# Thread: System of linear algebraic equations

1. ## System of linear algebraic equations

Hi, i've been trying to solve following system for last two hours and cant find a solution. Can anyone help?

Problem: Detremnie the values of k, for which the following system has a nontrivial solution:

9x-3y=kx
-3x+12y-3z=ky
-3y+9z=kz

Find a nontrivial solution in each case.

2. ## Re: System of linear algebraic equations

Solve the first and last equation for y.

From the first equation: $\displaystyle y = \dfrac{9-k}{3}x$

From the last equation: $\displaystyle y = \dfrac{9-k}{3}z$

Either $\displaystyle y=0,k=9$ or $\displaystyle x=z$. If $\displaystyle y=0,k=9$ then $\displaystyle -3x-3z=0$ implies $\displaystyle x=-z$, so the solution is nontrivial (although there are an infinite number of solutions where $\displaystyle k=9$).

Next, suppose $\displaystyle k\neq 9$ and $\displaystyle x=z$. Then plug in what you know $\displaystyle y=\dfrac{9-k}{3}x, z=x$ into the second equation:
$\displaystyle -3x + (12-k)\dfrac{9-k}{3}x-3x = 0$

$\displaystyle (12-k)\dfrac{9-k}{3}x = 6x$

If $\displaystyle x=z=0$, then $\displaystyle y=0$, and this solution is trivial. So, suppose $\displaystyle x\neq 0$. Then multiply both sides by $\displaystyle \dfrac{3}{x}$. Hence $\displaystyle (12-k)(9-k) = 18$. Solve this quadratic: $\displaystyle 108 - 21k + k^2 = 18$. So $\displaystyle k^2-21k + 90 = (k-6)(k-15)=0$. So, you get $\displaystyle k=6$ or $\displaystyle k=15$.

Hence, the three values for $\displaystyle k$ that give nontrivial solutions for the system are $\displaystyle k=6, k=9, k=15$. Note that for any of those three values for $\displaystyle k$, there are an infinite number of solutions for $\displaystyle x,y,z$.