How to find solutions of a matrix

Question

I need to find the solutions of a matrix.

So I have this system of equations

$2x+ky+2z=0$

$x-y+z=1$

$y-z=k$

I augment it as to get:

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

If i reduced the latter in reduced echelon form I get:

$\begin{pmatrix} 1 & 0 &0 &k+1\\ 0 &1 &-1& k\\ 0 &0 &-k-2 & k²+2(k+1) \end{pmatrix}$

I was thinking that the solutions for x,y,z are

x=k+1

y= -2/k+2

z= -(k²+2k+2) / k+2

Are my solutions correct?

Answer 1

I was thinking that the solutions for x,y,z are

x=k+1

y= -2/k+2

z= -(k²+2k+2) / k+2

Are my solutions correct?

Apart from sloppy notation (don't forget brackets!), this is correct. At least for $k \ne -2$; what happens for $k=-2$? You can substitute into the reduced echelon form to check. So you get:

• For $k \ne -2$: $$x = k+1 \quad,\quad y = -\frac{2}{k+2}\quad,\quad z = -\frac{k^2+2k+2}{k+2}$$
• For $k = -2$: $$\ldots$$