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

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(D) Given $A^2 = I$. Taking the determinant on both sides,we get $\det(A^2) = \det(I) = 1$.Since $\det(A^2) = (\det(A))^2$,we have $(\det(A))^2 = 1$,w

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Statement-$1$ is true,Statement-$2$ is false.

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(D) Given $A^2 = I$. Taking the determinant on both sides,we get $\det(A^2) = \det(I) = 1$.Since $\det(A^2) = (\det(A))^2$,we have $(\det(A))^2 = 1$,which implies $\det(A) = 1$ or $\det(A) = -1$.If $\det(A) = 1$,the characteristic equation is $\lambda^2 - tr(A)\lambda + 1 = 0$. Since $A^2 = I$,the eigenvalues are $\pm 1$. If $\det(A) = 1$,the eigenvalues are either $(1, 1)$ or $(-1, -1)$.If eigenvalues are $(1, 1)$,then $A = I$. If eigenvalues are $(-1, -1)$,then $A = -I$.Thus,if $A 
eq I$ and $A 
eq -I$,we must have $\det(A) = -1$. So,Statement-$1$ is true.For Statement-$2$,if $A^2 = I$,the eigenvalues are $\lambda_1, \lambda_2 \in \{1, -1\}$.If $A 
eq I$ and $A 
eq -I$,the eigenvalues must be $1$ and $-1$.The trace of $A$ is the sum of eigenvalues,so $tr(A) = 1 + (-1) = 0$.Therefore,Statement-$2$ is false because $tr(A)$ must be $0$ in this case.

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