System of linear equation with one parameter

Question

I'm trying to understand and solve a linear equation but i'm not sure how to go about it next, I was trying to reduce it with row operations but I can't seem to get all zero's under the first 'pivot' / leading one ...

So this is the matrix of equations :

1 b b 1
1 1 b b
b b 1 1 
b 1 1 b

with the first colum x, second colum y third colum z and last colum the 'solution'

I tried to reduce this and got to this form :

 1   b   b   1
 0  1-b  0   b
b-1  0  1-b  0 
 0   0   0   0

and I can't seem to get rid of that b-1 in the first column ...

the original question is " for which value of b does the system of linear equantions have : exactly one solution no solutions infinite solutions

Answer 1

If you eliminate elements in the first column below the first one, you should get:

$$ \begin{bmatrix} 1 & b & b & 1\\ 0 & 1-b & 0 & b-1\\ 0 & b-b^{2} & 1-b^{2} & 1-b\\ 0 & 1-b^{2} & 1-b^{2} & 0\\ \end{bmatrix} $$

Then you can discuss the case $b=1,b=-1$ and others. In each case you should reduce the matrix again to row echelon form.