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(C) The characteristic equation of matrix $A$ is given by $|A - \lambda I| = 0$. For a $3 	imes 3$ matrix,the characteristic equation is $\lambda^3 -

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$k = -1, \beta = 22$

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(C) The characteristic equation of matrix $A$ is given by $|A - \lambda I| = 0$. For a $3 	imes 3$ matrix,the characteristic equation is $\lambda^3 - 	ext{tr}(A)\lambda^2 + (A_{11} + A_{22} + A_{33})\lambda - |A| = 0$,where $A_{ii}$ are the principal minors of order $2$. Given $f(A) = 0$,the characteristic equation is $\lambda^3 - 2\lambda^2 - \alpha\lambda + \beta = 0$. Comparing the trace: $	ext{tr}(A) = 1 + 2 + k = 2 \Rightarrow 3 + k = 2 \Rightarrow k = -1$. Now,calculate the determinant $|A|$: $|A| = 1(2k - 0) - 2(2k + 3) + 3(0 - 6) = 2k - 4k - 6 - 18 = -2k - 24$. Substituting $k = -1$: $|A| = -2(-1) - 24 = 2 - 24 = -22$. Since the constant term in the characteristic equation is $-|A|$,we have $-\beta = -|A| \Rightarrow \beta = |A| = -22$. Wait,re-evaluating the characteristic equation form: $\lambda^3 - 	ext{tr}(A)\lambda^2 + (	ext{sum of principal minors})\lambda - |A| = 0$. Given $f(x) = x^3 - 2x^2 - \alpha x + \beta = 0$,we have $	ext{tr}(A) = 2$ (which gives $k = -1$) and $|A| = -\beta$. $|A| = \begin{vmatrix} 1 & 2 & 3 \ 2 & 2 & -1 \ 3 & 0 & -1 \end{vmatrix} = 1(-2) - 2(-2 + 3) + 3(0 - 6) = -2 - 2 - 18 = -22$. Thus,$-\beta = -22 \Rightarrow \beta = 22$. Therefore,$k = -1$ and $\beta = 22$.

Let $A = \begin{bmatrix} 1 & 2 & 3 \\ 2 & 2 & -1 \\ 3 & 0 & k \end{bmatrix}$ and $f(x) = x^3 - 2x^2 - \alpha x + \beta = 0$. If $A$ satisfies $f(A) = 0$,then:

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