Let $A=\left[\begin{array}{ll}{a} & {b} \\ {c} & {d}\end{array}\right]$

It is given that,

$A^{2}=I$

${\therefore \left[ {\begin{array}{*{20}{c}}
a&b\\
c&d
\end{array}} \right]\left[ {\begin{array}{*{20}{c}}
a&b\\
c&d
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
1&0\\
0&1
\end{array}} \right]}$

${ \Rightarrow \left[ {\begin{array}{*{20}{c}}
{{a^2} + bc}&{ab + bd}\\
{ac + cd}&{bc + {d^2}}
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
1&0\\
0&1
\end{array}} \right]}$

${ \Rightarrow {a^2} + bc = 1 \to 1}$

${ \Rightarrow ab + bd = 0 \Rightarrow b(a + d) = 0 \Rightarrow a =  – d \to (2) \ldots [Asb \ne 0]}$

So, we can write

$A=\left[\begin{array}{cc}{a} & {b} \\ {c} & {-a}\end{array}\right]$

$\therefore \operatorname{Tr}(A)=a+(-a)=0$

$|A|=-a^{2}-b c=-\left(a^{2}+b c\right)=-1$

So, first statement is true but second statement is false.