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(A) The characteristic equation of matrix $A$ is given by $|A - \lambda I| = 0$.$|A - \lambda I| = \begin{vmatrix} 1-\lambda & 2 & 3 \ 0 & 5-\lambda

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$(5, 2)$

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(A) The characteristic equation of matrix $A$ is given by $|A - \lambda I| = 0$.$|A - \lambda I| = \begin{vmatrix} 1-\lambda & 2 & 3 \ 0 & 5-\lambda & 4 \ 0 & 3 & 2-\lambda \end{vmatrix} = 0$Expanding along the first column:$(1-\lambda) [(5-\lambda)(2-\lambda) - 12] = 0$$(1-\lambda) [10 - 5\lambda - 2\lambda + \lambda^2 - 12] = 0$$(1-\lambda) [\lambda^2 - 7\lambda - 2] = 0$$\lambda^2 - 7\lambda - 2 - \lambda^3 + 7\lambda^2 + 2\lambda = 0$$-\lambda^3 + 8\lambda^2 - 5\lambda - 2 = 0$$\lambda^3 - 8\lambda^2 + 5\lambda + 2 = 0$By the Cayley-Hamilton theorem,every square matrix satisfies its characteristic equation:$A^3 - 8A^2 + 5A + 2I = O$Comparing this with $A^3 - 8A^2 + \alpha A + \beta I = O$,we get $\alpha = 5$ and $\beta = 2$.Thus,the ordered pair is $(\alpha, \beta) = (5, 2)$.

Let matrix $A = \begin{bmatrix} 1 & 2 & 3 \\ 0 & 5 & 4 \\ 0 & 3 & 2 \end{bmatrix}$ and $A^3 - 8A^2 + \alpha A + \beta I = O$,then the ordered pair $(\alpha, \beta)$ is

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