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