Answer
The solution is $(1,-3,4,-4,2)$
Work Step by Step
The augmented matrix of the system is:
$\begin{bmatrix}
2&-1 &2 &1 &-1 |11\\
1 &-3 &-2 &-1 &-2| 2\\
3& 1 & -2 & -1&1 |-2\\
1 & 2& 1 & 2 & 3|-3\\
5 & -3 & -3 & 1 &2 |2
\end{bmatrix}$
with reduced row-echelon form:
$\begin{bmatrix}
2&-1 &2 &1 &-1 |11\\
1 &-3 &-2 &-1 &-2| 2\\
3& 1 & -2 & -1&1 |-2\\
1 & 2& 1 & 2 & 3|-3\\
5 & -3 & -3 & 1 &2 |2
\end{bmatrix} \approx^1\begin{bmatrix}
1 &-3 &-2 &-1 &-2| 2\\
2&-1 &2 &1 &-1 |11\\
3& 1 & -2 & -1&1 |-2\\
1 & 2& 1 & 2 & 3|-3\\
5 & -3 & -3 & 1 &2 |2
\end{bmatrix}\approx^2\begin{bmatrix}
1 &-3 &-2 &-1 &-2| 2\\
2&5 &7 &3 &3 |7\\
0& 10 & 4 & 2&7 |-8\\
0 & 5&3 & 3 & 5|-5\\
0 & 12 &7 &6 &12 |-8
\end{bmatrix}\approx^3\begin{bmatrix}
1 &-3 &-2 &-1 &-2| 2\\
0&1 &\frac{7}{5} &\frac{3}{5} &\frac{3}{5} |\frac{7}{5}\\
0& 10 & 4 & 2&7 |-8\\
0 & 5&3 & 3 & 5|-5\\
0 & 12 &7 &6 &12 |-8
\end{bmatrix} \approx^4\begin{bmatrix}
1 &0 &\frac{11}{5} &\frac{4}{5} &\frac{-1}{5}| \frac{31}{5}\\
0&1 &\frac{7}{5} &\frac{3}{5} &\frac{3}{5} |\frac{7}{5}\\
0& 0&-10 & -4 & 1 |-22\\
0 & 0&-4 & 0 & 2|-12\\
0 & 0 &\frac{-49}{5} &\frac{-6}{5} &\frac{24}{5} |\frac{-124}{5}
\end{bmatrix} \approx^5\begin{bmatrix}
1 &0 &\frac{11}{5} &\frac{4}{5} &\frac{-1}{5}| \frac{31}{5}\\
0&1 &\frac{7}{5} &\frac{3}{5} &\frac{3}{5} |\frac{7}{5}\\
0& 0&-10 & -4 & 1 |-22\\
0 & 0&0 & 1 & 1|-2\\
0 & 0 &\frac{-49}{5} &\frac{-6}{5} &\frac{24}{5} |\frac{-124}{5}
\end{bmatrix} \approx^6 \begin{bmatrix}
1 &0 &0 &\frac{-2}{25} &\frac{-1}{50}| \frac{34}{25}\\
0&1 &0 &\frac{1}{25} &\frac{37}{50} |\frac{-42}{25}\\
0& 0&1&\frac{2}{5} & -\frac{1}{10} |\frac{11}{5}\\
0 & 0&0 & \frac{8}{5} & \frac{8}{5} |-\frac{16}{5} \\
0 & 0 &0&\frac{68}{25} &\frac{191}{50} |\frac{-81}{25}
\end{bmatrix} \approx^7 \begin{bmatrix}
1 &0 &0 &\frac{-2}{25} &\frac{-1}{50}| \frac{34}{25}\\
0&1 &0 &\frac{1}{25} &\frac{37}{50} |\frac{-42}{25}\\
0& 0&1&\frac{2}{5} & -\frac{1}{10} |\frac{11}{5}\\
0 & 0&0 &1 &1 |-2 \\
0 & 0 &0&\frac{68}{25} &\frac{191}{50} |\frac{-81}{25}
\end{bmatrix} \approx^8 \begin{bmatrix}
1 &0 &0 &0 &\frac{1}{10}| \frac{6}{5}\\
0&1 &0 &0 &\frac{7}{10} |\frac{-8}{5}\\
0& 0&1&0& -\frac{1}{2} |3\\
0 & 0&0 &1 &1 |-2 \\
0 & 0 &0&0&\frac{11}{10} |\frac{11}{5}
\end{bmatrix} \approx^9 \begin{bmatrix}
1 &0 &0 &0 &\frac{1}{10}| \frac{6}{5}\\
0&1 &0 &0 &\frac{7}{10} |\frac{-8}{5}\\
0& 0&1&0& -\frac{1}{2} |3\\
0 & 0&0 &1 &1 |-2 \\
0 & 0 &0&0&1 |2
\end{bmatrix} \approx^10 \begin{bmatrix}
1 &0 &0 &0 &0| 1\\
0&1 &0 &0 &0 |-3\\
0& 0&1&0&0|4\\
0 & 0&0 &1 &0 |-4 \\
0 & 0 &0&0&1 |2
\end{bmatrix}$
$1.P_{12}$
$2.A_{12}(-2), A_{13}(-3),A_{14}(-1),A_{15}(-5)$
$3.M_2(-\frac{11}{5})$
$4.A_{21}(3),A_{23}(-10),A_{25}(-12)$
$5.M_2(-\frac{1}{10})$
$6.A_{31}(-\frac{11}{5}),A_{32}(-\frac{7}{5}),A_{34}(4),A_{35}(\frac{49}{5})$
$7.M_4(\frac{5}{8})$
$8.A_{41}(\frac{2}{25}),A_{42}(-\frac{1}{25}),A_{43}(-\frac{2}{5}),A_{45}(\frac{-68}{25}),M_5(\frac{10}{11})$
$9.A_{51}(-\frac{1}{10}),A_{52}(-\frac{7}{10}),A_{53}(\frac{1}{2}),A_{54}(-1)$
This matrix is now in row-echelon form. The equivalent system is:
$x_1=1$
$x_2=-3$
$x_3=4$
$x_4=-4$
$x_5=2$
The solution is $(1,-3,4,-4,2)$
