Let the matrix $A = \begin{bmatrix}2&3\\1&4\end{bmatrix}$. Then which of the following are true?

(A) $adj\, A =\begin{bmatrix}4&-3\\-1&2\end{bmatrix}$
(B) $det (A) = 5$
(C) $det (adjA) = 25$
(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:

(A), (B) and (D) only

The correct answer is Option (4) → (A), (B) and (D) only

Given matrix:

$A=\begin{pmatrix}2 & 3 \\ 1 & 4\end{pmatrix}$

(A) adj A

For a $2\times 2$ matrix $\begin{pmatrix}a & b \\ c & d\end{pmatrix}$,

$\text{adj }A=\begin{pmatrix}d & -b \\ -c & a\end{pmatrix}$

So,

$\text{adj }A=\begin{pmatrix}4 & -3 \\ -1 & 2\end{pmatrix}$

Therefore, (A) is True.

(B) $\det(A)=$

$\det(A)=2\cdot 4-3\cdot 1=8-3=5$

(B) is true.

(C) $\det(\text{adj }A)=$

For a $2\times 2$ matrix: $\det(\text{adj }A)=\det(A)^{1}$

$\det(\text{adj }A)=5^{1}=5$

(C) is False.

(D) Given $A^{3}=\begin{pmatrix}a & b \\ c & d\end{pmatrix}$, check whether $a+b=c+d$.

Compute $A^{2}$:

$A^{2} =\begin{pmatrix}2 & 3 \\ 1 & 4\end{pmatrix} \begin{pmatrix}2 & 3 \\ 1 & 4\end{pmatrix} =\begin{pmatrix}7 & 18 \\ 6 & 19\end{pmatrix}$

Now compute $A^{3}=A\cdot A^{2}$:

$A^{3} =\begin{pmatrix}2 & 3 \\ 1 & 4\end{pmatrix} \begin{pmatrix}7 & 18 \\ 6 & 19\end{pmatrix} =\begin{pmatrix}32 & 93 \\ 31 & 94\end{pmatrix}$

Thus: $a=32,\ b=93,\ c=31,\ d=94$

$a+b=32+93=125$

$c+d=31+94=125$

So $a+b=c+d$

(D) is true.

Therefore, the true statements are (B), (A), and (D).