If \(A = \begin{bmatrix}1&0\\0&-1\end{bmatrix}\) and \(B = \begin{bmatrix}0&2\\2&0\end{bmatrix}\), then the matrix AB is equal to
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \(\begin{bmatrix}0&2\\-2&0\end{bmatrix}\)
Each entry of AB is a row of A times a column of B:
\( \displaystyle AB = \begin{bmatrix}1&0\\0&-1\end{bmatrix}\begin{bmatrix}0&2\\2&0\end{bmatrix} \)
\( \displaystyle = \begin{bmatrix}1 \cdot 0 + 0 \cdot 2&1 \cdot 2 + 0 \cdot 0\\0 \cdot 0 + (-1) \cdot 2&0 \cdot 2 + (-1) \cdot 0\end{bmatrix} \)
\( \displaystyle = \begin{bmatrix}0&2\\-2&0\end{bmatrix} \)
Why the other options are incorrect
Option (1) \(\begin{bmatrix}0&-2\\2&0\end{bmatrix}\) — Incorrect. This is BA, the product in the reverse order. Matrix multiplication is not commutative, and here \(BA = -AB\).
Option (2) \(\begin{bmatrix}0&0\\0&0\end{bmatrix}\) — Incorrect. AB is not the zero matrix; its first-row, second-column entry is 2.
Option (4) \(\begin{bmatrix}2&-2\\-2&2\end{bmatrix}\) — Incorrect. The diagonal entries of AB are 0, not 2.