Answer
$(1,0,-1,2)$
Work Step by Step
Step 1. Set up the augmented matrix of the system of equations:
$\begin{array} \\A=\\ \end{array}
\begin{bmatrix} 1&1&1&1&2\\ 2&0&-3&0&5\\ 1&-2&0&4&9\\1&1&2&3&5 \end{bmatrix}
\begin{array} \\ \\-R_2+2R_1\to R_2\\-R_3+R_1\to R_3\\R_4-R_1\to R_4 \\ \end{array}$
Step 2. Do the row operations indicated on the right side of the matrix:
$\begin{array} \\A=\\ \end{array}
\begin{bmatrix} 1&1&1&1&2\\ 0&2&5&2&-1\\ 0&3&1&-3&-7\\0&0&1&2&3 \end{bmatrix}
\begin{array} \\ \\ \\-2R_3+3R_2\to R_3\\R_4-R_3\to R_4 \\ \end{array}$
Step 3. Do the row operations indicated on the right side of the matrix:
$\begin{array} \\A=\\ \end{array}
\begin{bmatrix} 1&1&1&1&2\\ 0&2&5&2&-1\\ 0&0&13&12&11\\0&0&0&5&10 \end{bmatrix}$
Step 4. The last row gives $w=2$. Back substitute into the third row to get $13z+12\times2=11$ or $z=-1$.
Step 5. Continue the back substitution process to get $y=0,x=1$ and the solution as $(1,0,-1,2)$
