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(D) The characteristic equation is given by $|A - xI| = 0$.Given $A = \begin{bmatrix} 2 & 3 & 0 \ 1 & 2 & 5 \ 3 & -1 & 2 \end{bmatrix}$,then $A - xI

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$x^3 - 6x^2 + 14x - 57 = 0$

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(D) The characteristic equation is given by $|A - xI| = 0$.Given $A = \begin{bmatrix} 2 & 3 & 0 \ 1 & 2 & 5 \ 3 & -1 & 2 \end{bmatrix}$,then $A - xI = \begin{bmatrix} 2-x & 3 & 0 \ 1 & 2-x & 5 \ 3 & -1 & 2-x \end{bmatrix}$.The determinant $|A - xI| = (2-x) \begin{vmatrix} 2-x & 5 \ -1 & 2-x \end{vmatrix} - 3 \begin{vmatrix} 1 & 5 \ 3 & 2-x \end{vmatrix} + 0 \begin{vmatrix} 1 & 2-x \ 3 & -1 \end{vmatrix} = 0$.$(2-x)((2-x)^2 + 5) - 3(2-x - 15) = 0$.$(2-x)(4 - 4x + x^2 + 5) - 3(-x - 13) = 0$.$(2-x)(x^2 - 4x + 9) + 3x + 39 = 0$.$2x^2 - 8x + 18 - x^3 + 4x^2 - 9x + 3x + 39 = 0$.$-x^3 + 6x^2 - 14x + 57 = 0$.Multiplying by $-1$,we get $x^3 - 6x^2 + 14x - 57 = 0$.

The characteristic equation of the matrix $A = \begin{bmatrix} 2 & 3 & 0 \\ 1 & 2 & 5 \\ 3 & -1 & 2 \end{bmatrix}$ is:

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