If \(A = \begin{bmatrix} 2 & 1 \\ 1 & 1 \end{bmatrix}\), then \((A^2)^{-1}\) is equal to
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \(\begin{bmatrix} 2 & -3 \\ -3 & 5 \end{bmatrix}\)
By the reversal law, \((A^2)^{-1} = (AA)^{-1} = A^{-1}A^{-1}\), so \((A^2)^{-1} = (A^{-1})^2\).
\(|A| = 2 - 1 = 1\)
\( \displaystyle A^{-1} = \begin{bmatrix} 1 & -1 \\ -1 & 2 \end{bmatrix} \)
\( \displaystyle (A^{-1})^2 = \begin{bmatrix} 1 + 1 & -1 - 2 \\ -1 - 2 & 1 + 4 \end{bmatrix} = \begin{bmatrix} 2 & -3 \\ -3 & 5 \end{bmatrix} \)
Check: \(A^2 = \begin{bmatrix} 5 & 3 \\ 3 & 2 \end{bmatrix}\), and \(\begin{bmatrix} 5 & 3 \\ 3 & 2 \end{bmatrix}\begin{bmatrix} 2 & -3 \\ -3 & 5 \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}\).
Why the other options are incorrect
Option (2) \(\begin{bmatrix} 5 & 3 \\ 3 & 2 \end{bmatrix}\) — Incorrect. This is \(A^2\) itself, not its inverse.
Option (3) \(\begin{bmatrix} 1 & -1 \\ -1 & 2 \end{bmatrix}\) — Incorrect. This is \(A^{-1}\); it still has to be squared.
Option (4) \(\begin{bmatrix} 2 & 3 \\ 3 & 5 \end{bmatrix}\) — Incorrect. The signs of the off-diagonal elements are wrong; its product with \(A^2\) has 19, not 1, in row 1, column 1.