Question-66
Which of the following statements are true for square matrices \(\displaystyle A\) and \(\displaystyle B\) of order \(\displaystyle n\)?
Option-1
We have the following identity:
\[ \begin{equation*} \text{det}( cA) =c^{n}\text{det}( A) \end{equation*} \]
Plugging \(\displaystyle c=-1\), we see that \(\displaystyle \text{det}( -A)\) being equal to \(\displaystyle \text{det}( A)\) is dependent on the parity of \(\displaystyle n\).
Option-2
We have the following identity:
\[ \begin{equation*} \text{det}( AB) =\text{det}( A)\text{det}( B) \end{equation*} \]
It follows that \(\displaystyle \text{det}( AB) =\text{det}( BA)\).
Option-3
As a counter example, consider \(\displaystyle A=B=I\) with \(\displaystyle n=2\).
\[ \begin{equation*} \text{det}( A+B) =4,\ \text{det}( A) =\text{det}( B) =1 \end{equation*} \]
Option-4
If the reduced row echelon form of \(\displaystyle A\) is \(\displaystyle I\), all that we can say is that \(\displaystyle \text{det}( A) \neq 0\) since \(\displaystyle A\) is invertible. We can furnish a counter example here:
\[ \begin{equation*} A=\begin{bmatrix} 1 & 0\\ 0 & -1 \end{bmatrix} \end{equation*} \]
\(\displaystyle \text{det}( A) =-1\) even though its RREF is \(\displaystyle I\).
Option-5
Without loss of generality, let the first column be expressible as a linear combination of the remaining \(\displaystyle n-1\) columns:
\[ \begin{equation*} C_{1} =a_{2} C_{2} +\cdots +a_{n} C_{n} \end{equation*} \]
We can now use the following column operation to get zeros in the first column:
\[ \begin{equation*} C_{1}\rightarrow C_{1} -a_{2} C_{2} -\cdots -a_{n} C_{n} \end{equation*} \]
The determinant of the resulting matrix is zero. Since this column operation doesn’t change the value of the determinant, we conclude that the original matrix also has determinant zero.