For option $(A):$ The determinant of a matrix remains unaltered if its rows are changed with columns. That is, $\mid A \mid = \mid A^{T} \mid .$
Since the row-wise expansion is same as the column-wise expansion, the result holds good.
For option $(B):$ If two rows or columns are interchanged, then the value of determinant changes its sign.
For example,
Let $\mid A \mid = \begin{vmatrix} a & b \\ c & d \end{vmatrix} = ad - bc$
Interchanging rows
$\mid B \mid = \begin{vmatrix} c & d \\ a & b \end{vmatrix} = bc - ad = -(ad - bc)$
Interchanging columns
$\mid C \mid = \begin{vmatrix} b & a \\ d & c \end{vmatrix} = bc - ad = -(ad - bc)$
For option $(C):$ Let the determinant of a matrix be $x.$ If we interchange the two identical rows of this matrix, then by the property in option B, the determinant of the new matrix is $-x,$ but overall the matrix will be same as we have interchanged only the two identical rows.
So, $x = -x, \implies x = 0.$
Hence, the determinant is zero.
Option $(D)$ is also TRUE. Some examples:
Let $A = \begin{bmatrix} a & b \\ c & d \end{bmatrix}$
$\implies \mid A \mid = \begin{vmatrix} a & b \\ c & d \end{vmatrix} = ad - bc$
Now, $\mid B \mid = \begin{vmatrix} ka & kb \\ c & d \end{vmatrix} = k ad - k bc = k(ad - bc)$
$\text{(OR)}$
Now, $\mid C \mid = \begin{vmatrix} ka & b \\ kc & d \end{vmatrix} = k ad - k bc = k(ad - bc)$
So, the correct answer is $A; B; C;D.$
Ref:
https://www.math.fsu.edu/~fusaro/EngMath/Ch6/SPDR.html