If A+B=left[begin{array}{lll}2 & 1 & 2 1 & 2 | vedclass

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(A) Given, $A+B=\left[\begin{array}{lll} 2 & 1 & 2 \ 1 & 2 & 0 \ 0 & 2 & 2\end{array}ight]$ and $AB=\left[\begin{array}{lll}1 & 2 & 2 \ 1 & 1 & 0

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$\left[\begin{array}{lll}4 & 6 & 6 \ 3 & 4 & 2 \ 1 & 6 & 3\end{array}ight]$

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(A) Given, $A+B=\left[\begin{array}{lll} 2 & 1 & 2 \ 1 & 2 & 0 \ 0 & 2 & 2\end{array}ight]$ and $AB=\left[\begin{array}{lll}1 & 2 & 2 \ 1 & 1 & 0 \ 1 & 2 & 1\end{array}ight]$.We need to find $A^2+B(A+B)$.Note that $A^2+B(A+B) = A^2+BA+B^2$.We know that $(A+B)^2 = (A+B)(A+B) = A^2+AB+BA+B^2$.Thus, $A^2+BA+B^2 = (A+B)^2 - AB$.First, calculate $(A+B)^2 = \left[\begin{array}{lll} 2 & 1 & 2 \ 1 & 2 & 0 \ 0 & 2 & 2\end{array}ight] 	imes \left[\begin{array}{lll} 2 & 1 & 2 \ 1 & 2 & 0 \ 0 & 2 & 2\end{array}ight] = \left[\begin{array}{lll} 4+1+0 & 2+2+4 & 4+0+4 \ 2+2+0 & 1+4+0 & 2+0+0 \ 0+2+0 & 0+4+4 & 0+0+4\end{array}ight] = \left[\begin{array}{lll} 5 & 8 & 8 \ 4 & 5 & 2 \ 2 & 8 & 4\end{array}ight]$.Now, $A^2+B(A+B) = \left[\begin{array}{lll} 5 & 8 & 8 \ 4 & 5 & 2 \ 2 & 8 & 4\end{array}ight] - \left[\begin{array}{lll} 1 & 2 & 2 \ 1 & 1 & 0 \ 1 & 2 & 1\end{array}ight] = \left[\begin{array}{lll} 4 & 6 & 6 \ 3 & 4 & 2 \ 1 & 6 & 3\end{array}ight]$.

If $A+B=\left[\begin{array}{lll}2 & 1 & 2 \\ 1 & 2 & 0 \\ 0 & 2 & 2\end{array}\right]$ and $AB=\left[\begin{array}{lll}1 & 2 & 2 \\ 1 & 1 & 0 \\ 1 & 2 & 1\end{array}\right]$, then $A^2+B(A+B)=$

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