Answer:
[tex]x=\frac{9}{17},y=-\frac{14}{17},z=-\frac{55}{17}[/tex]
Step-by-step explanation:
we are given
system of equations as
[tex]2x+y+z=-3[/tex]
[tex]3x-5y+3z=-4[/tex]
[tex]5x-y+2z=-3[/tex]
Firstly, we will find augmented matrix
[tex]A=\begin{pmatrix}2&1&1&-3\\ 3&-5&3&-4\\ 5&-1&2&-3\end{pmatrix}[/tex]
now, we can change it into reduced row-echelon form
[tex]\mathrm{Swap\:matrix\:rows:}\:R_1\:\leftrightarrow \:R_3[/tex]
[tex]=\begin{pmatrix}5&-1&2&-3\\ 3&-5&3&-4\\ 2&1&1&-3\end{pmatrix}[/tex]
[tex]\mathrm{Cancel\:leading\:coefficient\:in\:row\:}\:R_2\:\mathrm{\:by\:performing}\:R_2\:\leftarrow \:R_2-\frac{3}{5}\cdot \:R_1[/tex]
[tex]=\begin{pmatrix}5&-1&2&-3\\ 0&-\frac{22}{5}&\frac{9}{5}&-\frac{11}{5}\\ 2&1&1&-3\end{pmatrix}[/tex]
[tex]\mathrm{Cancel\:leading\:coefficient\:in\:row\:}\:R_3\:\mathrm{\:by\:performing}\:R_3\:\leftarrow \:R_3-\frac{2}{5}\cdot \:R_1[/tex]
[tex]=\begin{pmatrix}5&-1&2&-3\\ 0&-\frac{22}{5}&\frac{9}{5}&-\frac{11}{5}\\ 0&\frac{7}{5}&\frac{1}{5}&-\frac{9}{5}\end{pmatrix}[/tex]
[tex]\mathrm{Cancel\:leading\:coefficient\:in\:row\:}\:R_3\:\mathrm{\:by\:performing}\:R_3\:\leftarrow \:R_3+\frac{7}{22}\cdot \:R_2[/tex]
[tex]=\begin{pmatrix}5&-1&2&-3\\ 0&-\frac{22}{5}&\frac{9}{5}&-\frac{11}{5}\\ 0&0&\frac{17}{22}&-\frac{5}{2}\end{pmatrix}[/tex]
[tex]\mathrm{Multiply\:matrix\:row\:by\:constant:}\:R_3\:\leftarrow \frac{22}{17}\cdot \:R_3[/tex]
[tex]=\begin{pmatrix}5&-1&2&-3\\ 0&-\frac{22}{5}&\frac{9}{5}&-\frac{11}{5}\\ 0&0&1&-\frac{55}{17}\end{pmatrix}[/tex]
[tex]\mathrm{Cancel\:leading\:coefficient\:in\:row\:}\:R_2\:\mathrm{\:by\:performing}\:R_2\:\leftarrow \:R_2-\frac{9}{5}\cdot \:R_3[/tex]
[tex]=\begin{pmatrix}5&-1&2&-3\\ 0&-\frac{22}{5}&0&\frac{308}{85}\\ 0&0&1&-\frac{55}{17}\end{pmatrix}[/tex]
[tex]\mathrm{Cancel\:leading\:coefficient\:in\:row\:}\:R_1\:\mathrm{\:by\:performing}\:R_1\:\leftarrow \:R_1-2\cdot \:R_3[/tex]
[tex]=\begin{pmatrix}5&-1&0&\frac{59}{17}\\ 0&-\frac{22}{5}&0&\frac{308}{85}\\ 0&0&1&-\frac{55}{17}\end{pmatrix}[/tex]
[tex]\mathrm{Multiply\:matrix\:row\:by\:constant:}\:R_2\:\leftarrow \:-\frac{5}{22}\cdot \:R_2[/tex]
[tex]=\begin{pmatrix}5&-1&0&\frac{59}{17}\\ 0&1&0&-\frac{14}{17}\\ 0&0&1&-\frac{55}{17}\end{pmatrix}[/tex]
[tex]\mathrm{Cancel\:leading\:coefficient\:in\:row\:}\:R_1\:\mathrm{\:by\:performing}\:R_1\:\leftarrow \:R_1+1\cdot \:R_2[/tex]
[tex]=\begin{pmatrix}5&0&0&\frac{45}{17}\\ 0&1&0&-\frac{14}{17}\\ 0&0&1&-\frac{55}{17}\end{pmatrix}[/tex]
[tex]\mathrm{Multiply\:matrix\:row\:by\:constant:}\:R_1\:\leftarrow \frac{1}{5}\cdot \:R_1[/tex]
[tex]=\begin{pmatrix}1&0&0&\frac{9}{17}\\ 0&1&0&-\frac{14}{17}\\ 0&0&1&-\frac{55}{17}\end{pmatrix}[/tex]
so, we got solution as
[tex]x=\frac{9}{17},y=-\frac{14}{17},z=-\frac{55}{17}[/tex]
