Let A, other than I or - I, be a 2 × 2 real matrix such that A^2= I, I being the unit matrix. Let Tr( A) be the sum of diagonal elements of A.…

$\left[\begin{array}{ll}a & b \\ c & d\end{array}\right]\left[\begin{array}{ll}a & b \\ c & d\end{array}\right]=\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]$[ac​bd​][ac​bd​]=[10​01​] $$\begin{aligned} & {\left[\begin{array}{ll} a^2+b c & a b+b d \\ a c+c d & b c+d^2 \end{array}\right]=\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right]} \\ & b(a+d)=0, b=0 \text { or } a=-d \\ & c(a+d)=0, c=0 \text { or } a=-d \\ & a^2+b c=1, b c+d^2=1 \end{aligned}$$​[a2+bcac+cd​ab+bdbc+d2​]=[10​01​]b(a+d)=0,b=0 or a=−dc(a+d)=0,c=0 or a=−da2+bc=1,bc+d2=1​ ' $a$a ' and ' $d$d ' are diagonal elements $a+d=0$a+d=0 statement- 1 is correct. Now, $\operatorname{det}(A)=a d-b c$det(A)=ad−bc Now, from (3) $a^2+b c=1$a2+bc=1 and $d^2+b c=1$d2+bc=1 So, $a^2-d^2=0$a2−d2=0 Adding $a^2+d^2+2 b c=2$a2+d2+2bc=2 $$=(a+d)^2-2 a d+2 b c=2$$=(a+d)2−2ad+2bc=2 or $0-2(a d-b c)=2$0−2(ad−bc)=2 So, $a d-b c=1 \Rightarrow \operatorname{det}(A)=-1$ad−bc=1⇒det(A)=−1 So, statement $-2$−2 is also true. But statement $-2$−2 is not the correct explanation of statement-I