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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

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