# System of linear equations

#### Yankel

##### Active member
Hello again,

I have this system presented below:

$\left\{\begin{matrix} x+y+z+w=1 \\ -3x-3y+kz-kw=-2 \\ 2x+2y+2z+kw=4-k \end{matrix}\right.$

I need to find for which values of k the system has an infinite number of solutions with 2 degrees of freedom, and to find a general solution for this case. I did two elementary row operations to get this:

$\begin{pmatrix} 1 &1 &1 &1 &1 \\ 0 &0 &k+3 &-k+3 &1 \\ 0 &0 &0 &k-2 &2-k \end{pmatrix}$

Then I said that infinite number of solutions with 2 DF will be when k=2, and my final solution was:

w=t
y=s
z=(1-t)/5
x=1-s-t-((1-t)/5)

Am I correct ?

Thanks !

Staff member
Looks good!