For which of the following pairs of matrices is AB = BA ?

The correct answer is Option (3) → \(A = \begin{bmatrix}1&2\\2&1\end{bmatrix}\), \(B = \begin{bmatrix}3&1\\1&3\end{bmatrix}\)

Matrix multiplication is not commutative in general, so both products must be worked out for each pair.

For option (3):

    \( \displaystyle AB = \begin{bmatrix}3 + 2&1 + 6\\6 + 1&2 + 3\end{bmatrix} = \begin{bmatrix}5&7\\7&5\end{bmatrix} \)

    \( \displaystyle BA = \begin{bmatrix}3 + 2&6 + 1\\1 + 6&2 + 3\end{bmatrix} = \begin{bmatrix}5&7\\7&5\end{bmatrix} \)

So \(AB = BA\) for this pair.

Why the other options are incorrect

• Option (1) \(A = \begin{bmatrix}1&2\\0&1\end{bmatrix}\), \(B = \begin{bmatrix}1&0\\2&1\end{bmatrix}\) — Incorrect. \(AB = \begin{bmatrix}5&2\\2&1\end{bmatrix}\) but \(BA = \begin{bmatrix}1&2\\2&5\end{bmatrix}\).

• Option (2) \(A = \begin{bmatrix}2&0\\0&3\end{bmatrix}\), \(B = \begin{bmatrix}1&1\\0&1\end{bmatrix}\) — Incorrect. \(AB = \begin{bmatrix}2&2\\0&3\end{bmatrix}\) but \(BA = \begin{bmatrix}2&3\\0&3\end{bmatrix}\).

• Option (4) \(A = \begin{bmatrix}0&1\\0&0\end{bmatrix}\), \(B = \begin{bmatrix}0&0\\1&0\end{bmatrix}\) — Incorrect. \(AB = \begin{bmatrix}1&0\\0&0\end{bmatrix}\) but \(BA = \begin{bmatrix}0&0\\0&1\end{bmatrix}\).