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(C) For a system of linear homogeneous equations to have non-trivial solutions, the determinant of the coefficient matrix must be equal to $0$.The coe

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$\frac{1+\sqrt{5}}{2}$

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(C) For a system of linear homogeneous equations to have non-trivial solutions, the determinant of the coefficient matrix must be equal to $0$.The coefficient matrix is $A = \begin{bmatrix} 1 & a & 1 \ a & 2 & -1 \ 2 & 3 & 1 \end{bmatrix}$.Setting the determinant $|A| = 0$:$1(2(1) - (-1)(3)) - a(a(1) - (-1)(2)) + 1(a(3) - 2(2)) = 0$$1(2 + 3) - a(a + 2) + 1(3a - 4) = 0$$5 - a^2 - 2a + 3a - 4 = 0$$-a^2 + a + 1 = 0$$a^2 - a - 1 = 0$Using the quadratic formula $a = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$:$a = \frac{1 \pm \sqrt{(-1)^2 - 4(1)(-1)}}{2(1)} = \frac{1 \pm \sqrt{1 + 4}}{2} = \frac{1 \pm \sqrt{5}}{2}$.Since we are looking for the positive value of $a$, we take $a = \frac{1+\sqrt{5}}{2}$.

The positive value of $a$ for which the system of linear homogeneous equations $x+ay+z=0$, $ax+2y-z=0$, and $2x+3y+z=0$ has non-trivial solutions is

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