# 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: $y = \dfrac{9-k}{3}x$

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

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

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

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

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

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