Let $A$ be a $2 \times 2$ real matrix with entries from $\{0,1\}$ and $|\mathrm{A}| \neq 0$. Consider the following two statements :
(P) If $A \neq I_{2}$, then $|A|=-1$
(Q) If $|\mathrm{A}|=1$, then $\operatorname{tr}(\mathrm{A})=2$,
where $\mathrm{I}_{2}$ denotes $2 \times 2$ identity matrix and tr(A) denotes the sum of the diagonal entries of A. Then:
Correct Option: , 4
$|\mathrm{A}| \neq 0$
For $(\mathrm{P}): \mathrm{A} \neq \mathrm{I}_{2}$
So, $A=\left[\begin{array}{ll}0 & 1 \\ 1 & 0\end{array}\right]$ or $\left[\begin{array}{ll}1 & 1 \\ 1 & 0\end{array}\right]$ or $\left[\begin{array}{ll}0 & 1 \\ 1 & 1\end{array}\right]$ or $\left[\begin{array}{ll}1 & 1 \\ 0 & 1\end{array}\right]$
or $\left[\begin{array}{ll}1 & 0 \\ 1 & 1\end{array}\right]$
$|\mathrm{A}|$ can be $-1$ or 1
So (P) is false.
For $(Q) ;|A|=1$
$A=\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]$ or $\left[\begin{array}{ll}1 & 1 \\ 0 & 1\end{array}\right]$ or $\left[\begin{array}{ll}1 & 0 \\ 1 & 1\end{array}\right]$
$\Rightarrow \operatorname{tr}(\mathrm{A})=2$
$\Rightarrow \mathrm{Q}$ is true