Question:medium

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

Show 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.

Updated On: Mar 30, 2026

0
1
$ \frac{1+\sqrt{5}}{2} $
$ \frac{\sqrt{5}-1}{2} $

Show Solution

The Correct Option is C

Solution and Explanation

Was this answer helpful?

Questions Asked in TS EAMCET exam

If $\omega \neq 1$ is a cube root of unity, then one root among the $7^{th}$ roots of $(1+\omega)$ is

TS EAMCET - 2025
Complex numbers

View Solution

If $A = \begin{pmatrix} 1 & 2 & 3 \\ 2 & 1 & 1 \\ 1 & 3 & 1 \end{pmatrix}$ and $B = \begin{pmatrix} 2 & 3 & 4 \\ 3 & 2 & 2 \\ 2 & 4 & 2 \end{pmatrix}$, then $\sqrt{|\text{Adj}(AB)|} =$

TS EAMCET - 2025
Matrices and Determinants

View Solution

If $ p_1, p_2, p_3 $ are the altitudes and $ a=4, b=5, c=6 $ are the sides of a triangle ABC, then $ \frac{1}{p_1^2} + \frac{1}{p_2^2} + \frac{1}{p_3^2} = $

TS EAMCET - 2025
Geometry

View Solution

If $\cos\alpha = \frac{l\cos\beta+m}{l+m\cos\beta}$, then $\frac{\tan^2(\alpha/2)}{\tan^2(\beta/2)} =$

TS EAMCET - 2025
Trigonometry

View Solution

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

Show Hint

The Correct Option is C

Solution and Explanation

Top Questions on Algebra

Questions Asked in TS EAMCET exam