For $z, a, b, c \in \mathbb{R}$ solve this system of linear equations

Question

For $z, a, b, c \in\mathbb{R}$ solve this system of linear equations using matrix methods, $$ \left[ \begin{array}{ccc|c} 1 & z & 1 & a \\ z & 2 & 2 & b \\ 2 & 1 & z & c \\ \end{array} \right] $$ I just added first and second row to the third row and got: $$ \left[ \begin{array}{ccc|c} 1 & z & 1 & a \\ z & 2 & 2 & b \\ z + 3 & z + 3 & z + 3 & a + b + c \\ \end{array} \right] $$ But I don't know what should I do in next steps to solve this system for $z, a, b, c \in \mathbb{R}$

Answer 1

The next steps could be as follows: $$ \left( \begin{array}{ccc|c} 1 & z & 1 & a \\ z & 2 & 2 & b\\ z+3 & z+3 & z+3 & a+b+c \\ \end{array} \right)\sim \left( \begin{array}{ccc|c} 1 & z & 1 & a \\ z & 2 & 2 & b\\ 1 & 1 & 1 & \Delta \\ \end{array} \right),\ \Delta=\frac{a+b+c}{z+3} $$ $$ \left( \begin{array}{ccc|c} 0 & z-1 & 0 & a-\Delta \\ z-2 & 0 & 0 & b-2\Delta\\ 1 & 1 & 1 & \Delta \\ \end{array} \right). $$ It follows that if $z\neq1,2,-3$, then $$ x_1=\frac{b-2\Delta}{z-2},\ x_2=\frac{a-\Delta}{z-1},\ x_3=\Delta-\frac{b-2\Delta}{z-2}-\frac{a-\Delta}{z-1}. $$ Now each of the three cases must be examined separately: $z=1,2,-3$.