Let \(A =\begin{bmatrix}2&3\\1&2\end{bmatrix}\) and \(B = \begin{bmatrix}4&-6\\-2&4\end{bmatrix}\), then
(A) \(\det (A^T) = 1\)
(B) \(AB = I\), where I is the identity matrix of order 2.
(C) \(A^{-1} =\begin{bmatrix}2&-3\\-1&2\end{bmatrix}\)
(D) \(\mathrm{adj}\,(B) = \begin{bmatrix}4&2\\6&4\end{bmatrix}\)
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → (A) and (C) only
For a 2 × 2 matrix, the adjoint swaps the diagonal entries and changes the signs of the other two entries.
\(|A| = 4 - 3 = 1\)
\(|A^T| = |A| = 1\)
\( \displaystyle AB = \begin{bmatrix}8 - 6&-12 + 12\\4 - 4&-6 + 8\end{bmatrix} \)
\( \displaystyle = \begin{bmatrix}2&0\\0&2\end{bmatrix} = 2I \)
\( \displaystyle A^{-1} = \frac{1}{|A|}\,\mathrm{adj}\,A = \begin{bmatrix}2&-3\\-1&2\end{bmatrix} \)
\( \displaystyle \mathrm{adj}\,B = \begin{bmatrix}4&6\\2&4\end{bmatrix} \)
So \(|A^T| = 1\), \(AB = 2I\), \(A^{-1} = \begin{bmatrix}2&-3\\-1&2\end{bmatrix}\) and \(\mathrm{adj}\,B = \begin{bmatrix}4&6\\2&4\end{bmatrix}\).
Checking the statements
(A) \(\det (A^T) = 1\) — Correct. \(|A^T| = |A| = 1\).
(B) \(AB = I\), where I is the identity matrix of order 2. — Incorrect. \(AB = 2I\), not I.
(C) \(A^{-1} =\begin{bmatrix}2&-3\\-1&2\end{bmatrix}\) — Correct. Since \(|A| = 1\), \(A^{-1} = \mathrm{adj}\,A\), which is this matrix.
(D) \(\mathrm{adj}\,(B) = \begin{bmatrix}4&2\\6&4\end{bmatrix}\) — Incorrect. \(\mathrm{adj}\,B = \begin{bmatrix}4&6\\2&4\end{bmatrix}\); the entries 6 and 2 are interchanged.