If \(A = \begin{bmatrix}2 & -1\\ 1 & 2\end{bmatrix}\), then \(A^{T}A^{-1}\) is
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \(\displaystyle\frac{1}{5}\begin{bmatrix}3 & 4\\ -4 & 3\end{bmatrix}\)
\(|A| = 2 \times 2 - (-1) \times 1 = 5\)
\(\displaystyle A^{-1} = \frac{1}{5}\begin{bmatrix}2 & 1\\ -1 & 2\end{bmatrix}\)
The transpose turns rows into columns:
\(\displaystyle A^{T} = \begin{bmatrix}2 & 1\\ -1 & 2\end{bmatrix}\)
\(\displaystyle A^{T}A^{-1} = \frac{1}{5}\begin{bmatrix}2 & 1\\ -1 & 2\end{bmatrix}\begin{bmatrix}2 & 1\\ -1 & 2\end{bmatrix}\)
\(\displaystyle A^{T}A^{-1} = \frac{1}{5}\begin{bmatrix}4 - 1 & 2 + 2\\ -2 - 2 & -1 + 4\end{bmatrix}\)
\(\displaystyle A^{T}A^{-1} = \frac{1}{5}\begin{bmatrix}3 & 4\\ -4 & 3\end{bmatrix}\)
Why the other options are incorrect
Option (1) I — Incorrect. \(AA^{-1} = I\), but \(A^{T}\) is not A here, so \(A^{T}A^{-1} \ne I\).
Option (2) \(\displaystyle\frac{1}{5}\begin{bmatrix}3 & -4\\ 4 & 3\end{bmatrix}\) — Incorrect. This is the transpose of the correct product; the (1, 2) entry is \(\frac{1}{5}(2 + 2) = \frac{4}{5}\), not \(-\frac{4}{5}\).
Option (3) \(\begin{bmatrix}3 & 4\\ -4 & 3\end{bmatrix}\) — Incorrect. The factor \(\frac{1}{5}\) from \(A^{-1}\) is left out.