If M = begin{bmatrix} 1 & 2 2 & 3 end{bmatri | vedclass

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(D) Given the equation $M^2 - \lambda M - I_2 = 0$,where $M = \begin{bmatrix} 1 & 2 \ 2 & 3 \end{bmatrix}$.First,calculate $M^2$:$M^2 = \begin{bmatri

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$4$

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(D) Given the equation $M^2 - \lambda M - I_2 = 0$,where $M = \begin{bmatrix} 1 & 2 \ 2 & 3 \end{bmatrix}$.First,calculate $M^2$:$M^2 = \begin{bmatrix} 1 & 2 \ 2 & 3 \end{bmatrix} \begin{bmatrix} 1 & 2 \ 2 & 3 \end{bmatrix} = \begin{bmatrix} (1)(1) + (2)(2) & (1)(2) + (2)(3) \ (2)(1) + (3)(2) & (2)(2) + (3)(3) \end{bmatrix} = \begin{bmatrix} 5 & 8 \ 8 & 13 \end{bmatrix}$.Substitute $M^2$,$M$,and $I_2$ into the equation:$\begin{bmatrix} 5 & 8 \ 8 & 13 \end{bmatrix} - \lambda \begin{bmatrix} 1 & 2 \ 2 & 3 \end{bmatrix} - \begin{bmatrix} 1 & 0 \ 0 & 1 \end{bmatrix} = \begin{bmatrix} 0 & 0 \ 0 & 0 \end{bmatrix}$.This simplifies to:$\begin{bmatrix} 5 - \lambda - 1 & 8 - 2\lambda - 0 \ 8 - 2\lambda - 0 & 13 - 3\lambda - 1 \end{bmatrix} = \begin{bmatrix} 0 & 0 \ 0 & 0 \end{bmatrix}$.Equating the elements:$4 - \lambda = 0 \Rightarrow \lambda = 4$.$8 - 2\lambda = 0 \Rightarrow 2\lambda = 8 \Rightarrow \lambda = 4$.$12 - 3\lambda = 0 \Rightarrow 3\lambda = 12 \Rightarrow \lambda = 4$.Since $\lambda = 4$ satisfies all equations,the correct option is $D$.

If $M = \begin{bmatrix} 1 & 2 \\ 2 & 3 \end{bmatrix}$ and $M^2 - \lambda M - I_2 = 0$,then $\lambda = $

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