrix}\) — Incorrect. This is AB itself, before taking the transpose.
Option (3) \(\begin{bmatrix}3&2\\5&0\end{bmatrix}\) — Incorrect. This is \(A'B'\); the transposes must be multiplied in the reverse order, \(B'A'\).
Option (4) \(\begin{bmatrix}3&5\\2&0\end{bmatrix}\) — Incorrect. This is BA, a different product.