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

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

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

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(D) Given the matrix $A = \begin{bmatrix} 0 & -1 \ 1 & 0 \end{bmatrix}$.First,calculate $A^2$: $A^2 = \begin{bmatrix} 0 & -1 \ 1 & 0 \end{bmatrix} \begin{bmatrix} 0 & -1 \ 1 & 0 \end{bmatrix} = \begin{bmatrix} -1 & 0 \ 0 & -1 \end{bmatrix} = -I$.This implies $A^2 + I = 0$,or $A^2 = -I$.From this,$A^3 = A^2 \cdot A = -I \cdot A = -A$.Also,$A^4 = (A^2)^2 = (-I)^2 = I$.Now,let us check each option:Option $A$: $A^3 - I = -A - I$ and $A(A - I) = A^2 - A = -I - A$. Thus,$A^3 - I = A(A - I)$ is correct.Option $B$: $A^3 + I = -A + I$ and $A(A^3 - I) = A(-A - I) = -A^2 - A = I - A$. Thus,$A^3 + I = A(A^3 - I)$ is correct.Option $C$: $A^4 - I = I - I = 0$ and $A^2 + I = -I + I = 0$. Thus,$A^4 - I = A^2 + I$ is correct.Option $D$: $A^2 + I = -I + I = 0$ and $A(A^2 - I) = A(-I - I) = A(-2I) = -2A$. Since $0 
eq -2A$,this option is incorrect.

If $A = \begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix}$,then the incorrect option among the following is

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