Transpose of the matrix is obtained by changing the rows to columns and columns to rows.
Let’s consider $2$ matrix $A$ and $B$.
$A=\begin{bmatrix} 1&2 \\ 0&1 \end{bmatrix}$
$B=\begin{bmatrix} 1&1 \\ 2&3 \end{bmatrix}$
1) $(A^T)^T=A$:
$A^T=\begin{bmatrix} 1&0 \\ 2&1 \end{bmatrix}$
$ (A^T)^T=\begin{bmatrix} 1&2 \\ 0&1 \end{bmatrix}\equiv A$
$\therefore (A^T)^T=A$:
2) $|A^T|=|A|$
$|A^T|=1-0=1$
$|A|=1-0=1$
$\therefore |A^T|=|A|$
3) $(AB)^T=A^TB^T$
$AB=\begin{bmatrix} 5&7 \\ 2&3 \end{bmatrix},(AB)^T=\begin{bmatrix} 5&2 \\ 7&3 \end{bmatrix}$
$A^T=\begin{bmatrix} 1&0 \\ 2&1 \end{bmatrix},B^T=\begin{bmatrix} 1&2 \\ 1&3 \end{bmatrix}$
$A^TB^T=\begin{bmatrix} 3&7 \\ 1&3 \end{bmatrix}$
$\therefore (AB)^T\neq(A^T)(B^T)$
4) $(A+B)^T=A^T+B^T$
$A+B=\begin{bmatrix} 2&3 \\ 2&4 \end{bmatrix}$
$(A+B)^T=\begin{bmatrix} 2&2 \\ 3&4 \end{bmatrix}$
$A^T=\begin{bmatrix} 1&0 \\ 2&1 \end{bmatrix}$
$B^T=\begin{bmatrix} 1&2 \\ 1&3 \end{bmatrix}$
$A^T+B^T=\begin{bmatrix} 2&2 \\ 3&4 \end{bmatrix}$
$\therefore (A+B)^T=A^T+B^T$
$\therefore$ Option $A,B, D$ are true.
Transpose of Matrix