If \(A = \begin{bmatrix}1&-2\\3&0\end{bmatrix}\) and \(B = \begin{bmatrix}2&1\\1&-1\end{bmatrix}\), then \(AB - BA\) is equal to
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \(\begin{bmatrix}-5&7\\8&5\end{bmatrix}\)
Matrix multiplication is not commutative, so AB and BA must be worked out separately, row by column.
\( \displaystyle AB = \begin{bmatrix}2 - 2&1 + 2\\6 + 0&3 + 0\end{bmatrix} = \begin{bmatrix}0&3\\6&3\end{bmatrix} \)
\( \displaystyle BA = \begin{bmatrix}2 + 3&-4 + 0\\1 - 3&-2 + 0\end{bmatrix} = \begin{bmatrix}5&-4\\-2&-2\end{bmatrix} \)
Subtracting:
\( \displaystyle AB - BA = \begin{bmatrix}0 - 5&3 + 4\\6 + 2&3 + 2\end{bmatrix} = \begin{bmatrix}-5&7\\8&5\end{bmatrix} \)
Since \(AB - BA \ne O\), here \(AB \ne BA\).
Why the other options are incorrect
Option (2) \(\begin{bmatrix}5&-7\\-8&-5\end{bmatrix}\) — Incorrect. This is \(BA - AB\), the products taken in the wrong order.
Option (3) \(\begin{bmatrix}0&0\\0&0\end{bmatrix}\) — Incorrect. This would need AB = BA, but the first elements of AB and BA are 0 and 5.
Option (4) \(\begin{bmatrix}5&-1\\4&1\end{bmatrix}\) — Incorrect. This is \(AB + BA\), not the difference.