If A = begin{bmatrix} 0 & -1 1 & 0 end{bmatr | vedclass

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(A) Given $A = \begin{bmatrix} 0 & -1 \ 1 & 0 \end{bmatrix}$.First,calculate $A^2$:$A^2 = \begin{bmatrix} 0 & -1 \ 1 & 0 \end{bmatrix} \begin{bmatri

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$A^2 + I = A(A^2 - I)$

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(A) Given $A = \begin{bmatrix} 0 & -1 \ 1 & 0 \end{bmatrix}$.First,calculate $A^2$:$A^2 = \begin{bmatrix} 0 & -1 \ 1 & 0 \end{bmatrix} \begin{bmatrix} 0 & -1 \ 1 & 0 \end{bmatrix} = \begin{bmatrix} -1 & 0 \ 0 & -1 \end{bmatrix} = -I$.Now,calculate higher powers:$A^3 = A^2 \cdot A = -I \cdot A = -A = \begin{bmatrix} 0 & 1 \ -1 & 0 \end{bmatrix}$.$A^4 = (A^2)^2 = (-I)^2 = I$.Check the options:$A$. $A^2 + I = -I + I = 0$. $A(A^2 - I) = A(-I - I) = A(-2I) = -2A$. Since $0 
eq -2A$,this is incorrect.$B$. $A^4 - I = I - I = 0$. $A^2 + I = -I + I = 0$. Thus $0 = 0$ (Correct).$C$. $A^3 + I = -A + I$. $A(A^3 - I) = A(-A - I) = -A^2 - A = I - A$. Since $-A + I = I - A$,this is correct.$D$. $A^3 - I = -A - I$. $A(A - I) = A^2 - A = -I - A$. Since $-A - I = -I - A$,this is correct.Therefore,the statement in option $A$ is not correct.

If $A = \begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix}$,then which one of the following statements is not correct?

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