If \( A^T=\begin{bmatrix}a & b\\c & d\end{bmatrix} \) and \( A^{-1}=\begin{bmatrix}p & q\\r & s\end{bmatrix} \) , then
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \( \begin{bmatrix}a & b\\c & d\end{bmatrix}^{-1}=\begin{bmatrix}p & q\\r & s\end{bmatrix}^T \)
For an invertible matrix, the inverse of the transpose equals the transpose of the inverse:
\( \displaystyle (A^T)^{-1}=(A^{-1})^T \)
Since \( A^T=\begin{bmatrix}a & b\\c & d\end{bmatrix} \) and \( A^{-1}=\begin{bmatrix}p & q\\r & s\end{bmatrix} \),
\( \displaystyle \begin{bmatrix}a & b\\c & d\end{bmatrix}^{-1}=\begin{bmatrix}p & q\\r & s\end{bmatrix}^T \)
(This follows from \( A^T(A^{-1})^T=(A^{-1}A)^T=I^T=I \).)
Why the other options are incorrect
Option (1) \( \begin{bmatrix}a & b\\c & d\end{bmatrix}^{-1}=\begin{bmatrix}p & q\\r & s\end{bmatrix} \) — Incorrect. This says \( (A^T)^{-1}=A^{-1} \), which holds only when A is symmetric.
Option (2) \( \begin{bmatrix}a & b\\c & d\end{bmatrix}=\begin{bmatrix}p & q\\r & s\end{bmatrix}^{-1} \) — Incorrect. This says \( A^T=(A^{-1})^{-1}=A \), which again holds only when A is symmetric.
Option (3) \( \begin{bmatrix}a & b\\c & d\end{bmatrix}^T=\begin{bmatrix}p & -q\\-r & s\end{bmatrix}^{-1} \) — Incorrect. The left side is A, which equals \( \begin{bmatrix}p & q\\r & s\end{bmatrix}^{-1} \); changing the signs of q and r gives a different matrix.