Question

Solve the system of equations by finding the reduced row-echelon form of the augmentee matrix for the system of equations..
2x+y+z=-3
3x-5y+3z=-4
5x-y+2z=-2

Answer 1

Answer:

$x=1,y=-1,z=-4$

Step-by-step explanation:

we are given system of equations as

2x+y+z=-3

3x-5y+3z=-4

5x-y+2z=-2

Firstly, we can write augmented matrix

$A=\begin{pmatrix}2&1&1&-3\\ 3&-5&3&-4\\ 5&-1&2&-2\end{pmatrix}$

now, we can change it into reduced- row echelon form

$\mathrm{Swap\:matrix\:rows:}\:R_1\:\leftrightarrow \:R_3$

$=\begin{pmatrix}5&-1&2&-2\\ 3&-5&3&-4\\ 2&1&1&-3\end{pmatrix}$

$\mathrm{Cancel\:leading\:coefficient\:in\:row\:}\:R_2\:\mathrm{\:by\:performing}\:R_2\:\leftarrow \:R_2-\frac{3}{5}\cdot \:R_1$

$=\begin{pmatrix}5&-1&2&-2\\ 0&-\frac{22}{5}&\frac{9}{5}&-\frac{14}{5}\\ 2&1&1&-3\end{pmatrix}$

$\mathrm{Cancel\:leading\:coefficient\:in\:row\:}\:R_3\:\mathrm{\:by\:performing}\:R_3\:\leftarrow \:R_3-\frac{2}{5}\cdot \:R_1$

$=\begin{pmatrix}5&-1&2&-2\\ 0&-\frac{22}{5}&\frac{9}{5}&-\frac{14}{5}\\ 0&\frac{7}{5}&\frac{1}{5}&-\frac{11}{5}\end{pmatrix}$

$\mathrm{Cancel\:leading\:coefficient\:in\:row\:}\:R_3\:\mathrm{\:by\:performing}\:R_3\:\leftarrow \:R_3+\frac{7}{22}\cdot \:R_2$

$=\begin{pmatrix}5&-1&2&-2\\ 0&-\frac{22}{5}&\frac{9}{5}&-\frac{14}{5}\\ 0&0&\frac{17}{22}&-\frac{34}{11}\end{pmatrix}$

$\mathrm{Multiply\:matrix\:row\:by\:constant:}\:R_3\:\leftarrow \frac{22}{17}\cdot \:R_3$

$=\begin{pmatrix}5&-1&2&-2\\ 0&-\frac{22}{5}&\frac{9}{5}&-\frac{14}{5}\\ 0&0&1&-4\end{pmatrix}$

$\mathrm{Cancel\:leading\:coefficient\:in\:row\:}\:R_2\:\mathrm{\:by\:performing}\:R_2\:\leftarrow \:R_2-\frac{9}{5}\cdot \:R_3$

$=\begin{pmatrix}5&-1&2&-2\\ 0&-\frac{22}{5}&0&\frac{22}{5}\\ 0&0&1&-4\end{pmatrix}$

$\mathrm{Cancel\:leading\:coefficient\:in\:row\:}\:R_1\:\mathrm{\:by\:performing}\:R_1\:\leftarrow \:R_1-2\cdot \:R_3$

$=\begin{pmatrix}5&-1&0&6\\ 0&-\frac{22}{5}&0&\frac{22}{5}\\ 0&0&1&-4\end{pmatrix}$

$\mathrm{Multiply\:matrix\:row\:by\:constant:}\:R_2\:\leftarrow \:-\frac{5}{22}\cdot \:R_2$

$=\begin{pmatrix}5&-1&0&6\\ 0&1&0&-1\\ 0&0&1&-4\end{pmatrix}$

$\mathrm{Cancel\:leading\:coefficient\:in\:row\:}\:R_1\:\mathrm{\:by\:performing}\:R_1\:\leftarrow \:R_1+1\cdot \:R_2$

$=\begin{pmatrix}5&0&0&5\\ 0&1&0&-1\\ 0&0&1&-4\end{pmatrix}$

$\mathrm{Multiply\:matrix\:row\:by\:constant:}\:R_1\:\leftarrow \frac{1}{5}\cdot \:R_1$

$=\begin{pmatrix}1&0&0&1\\ 0&1&0&-1\\ 0&0&1&-4\end{pmatrix}$

so, we will get solution as

$x=1,y=-1,z=-4$