Question

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

• A. 0
• B. 1
• C. \( \frac{1+\sqrt{5}}{2} \)
• D. \( \frac{\sqrt{5}-1}{2} \)

Hint:

Remember the key condition for non-trivial solutions in a homogeneous system \( AX = 0 \): the determinant of the matrix \(A\) must be zero. This transforms the problem into solving a polynomial equation.

Answer 1

The Correct Option is C

A system of homogeneous linear equations has a non-trivial solution if and only if the determinant of the coefficient matrix is equal to zero.
The given system is:
\( x+ay+z=0 \)
\( ax+2y-z=0 \)
\( 2x+3y+z=0 \)
The coefficient matrix is \( A = \begin{pmatrix} 1 & a & 1\\ a & 2 & -1 \\ 2 & 3 & 1 \end{pmatrix} \).
We set the determinant of A to zero: \( \det(A) = 0 \).
\( 1 \begin{vmatrix} 2 & -1 \\ 3 & 1 \end{vmatrix} - a \begin{vmatrix} a & -1 \\ 2 & 1 \end{vmatrix} + 1 \begin{vmatrix} a & 2 \\ 2 & 3 \end{vmatrix} = 0 \)
\( 1((2)(1) - (-1)(3)) - a((a)(1) - (-1)(2)) + 1((a)(3) - (2)(2)) = 0 \)
\( 1(2 + 3) - a(a + 2) + (3a - 4) = 0 \)
\( 5 - a^2 - 2a + 3a - 4 = 0 \)
\( -a^2 + a + 1 = 0 \)
Multiplying by -1, we get the quadratic equation: \( a^2 - a - 1 = 0 \).
We use the quadratic formula to solve for \( a \): \( a = \frac{-b \pm \sqrt{b^2-4ac}}{2a} \).
\( a = \frac{-(-1) \pm \sqrt{(-1)^2 - 4(1)(-1)}}{2(1)} \)
\( a = \frac{1 \pm \sqrt{1 + 4}}{2} = \frac{1 \pm \sqrt{5}}{2} \).
The two possible values for \(a\) are \( \frac{1 + \sqrt{5}}{2} \) and \( \frac{1 - \sqrt{5}}{2} \).
The question asks for the positive value of 'a'. Since \( \sqrt{5} \approx 2.23 \), the value \( \frac{1 - \sqrt{5}}{2} \) is negative.
Therefore, the required positive value is \( a = \frac{1 + \sqrt{5}}{2} \).