If A = begin{bmatrix} ab & b^2 -a^2 & -ab en | vedclass

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(A) Given $A = \begin{bmatrix} ab & b^2 \ -a^2 & -ab \end{bmatrix}$.We calculate $A^2$:$A^2 = A \cdot A = \begin{bmatrix} ab & b^2 \ -a^2 & -ab \end

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$2$

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(A) Given $A = \begin{bmatrix} ab & b^2 \ -a^2 & -ab \end{bmatrix}$.We calculate $A^2$:$A^2 = A \cdot A = \begin{bmatrix} ab & b^2 \ -a^2 & -ab \end{bmatrix} \begin{bmatrix} ab & b^2 \ -a^2 & -ab \end{bmatrix}$$A^2 = \begin{bmatrix} (ab)(ab) + (b^2)(-a^2) & (ab)(b^2) + (b^2)(-ab) \ (-a^2)(ab) + (-ab)(-a^2) & (-a^2)(b^2) + (-ab)(-ab) \end{bmatrix}$$A^2 = \begin{bmatrix} a^2b^2 - a^2b^2 & ab^3 - ab^3 \ -a^3b + a^3b & -a^2b^2 + a^2b^2 \end{bmatrix} = \begin{bmatrix} 0 & 0 \ 0 & 0 \end{bmatrix} = O$.Since $A^2 = O$,it follows that $A^n = O$ for all $n \ge 2$.Therefore,the minimum value of $n$ is $2$.

If $A = \begin{bmatrix} ab & b^2 \\ -a^2 & -ab \end{bmatrix}$ and $A^n = O$,then the minimum value of $n$ is

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