If \(A^{-1} = \begin{bmatrix} 4 & -3 \\ -1 & 1 \end{bmatrix}\), then \((A^{T})^{-1}\) is
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \(\begin{bmatrix} 4 & -1 \\ -3 & 1 \end{bmatrix}\)
For an invertible matrix, the inverse of the transpose is the transpose of the inverse:
\((A^{T})^{-1} = (A^{-1})^{T}\)
\( \displaystyle (A^{-1})^{T} = \begin{bmatrix} 4 & -1 \\ -3 & 1 \end{bmatrix} \)
So A itself is never needed: the rows of \(A^{-1}\) simply become columns.
Why the other options are incorrect
Option (1) \(\begin{bmatrix} 1 & 3 \\ 1 & 4 \end{bmatrix}\) — Incorrect. This is A, the inverse of \(A^{-1}\).
Option (2) \(\begin{bmatrix} 4 & 3 \\ 1 & 1 \end{bmatrix}\) — Incorrect. Its elements 3 and 1 in the off-diagonal places have the wrong sign; the transpose of \(A^{-1}\) has \(-1\) and \(-3\) there.
Option (4) \(\begin{bmatrix} 1 & 1 \\ 3 & 4 \end{bmatrix}\) — Incorrect. This is \(A^{T}\), not its inverse.