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

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(B) Given $A = \begin{bmatrix} 0 & 1 \ 0 & 0 \end{bmatrix}$ and $I = \begin{bmatrix} 1 & 0 \ 0 & 1 \end{bmatrix}$.First,calculate $aI + bA$:$aI + bA

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$a^2I + 2abA$

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(B) Given $A = \begin{bmatrix} 0 & 1 \ 0 & 0 \end{bmatrix}$ and $I = \begin{bmatrix} 1 & 0 \ 0 & 1 \end{bmatrix}$.First,calculate $aI + bA$:$aI + bA = a \begin{bmatrix} 1 & 0 \ 0 & 1 \end{bmatrix} + b \begin{bmatrix} 0 & 1 \ 0 & 0 \end{bmatrix} = \begin{bmatrix} a & 0 \ 0 & a \end{bmatrix} + \begin{bmatrix} 0 & b \ 0 & 0 \end{bmatrix} = \begin{bmatrix} a & b \ 0 & a \end{bmatrix}$.Now,calculate $(aI + bA)^2$:$(aI + bA)^2 = \begin{bmatrix} a & b \ 0 & a \end{bmatrix} \begin{bmatrix} a & b \ 0 & a \end{bmatrix} = \begin{bmatrix} a^2 + 0 & ab + ba \ 0 + 0 & 0 + a^2 \end{bmatrix} = \begin{bmatrix} a^2 & 2ab \ 0 & a^2 \end{bmatrix}$.This can be written as:$a^2 \begin{bmatrix} 1 & 0 \ 0 & 1 \end{bmatrix} + 2ab \begin{bmatrix} 0 & 1 \ 0 & 0 \end{bmatrix} = a^2I + 2abA$.

If $A = \begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix}$,$I$ is the identity matrix of order $2$,and $a, b$ are arbitrary constants,then $(aI + bA)^2$ is equal to

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