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(D) For the system of equations to be consistent,the determinant of the augmented matrix must be zero:$\Delta = \begin{vmatrix} 1 & k & 1 \ k & 1 & 2

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$7$

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(D) For the system of equations to be consistent,the determinant of the augmented matrix must be zero:$\Delta = \begin{vmatrix} 1 & k & 1 \ k & 1 & 2 \ 1 & 1 & k \end{vmatrix} = 0$Expanding the determinant:$1(k - 2) - k(k^2 - 2) + 1(k - 1) = 0$$k - 2 - k^3 + 2k + k - 1 = 0$$-k^3 + 4k - 3 = 0$$k^3 - 4k + 3 = 0$By inspection,$k = 1$ is a root. Dividing by $(k - 1)$:$(k - 1)(k^2 + k - 3) = 0$If $k = 1$,the equations become $x + y = 1$,$x + y = 2$,and $x + y = 1$,which is inconsistent.Thus,$k^2 + k - 3 = 0$. The roots are $k = \frac{-1 \pm \sqrt{1 - 4(1)(-3)}}{2} = \frac{-1 \pm \sqrt{13}}{2}$.Let $k_1 = \frac{-1 + \sqrt{13}}{2}$ and $k_2 = \frac{-1 - \sqrt{13}}{2}$.Then $k_1^2 + k_2^2 = (k_1 + k_2)^2 - 2k_1k_2 = (-1)^2 - 2(-3) = 1 + 6 = 7$.

Let $k_1$ and $k_2$ be the maximum and minimum values of $k$ for which the system of equations $x + ky = 1$,$kx + y = 2$,and $x + y = k$ are consistent. Then $k_1^2 + k_2^2$ is equal to:

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