Let $\mathrm{A}$, other than $\mathrm{I}$ or $-\mathrm{I}$, be a $2 \times 2$ real matrix such that…
Let $\mathrm{A}$, other than $\mathrm{I}$ or $-\mathrm{I}$, be a $2 \times 2$ real matrix such that $\mathrm{A}^2=\mathrm{I}$, I being the unit matrix. Let $\operatorname{Tr}(\mathrm{A})$ be the sum of diagonal elements of A.
Statement-1: $\operatorname{Tr}(\mathrm{A})=0$
Statement-2: $\operatorname{det}(\mathrm{A})=-1$
Statement-1 is true; Statement- 2 is false.
Statement-1 is true; Statement-2 is true; Statement-2 is not a correct explanation for Statement-1.
Statement-1 is true; Statement-2 is true; Statement-2 is a correct explanation for Statement-1.
Statement-1 is false; Statement- 2 is true.
Solution
$\left[\begin{array}{ll}a & b \\ c & d\end{array}\right]\left[\begin{array}{ll}a & b \\ c & d\end{array}\right]=\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]$
$
\begin{aligned}
& {\left[\begin{array}{ll}
a^2+b c & a b+b d \\
a c+c d & b c+d^2
\end{array}\right]=\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right]} \\
& b(a+d)=0, b=0 \text { or } a=-d \\
& c(a+d)=0, c=0 \text { or } a=-d \\
& a^2+b c=1, b c+d^2=1
\end{aligned}
$
' $a$ ' and ' $d$ ' are diagonal elements $a+d=0$ statement- 1 is correct.
Now, $\operatorname{det}(A)=a d-b c$
Now, from (3) $a^2+b c=1$ and $d^2+b c=1$
So, $a^2-d^2=0$
Adding $a^2+d^2+2 b c=2$
$
=(a+d)^2-2 a d+2 b c=2
$
or $0-2(a d-b c)=2$
So, $a d-b c=1 \Rightarrow \operatorname{det}(A)=-1$
So, statement $-2$ is also true.
But statement $-2$ is not the correct explanation of statement-I