Let A = begin{bmatrix} 3 & 0 & 3 0 & 3 & 0 | vedclass

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(B) Given the matrix $A = \begin{bmatrix} 3 & 0 & 3 \ 0 & 3 & 0 \ 3 & 0 & 3 \end{bmatrix}$.The characteristic equation is given by $\operatorname{de

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$0, 3, 6$

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(B) Given the matrix $A = \begin{bmatrix} 3 & 0 & 3 \ 0 & 3 & 0 \ 3 & 0 & 3 \end{bmatrix}$.The characteristic equation is given by $\operatorname{det}(A - \lambda I_{3}) = 0$.$\begin{vmatrix} 3-\lambda & 0 & 3 \ 0 & 3-\lambda & 0 \ 3 & 0 & 3-\lambda \end{vmatrix} = 0$.Expanding along the second row:$(3-\lambda) \begin{vmatrix} 3-\lambda & 3 \ 3 & 3-\lambda \end{vmatrix} = 0$.$(3-\lambda) [(3-\lambda)^2 - 9] = 0$.$(3-\lambda) [9 + \lambda^2 - 6\lambda - 9] = 0$.$(3-\lambda) (\lambda^2 - 6\lambda) = 0$.$(3-\lambda) \lambda (\lambda - 6) = 0$.Thus, the roots are $\lambda = 0, 3, 6$.

Let $A = \begin{bmatrix} 3 & 0 & 3 \\ 0 & 3 & 0 \\ 3 & 0 & 3 \end{bmatrix}$. Then, the roots of the equation $\operatorname{det}(A - \lambda I_{3}) = 0$ (where $I_{3}$ is the identity matrix of order $3$) are

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