If \(A = \begin{bmatrix}2 & 1\\ 1 & 1\end{bmatrix}\), then \((A^{2})^{-1}\) is
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \(\begin{bmatrix}2 & -3\\ -3 & 5\end{bmatrix}\)
The square of A is
\(\displaystyle A^{2} = \begin{bmatrix}2 & 1\\ 1 & 1\end{bmatrix}\begin{bmatrix}2 & 1\\ 1 & 1\end{bmatrix} = \begin{bmatrix}5 & 3\\ 3 & 2\end{bmatrix}\)
\(|A^{2}| = 5 \times 2 - 3 \times 3 = 1\)
For a 2 × 2 matrix the inverse is the adjoint divided by the determinant: swap the diagonal entries and change the signs of the other two.
\(\displaystyle (A^{2})^{-1} = \frac{1}{1}\begin{bmatrix}2 & -3\\ -3 & 5\end{bmatrix}\)
Check: \(\displaystyle \begin{bmatrix}5 & 3\\ 3 & 2\end{bmatrix}\begin{bmatrix}2 & -3\\ -3 & 5\end{bmatrix} = \begin{bmatrix}1 & 0\\ 0 & 1\end{bmatrix}\).
The same answer comes from \((A^{-1})^{2}\), since \((A^{2})^{-1} = (A^{-1})^{2}\).
Why the other options are incorrect
Option (1) \(\begin{bmatrix}5 & -3\\ -3 & 2\end{bmatrix}\) — Incorrect. The signs are changed but the diagonal entries 5 and 2 are not swapped.
Option (2) \(\begin{bmatrix}1 & -1\\ -1 & 2\end{bmatrix}\) — Incorrect. This is \(A^{-1}\) itself, not the inverse of \(A^{2}\).
Option (3) \(\begin{bmatrix}2 & 3\\ 3 & 5\end{bmatrix}\) — Incorrect. The diagonal is swapped but the signs of the other entries are not changed; its product with \(A^{2}\) is not I.