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(C) To find $D$ and $A$,we expand the determinant $\Delta(\lambda) = \left|\begin{array}{ccc}1 & 2 & 3-\lambda \ 0 & -1-\lambda & 2 \ 1-\lambda & 1

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$-5$

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(C) To find $D$ and $A$,we expand the determinant $\Delta(\lambda) = \left|\begin{array}{ccc}1 & 2 & 3-\lambda \ 0 & -1-\lambda & 2 \ 1-\lambda & 1 & 3\end{array}ight|$.Step $1$: Find $D$ by setting $\lambda = 0$.$D = \Delta(0) = \left|\begin{array}{ccc}1 & 2 & 3 \ 0 & -1 & 2 \ 1 & 1 & 3\end{array}ight|$.Expanding along the first column: $D = 1((-1)(3) - (2)(1)) - 0 + 1((2)(2) - (-1)(3)) = 1(-3-2) + 1(4+3) = -5 + 7 = 2$.Step $2$: Find $A$ by looking at the coefficient of $\lambda^3$.The determinant is of the form $\left|\begin{array}{ccc}1 & 2 & -\lambda \ 0 & -\lambda & 2 \ -\lambda & 1 & 3\end{array}ight| + \dots$ (considering only terms with $\lambda^3$).The coefficient of $\lambda^3$ comes from the product of the diagonal elements: $(1)(-\lambda)(3) = -3\lambda^3$. Thus,$A = -3$.Step $3$: Calculate $D+A$.$D+A = 2 + (-3) = -1$.Wait,re-evaluating the expansion: $\Delta(\lambda) = 1((-1-\lambda)(3) - 2) - 2(0 - 2(1-\lambda)) + (3-\lambda)(0 - (-1-\lambda)(1-\lambda))$.$= (-3-3\lambda-2) - 2(2\lambda-2) + (3-\lambda)(-(1-\lambda^2)) = -5-3\lambda - 4\lambda + 4 - (3-\lambda)(1-\lambda^2) = -1-7\lambda - (3 - 3\lambda^2 - \lambda + \lambda^3) = -1-7\lambda - 3 + 3\lambda^2 + \lambda - \lambda^3 = -\lambda^3 + 3\lambda^2 - 6\lambda - 4$.So $A = -1$ and $D = -4$.$D+A = -4 + (-1) = -5$.

If $\left|\begin{array}{ccc}1 & 2 & 3-\lambda \\ 0 & -1-\lambda & 2 \\ 1-\lambda & 1 & 3\end{array}\right|=A \lambda^3+B \lambda^2+C \lambda+D$,then $D+A=$

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