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(D) The characteristic equation is given by $|A - xI| = 0$. For a $3 	imes 3$ matrix,this is $x^3 - (	ext{tr}(A))x^2 + (	ext{sum of principal minor

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$27$

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(D) The characteristic equation is given by $|A - xI| = 0$. For a $3 	imes 3$ matrix,this is $x^3 - (	ext{tr}(A))x^2 + (	ext{sum of principal minors})x - |A| = 0$. Here,$	ext{tr}(A) = 2 + 3 + 2 = 7$. The principal minors are: $M_{11} = \begin{vmatrix} 3 & 1 \ 2 & 2 \end{vmatrix} = 6 - 2 = 4$. $M_{22} = \begin{vmatrix} 2 & 1 \ 1 & 2 \end{vmatrix} = 4 - 1 = 3$. $M_{33} = \begin{vmatrix} 2 & 2 \ 1 & 3 \end{vmatrix} = 6 - 2 = 4$. Sum of principal minors $= 4 + 3 + 4 = 11$. Thus,the characteristic equation is $x^3 - 7x^2 + 11x - |A| = 0$. By the properties of roots,$\alpha + \beta + \gamma = 7$ and $\alpha\beta + \beta\gamma + \gamma\alpha = 11$. We know that $\alpha^2 + \beta^2 + \gamma^2 = (\alpha + \beta + \gamma)^2 - 2(\alpha\beta + \beta\gamma + \gamma\alpha)$. Substituting the values: $\alpha^2 + \beta^2 + \gamma^2 = (7)^2 - 2(11) = 49 - 22 = 27$.

If $A = \begin{bmatrix} 2 & 2 & 1 \\ 1 & 3 & 1 \\ 1 & 2 & 2 \end{bmatrix}$ and $\alpha, \beta, \gamma$ are the roots of the characteristic equation $|A - xI| = 0$,then $\alpha^2 + \beta^2 + \gamma^2 = $

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