Let \(A = \begin{bmatrix}4 & -1\\ 3 & 1\end{bmatrix}\). Consider the following statements:
(A) \(\mathrm{adj}\,A = \begin{bmatrix}1 & 1\\ -3 & 4\end{bmatrix}\)
(B) \(A^{-1} = \displaystyle\frac{1}{7}\begin{bmatrix}1 & -1\\ 3 & 4\end{bmatrix}\)
(C) \(|\mathrm{adj}\,A| = 49\)
(D) \(|A^{-1}| = \displaystyle\frac{1}{7}\)
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → (A) and (D) only
The determinant of A:
\( |A| = 4 \times 1 - (-1) \times 3 = 4 + 3 = 7 \)
For a 2 × 2 matrix the adjoint swaps the diagonal entries and changes the signs of the other two:
\( \displaystyle \mathrm{adj}\,A = \begin{bmatrix}1 & 1\\ -3 & 4\end{bmatrix} \)
\( \displaystyle A^{-1} = \frac{1}{7}\begin{bmatrix}1 & 1\\ -3 & 4\end{bmatrix} \)
For order n = 2, \(|\mathrm{adj}\,A| = |A|^{n-1} = |A|\):
\( |\mathrm{adj}\,A| = 1 \times 4 - 1 \times (-3) = 7 \)
\( \displaystyle |A^{-1}| = \frac{1}{|A|} = \frac{1}{7} \)
So \(|A| = 7\), \(\mathrm{adj}\,A = \begin{bmatrix}1 & 1\\ -3 & 4\end{bmatrix}\), \(A^{-1} = \frac{1}{7}\begin{bmatrix}1 & 1\\ -3 & 4\end{bmatrix}\), \(|\mathrm{adj}\,A| = 7\) and \(|A^{-1}| = \frac{1}{7}\).
Checking the statements
(A) \(\mathrm{adj}\,A = \begin{bmatrix}1 & 1\\ -3 & 4\end{bmatrix}\) — Correct. Diagonal 4, 1 swapped; −1 and 3 change sign.
(B) \(A^{-1} = \frac{1}{7}\begin{bmatrix}1 & -1\\ 3 & 4\end{bmatrix}\) — Incorrect. The off-diagonal entries must be 1 and −3, giving \(\frac{1}{7}\begin{bmatrix}1 & 1\\ -3 & 4\end{bmatrix}\).
(C) \(|\mathrm{adj}\,A| = 49\) — Incorrect. For order 2, \(|\mathrm{adj}\,A| = |A|^{1} = 7\).
(D) \(|A^{-1}| = \frac{1}{7}\) — Correct. \(|A^{-1}| = \frac{1}{|A|} = \frac{1}{7}\).