If \(A = \begin{bmatrix}1&0\\1&1\end{bmatrix}\) and \(B = \begin{bmatrix}1&1\\0&1\end{bmatrix}\), then \((AB)^2 - A^2B^2\) is equal to
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \(\begin{bmatrix}1&1\\1&0\end{bmatrix}\)
Here \((AB)^2 = ABAB\) while \(A^2B^2 = AABB\); they need not be equal, because matrix multiplication is not commutative. Working out each:
\( \displaystyle AB = \begin{bmatrix}1&1\\1&2\end{bmatrix} \)
\( \displaystyle (AB)^2 = \begin{bmatrix}1 + 1&1 + 2\\1 + 2&1 + 4\end{bmatrix} = \begin{bmatrix}2&3\\3&5\end{bmatrix} \)
\( \displaystyle A^2 = \begin{bmatrix}1&0\\2&1\end{bmatrix} \), \( \displaystyle B^2 = \begin{bmatrix}1&2\\0&1\end{bmatrix} \)
\( \displaystyle A^2B^2 = \begin{bmatrix}1&2\\2&4 + 1\end{bmatrix} = \begin{bmatrix}1&2\\2&5\end{bmatrix} \)
Hence
\( \displaystyle (AB)^2 - A^2B^2 = \begin{bmatrix}1&1\\1&0\end{bmatrix} \)
Why the other options are incorrect
Option (1) \(\begin{bmatrix}0&0\\0&0\end{bmatrix}\) — Incorrect. It assumes \((AB)^2 = A^2B^2\); here \(BA = \begin{bmatrix}2&1\\1&1\end{bmatrix} \ne AB\), and the two squares differ.
Option (2) \(\begin{bmatrix}-1&-1\\-1&0\end{bmatrix}\) — Incorrect. This is \(A^2B^2 - (AB)^2\), the subtraction in the wrong order.
Option (3) \(\begin{bmatrix}1&0\\0&1\end{bmatrix}\) — Incorrect. The difference has first row \(\begin{bmatrix}2 - 1&3 - 2\end{bmatrix} = \begin{bmatrix}1&1\end{bmatrix}\), not \(\begin{bmatrix}1&0\end{bmatrix}\).