If A=left[begin{array}{ccc}1 & 1 & 3 5 & 2 & | vedclass

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(B) Given $A=\left[\begin{array}{ccc}1 & 1 & 3 \ 5 & 2 & 6 \ -2 & -1 & -3\end{array}ight]$.The characteristic equation is given by $|A-\lambda I|=

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$\left[\begin{array}{ccc}4 & 1 & 3 \ 5 & 5 & 6 \ -2 & -1 & 0\end{array}ight]$

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(B) Given $A=\left[\begin{array}{ccc}1 & 1 & 3 \ 5 & 2 & 6 \ -2 & -1 & -3\end{array}ight]$.The characteristic equation is given by $|A-\lambda I|=0$.$\left|\begin{array}{ccc}1-\lambda & 1 & 3 \ 5 & 2-\lambda & 6 \ -2 & -1 & -3-\lambda\end{array}ight|=0$Expanding the determinant:$(1-\lambda)[(2-\lambda)(-3-\lambda) - (-6)] - 1[5(-3-\lambda) - (-12)] + 3[5(-1) - (-2)(2-\lambda)] = 0$$(1-\lambda)[\lambda^2+\lambda-6+6] - 1[-15-5\lambda+12] + 3[-5+4-2\lambda] = 0$$(1-\lambda)(\lambda^2+\lambda) - 1(-5\lambda-3) + 3(-2\lambda-1) = 0$$\lambda^2+\lambda-\lambda^3-\lambda^2 + 5\lambda+3 - 6\lambda-3 = 0$$-\lambda^3 = 0 \Rightarrow \lambda^3 = 0$.By the Cayley-Hamilton theorem,every matrix satisfies its characteristic equation,so $A^3 = 0$.Since $A^3 = 0$,it follows that $A^4 = A^3 \cdot A = 0$ and $A^5 = A^3 \cdot A^2 = 0$.Therefore,$A+A^3+A^4+A^5+3I = A + 0 + 0 + 0 + 3I = A + 3I$.$A+3I = \left[\begin{array}{ccc}1 & 1 & 3 \ 5 & 2 & 6 \ -2 & -1 & -3\end{array}ight] + \left[\begin{array}{ccc}3 & 0 & 0 \ 0 & 3 & 0 \ 0 & 0 & 3\end{array}ight] = \left[\begin{array}{ccc}4 & 1 & 3 \ 5 & 5 & 6 \ -2 & -1 & 0\end{array}ight]$.

If $A=\left[\begin{array}{ccc}1 & 1 & 3 \\ 5 & 2 & 6 \\ -2 & -1 & -3\end{array}\right]$,then $A+A^3+A^4+A^5+3 I=$

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