# Evaluate a linear system of three equations

Solve for $x, y\ \text{and}\ z\$:

$x-3z=10\\ -x+y+2z=7\\ 2x+2y-5z=-8$

My working: $$\left(\begin{array}{ccc|c} 1 & 0 & -3 & 10 \\ -1 & 1 & 2 & 7 \\ 2 & 2 & -5 & -8 \end{array}\right) = \left(\begin{array}{ccc|c} 1 & 0 & -3 & 10 \\ 0 & 1 & -1 & 17 \\ 0 & 3 & 0 & -11 \end{array}\right) \begin{array}{l} \\ R_1 + R_2 \\ R_3 + R_2 - R_1 \end{array}= \left(\begin{array}{ccc|c} 1 & 0 & -3 & 10 \\ 0 & 1 & -1 & 17 \\ 0 & 0 & 3 & -62 \end{array}\right) \begin{array}{l} \\ \\ R_3 - 3R_2 \end{array}= \left(\begin{array}{ccc|c} 1 & 0 & 0 & -52 \\ 0 & 1 & 0 & -7 \\ 0 & 0 & 1 & -24 \end{array}\right) \begin{array}{l} R_1 + R_3 \\ R_2 + R_3/3 \\ R_3/3 \end{array}$$ Yet plugging these solutions into the original equation does not work:

$-52+0-3 \times (-24)=20 \ne 10$ and

$-(-52)-7+2\times(-24)=-3\ne7$

What did I do wrong?

• Is it a typo in your second equation - $2$ or $2z$? – Jason Knapp Oct 13 '14 at 20:37
• OK one other thing I see - $62/3$ is not 24. – Jason Knapp Oct 13 '14 at 20:38
• @JasonKnapp is right, $62$ can be divide by $3$. – hlapointe Oct 13 '14 at 20:42

The last line should read $$\left(\begin{array}{ccc|c} 1 & 0 & -3 & 10 \\ 0 & 1 & -1 & 17 \\ 0 & 0 & 3 & -62 \end{array}\right) \begin{array}{l} \\ \\ R_3 - 3R_2 \end{array}= \left(\begin{array}{ccc|c} 1 & 0 & 0 & -52 \\ 0 & 1 & 0 & -11/3 \\ 0 & 0 & 1 & -62/3 \end{array}\right) \begin{array}{l} R_1 + R_3 \\ R_2 + R_3/3 \\ R_3/3 \end{array}$$