Let \(A = \begin{bmatrix}3 & 2\\ 4 & 3\end{bmatrix}\). Consider the following statements:
(A) \(A^{-1} = \begin{bmatrix}3 & -2\\ -4 & 3\end{bmatrix}\)
(B) \(\mathrm{adj}\,A = A^{-1}\)
(C) \(|\mathrm{adj}\,A| = 2\)
(D) \((A^{T})^{-1} = \begin{bmatrix}3 & -4\\ -2 & 3\end{bmatrix}\)
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → (A), (B) and (D) only
For A:
\(|A| = 3 \cdot 3 - 2 \cdot 4 = 1\)
\( \displaystyle \mathrm{adj}\,A = \begin{bmatrix}3 & -2\\ -4 & 3\end{bmatrix} \)
Since \( \displaystyle A^{-1} = \frac{1}{|A|}\,\mathrm{adj}\,A \) and \(|A| = 1\), the inverse equals the adjoint.
For a \(2 \times 2\) matrix, \(|\mathrm{adj}\,A| = |A|^{2-1} = |A| = 1\).
The inverse of the transpose is the transpose of the inverse: \((A^{T})^{-1} = (A^{-1})^{T}\).
So \(A^{-1} = \mathrm{adj}\,A = \begin{bmatrix}3 & -2\\ -4 & 3\end{bmatrix}\), \(|\mathrm{adj}\,A| = 1\) and \((A^{T})^{-1} = \begin{bmatrix}3 & -4\\ -2 & 3\end{bmatrix}\).
Checking the statements
(A) \(A^{-1} = \begin{bmatrix}3 & -2\\ -4 & 3\end{bmatrix}\) — Correct. \(|A| = 1\), so \(A^{-1}\) is the adjoint.
(B) \(\mathrm{adj}\,A = A^{-1}\) — Correct. \(A^{-1} = \frac{1}{1}\,\mathrm{adj}\,A\).
(C) \(|\mathrm{adj}\,A| = 2\) — Incorrect. \(|\mathrm{adj}\,A| = 3 \cdot 3 - (-2)(-4) = 1\).
(D) \((A^{T})^{-1} = \begin{bmatrix}3 & -4\\ -2 & 3\end{bmatrix}\) — Correct. It is the transpose of \(A^{-1}\).