$A =\left[ {\begin{array}{cc} 1&-2&1 \\ -2&1&1 \\ 1&1&-2 \\ \end{array} } \right] \ \ \ \ \left[ {\begin{array}{cc} 1&-2&1&a \\ -2&1&1&b \\ 1&1&-2&c \\ \end{array} } \right] \ \ \ \ \left[ {\begin{array}{cc} 1&-2&1&a \\ 0&-3&3&b+2a \\ 0&3&-3&c-a \\ \end{array} } \right] \ \ \ \ \left[ {\begin{array}{cc} 1&-2&1&a \\ 0&-3&3&b+2a \\ 0&0&0&a+b+c \\ \end{array} } \right] $
$A$ is the matrix. I concatenated $[A|b]$ and tried to reduce it. I end up with the last matrix, does it mean that the system is not compatible? Is there another way to solve it?
Or should I concatenate $[A|0]$ to make a homogeneous system?
UPDATE:
$A =\left[ {\begin{array}{cc} 1&-2&1 \\ -2&1&1 \\ 1&1&-2 \\ \end{array} } \right] \ \ \ \ \left[ {\begin{array}{cc} 1&-2&1&0 \\ -2&1&1&0 \\ 1&1&-2&0 \\ \end{array} } \right] \ \ \ \ \left[ {\begin{array}{cc} 1&-2&1&0 \\ 0&-3&3&0 \\ 0&0&0&0 \\ \end{array} } \right] \ \ \ \left[ {\begin{array}{cc} 1&0&-1&0 \\ 0&1&-1&0 \\ 0&0&0&0 \\ \end{array} } \right] $
After solving $[A|0]$ I get $a-c=0$ and $b-c=0$, the equations of the system seem to be: $a=c,b=c$, so $a=b=c$. Is this the correct solution?