Let A be a 2 × 2 matrix with non-zero entries and let $A^2=I$ where I is 2 × 2 identity matrix.

Statement-1 $tr (A) = 0$

Statement-2 $|A|=1$

Statement-1 is True, Statement-2 is False.

Let $A =\begin{bmatrix}a&b\\c&d\end{bmatrix}, a, b, c, d ≠0$. Then,

$A^2=\begin{bmatrix}a&b\\c&d\end{bmatrix}\begin{bmatrix}a&b\\c&d\end{bmatrix}=\begin{bmatrix}a^2+bc&ab+bd\\ac+dc&bc+d^2\end{bmatrix}$

$∴A^2 = I$

$⇒a^2 +bc=1, bc+d^2=1, ac+dc=0$ and $ab+bd=0$

$⇒(a^2+bc) -(d^2 + bc) = 0, c (a + d) = 0$ and $(a + d) b=0$

$⇒a^2-d^2 =0$ and $a + d=0$   $[∵ b≠0, c ≠ 0]$

$⇒a+d=0⇒ Tr (A) = 0$

So, statement-1 is true.

$|A|=\begin{vmatrix}a&b\\c&d\end{vmatrix}=ad-bc=-a^2-bc$   $[∵d=-a]$

$⇒|A|=-1$  $[∵a^2+bc=1]$

So, statement-2 is not true.