Evaluate A^2+2I if A=begin{bmatrix} 1 & 0 1 | vedclass

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(B) Given $A = \begin{bmatrix} 1 & 0 \ 1 & 2 \end{bmatrix}$ and $I = \begin{bmatrix} 1 & 0 \ 0 & 1 \end{bmatrix}$.First,calculate $A^2 = A 	imes A$

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$3$

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(B) Given $A = \begin{bmatrix} 1 & 0 \ 1 & 2 \end{bmatrix}$ and $I = \begin{bmatrix} 1 & 0 \ 0 & 1 \end{bmatrix}$.First,calculate $A^2 = A 	imes A$:$A^2 = \begin{bmatrix} 1 & 0 \ 1 & 2 \end{bmatrix} \begin{bmatrix} 1 & 0 \ 1 & 2 \end{bmatrix} = \begin{bmatrix} (1)(1)+(0)(1) & (1)(0)+(0)(2) \ (1)(1)+(2)(1) & (1)(0)+(2)(2) \end{bmatrix} = \begin{bmatrix} 1 & 0 \ 3 & 4 \end{bmatrix}$.Now,calculate $2I$:$2I = 2 \begin{bmatrix} 1 & 0 \ 0 & 1 \end{bmatrix} = \begin{bmatrix} 2 & 0 \ 0 & 2 \end{bmatrix}$.Finally,add $A^2$ and $2I$:$A^2 + 2I = \begin{bmatrix} 1 & 0 \ 3 & 4 \end{bmatrix} + \begin{bmatrix} 2 & 0 \ 0 & 2 \end{bmatrix} = \begin{bmatrix} 3 & 0 \ 3 & 6 \end{bmatrix}$.Since $3A = 3 \begin{bmatrix} 1 & 0 \ 1 & 2 \end{bmatrix} = \begin{bmatrix} 3 & 0 \ 3 & 6 \end{bmatrix}$,we conclude that $A^2 + 2I = 3A$.

Evaluate $A^2+2I$ if $A=\begin{bmatrix} 1 & 0 \\ 1 & 2 \end{bmatrix}$. (in $A$)

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