Let the matrix \(A = \begin{bmatrix}2&3\\1&4\end{bmatrix}\). Then which of the following are true?
(A) \(\mathrm{adj}\, A =\begin{bmatrix}4&-3\\-1&2\end{bmatrix}\)
(B) \(\det (A) = 5\)
(C) \(\det (\mathrm{adj}A) = 25\)
(D) If \(A^3 = \begin{bmatrix}a&b\\c&d\end{bmatrix}\) then \(a + b = c + d\)
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → (A), (B) and (D) only
(A) \(\mathrm{adj}\, A =\begin{bmatrix}4&-3\\-1&2\end{bmatrix}\) — Correct. For a 2 × 2 matrix, the adjoint swaps the diagonal entries and changes the signs of the other two:
\(\displaystyle \mathrm{adj}\begin{bmatrix}2&3\\1&4\end{bmatrix} = \begin{bmatrix}4&-3\\-1&2\end{bmatrix}\)
(B) \(\det (A) = 5\) — Correct. \(\det(A) = 2 \times 4 - 3 \times 1 = 5\).
(C) \(\det (\mathrm{adj}A) = 25\) — Incorrect. For a matrix of order n, \(|\mathrm{adj}\,A| = |A|^{n-1}\). Here n = 2:
\(\det(\mathrm{adj}\,A) = 5^{1} = 5\)
Directly: \(4 \times 2 - (-3)(-1) = 5\).
(D) If \(A^3 = \begin{bmatrix}a&b\\c&d\end{bmatrix}\) then \(a + b = c + d\) — Correct. Multiplying out:
\(\displaystyle A^2 = \begin{bmatrix}2&3\\1&4\end{bmatrix}\begin{bmatrix}2&3\\1&4\end{bmatrix} = \begin{bmatrix}7&18\\6&19\end{bmatrix}\)
\(\displaystyle A^3 = \begin{bmatrix}2&3\\1&4\end{bmatrix}\begin{bmatrix}7&18\\6&19\end{bmatrix} = \begin{bmatrix}32&93\\31&94\end{bmatrix}\)
\( \displaystyle a + b = 32 + 93 = 125 \)
\( \displaystyle c + d = 31 + 94 = 125 \)
Shortcut: each row of A adds up to 5, so A multiplied by the column \(\begin{bmatrix}1\\1\end{bmatrix}\) gives 5 times that column. Repeating, \(A^3\begin{bmatrix}1\\1\end{bmatrix} = 125\begin{bmatrix}1\\1\end{bmatrix}\), so both row sums of \(A^3\) are 125 with no multiplication.