Let A be a 2 times 2 matrix with real entries | vedclass

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(C) Given $A^2 = I$. Taking the determinant on both sides,we get $\operatorname{det}(A^2) = \operatorname{det}(I) = 1$.Since $\operatorname{det}(A^2)

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Statement $I$ is true,statement $II$ is false.

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(C) Given $A^2 = I$. Taking the determinant on both sides,we get $\operatorname{det}(A^2) = \operatorname{det}(I) = 1$.Since $\operatorname{det}(A^2) = (\operatorname{det}(A))^2$,we have $(\operatorname{det}(A))^2 = 1$,which implies $\operatorname{det}(A) = 1$ or $\operatorname{det}(A) = -1$.If $\operatorname{det}(A) = 1$,then $A$ is invertible and $A^2 = I$ implies $A = A^{-1}$. For a $2 	imes 2$ matrix,if $\operatorname{det}(A) = 1$ and $A^2 = I$,then $A$ must be $I$ or $-I$.Thus,if $A 
eq I$ and $A 
eq -I$,we must have $\operatorname{det}(A) = -1$. So,Statement $I$ is true.Now consider $A = \begin{bmatrix} 0 & 1 \ 1 & 0 \end{bmatrix}$. Here $A^2 = I$,$A 
eq I$,and $A 
eq -I$.The trace $\operatorname{Tr}(A) = 0 + 0 = 0$.Since we found a case where $\operatorname{Tr}(A) = 0$ while $A 
eq \pm I$,Statement $II$ is false.

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