To solve this problem, we need to determine the value of \( a \) for which the given system of linear homogeneous equations has non-trivial solutions. A system of homogeneous equations has non-trivial solutions if and only if the determinant of the coefficient matrix is zero.
Let's write the system of equations in matrix form:
| \[ \begin{bmatrix} 1 & a & 1 \\ a & 2 & -1 \\ 2 & 3 & 1 \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix} \] |
Next, compute the determinant of the coefficient matrix:
\[ \text{det} = \begin{vmatrix} 1 & a & 1 \\ a & 2 & -1 \\ 2 & 3 & 1 \end{vmatrix} \]
Using the rule of Sarrus for \(3 \times 3\) determinant, we get:
\[ \text{det} = 1 \cdot (2 \cdot 1 - (-1) \cdot 3) - a \cdot (a \cdot 1 - (-1) \cdot 2) + 1 \cdot (a \cdot 3 - 2 \cdot 2) \]
Simplify each term:
\[ = 1 \cdot (2 + 3) - a \cdot (a + 2) + 1 \cdot (3a - 4) \]
\[ = 5 - a^2 - 2a + 3a - 4 \]
\[ = 1 - a^2 + a \]
The system has a non-trivial solution for values of \( a \) that make the determinant zero:
\[ 1 - a^2 + a = 0 \]
This is a quadratic equation, which can be rewritten as:
\[ a^2 - a - 1 = 0 \]
To solve the quadratic equation, use the quadratic formula:
\[ a = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Substituting in the coefficients \( a = 1 \), \( b = -1 \), \( c = -1 \):
\[ a = \frac{1 \pm \sqrt{(-1)^2 - 4 \cdot 1 \cdot (-1)}}{2 \cdot 1} \]
\[ a = \frac{1 \pm \sqrt{1 + 4}}{2} \]
\[ a = \frac{1 \pm \sqrt{5}}{2} \]
The positive value for \( a \) is:
\[ a = \frac{1 + \sqrt{5}}{2} \]
Thus, the correct answer is \( \frac{1+\sqrt{5}}{2} \). This value of \( a \) ensures the system has non-trivial solutions.