# Thread: Solutions for a homogeneous system

1. ## Solutions for a homogeneous system

I need to determine for what values of "a" the system will have nontrivial solutions.

I know that nontrivial means that both values of x and y can not be zero.

I am able to, by trial and error, determine values of "a" which provide nontrivial solutions. But, I am sure that I am suppose to come up with a systematic method to determine this. Since I am in a matrices and power function class, I am sure I am not suppose to just use my current method.

My system I am currently working on is:

(1- a)x + 2y = 0

3x + (2 - a)y = 0

Any help would be much appreciated. Frostking

2. Originally Posted by Frostking
I need to determine for what values of "a" the system will have nontrivial solutions.

I know that nontrivial means that both values of x and y can not be zero.

I am able to, by trial and error, determine values of "a" which provide nontrivial solutions. But, I am sure that I am suppose to come up with a systematic method to determine this. Since I am in a matrices and power function class, I am sure I am not suppose to just use my current method.

My system I am currently working on is:

(1- a)x + 2y = 0

3x + (2 - a)y = 0

Any help would be much appreciated. Frostking
Writing the matrix equation gives.

$\begin{bmatrix}(1-a) & 2 \\ 3 & (2-a) \end {bmatrix} \begin{bmatrix} x \\ y \end{bmatrix}=\begin{bmatrix}0 \\ 0 \end{bmatrix}$

This equation will have a unique solution when the matrix is invertable. A matrix is invertable when its determinant is not zero.

$\begin{vmatrix}(1-a) & 2 \\ 3 & (2-a) \end {vmatrix}=(1-a)(2-a)-6$

$a^2-3a-4=0 \iff (a-4)(a+1)=0$

So you will get nontrivia(an infinite number of)l solutions when a=4 or a=-1