91Ó°ÊÓ

Consider the given system of equations,

$\left\{\begin{array}{l}2x-5y+3z=8\\ 3x-y+4z=7\\ x+3y+2z=-3\end{array}\right.$

The augmented matrix for the given system of equations is

$\left[\begin{array}{cccr}2& -5& 3& 8\\ 3& -1& 4& 7\\ 1& 3& 2& -3\end{array}\right]$

Apply $\frac{R_1}{2}→R_1$and $R_2-3×R_1→R_2$,

$\left[\begin{array}{cccr}1& -\frac{5}{2}& \frac{3}{2}& 4\\ 0& \frac{13}{2}& -\frac{1}{2}& -5\\ 1& 3& 2& -3\end{array}\right]$

Apply $R_3-R_1→R_3$and $R_2×\frac{2}{13}→R_2$,

$\left[\begin{array}{cccr}1& -\frac{5}{2}& \frac{3}{2}& 4\\ 0& 1& -\frac{1}{13}& -\frac{10}{13}\\ 0& \frac{11}{2}& \frac{1}{2}& -7\end{array}\right]$

Apply $R_3×\frac{13}{12}→R_3$,

$\left[\begin{array}{cccr}1& -\frac{5}{2}& \frac{3}{2}& 4\\ 0& 1& -\frac{1}{13}& -\frac{10}{13}\\ 0& 0& 1& -3\end{array}\right]$

Now, the matrix is in row-echelon form.

Writing the corresponding system of equations,

$$\begin{gathered} x-\frac{5}{2}y+\frac{3}{2}z=4......\left(i\right) \\ y-\frac{1}{13}z=-\frac{10}{13}......\left(ii\right) \\ z=-3......\left(iii\right) \end{gathered}$$

Substitute the value of $z$in equation (ii),

localid="1644517494244" $$\begin{gathered} y-\frac{1}{13}×\left(-3\right)=-\frac{10}{13} \\ y=-\frac{10}{13}-\frac{3}{13} \\ y=-\frac{13}{13} \\ y=-1 \end{gathered}$$

Substitute the value of $y,z$in equation (i),

localid="1644592366220" $$\begin{gathered} x-\frac{5}{2}×\left(-1\right)+\frac{3}{2}×\left(-3\right)=4 \\ x+\frac{5}{2}-\frac{9}{2}=4 \\ x-\frac{4}{2}=4 \\ x=6 \end{gathered}$$

Substitute the values $x,y,z$in equation (i),

role="math" localid="1644592410991" $$\begin{gathered} 6-\frac{5}{2}\left(-1\right)+\frac{3}{2}\left(-3\right)=4 \\ 6+\frac{5}{2}-\frac{9}{2}=4 \\ \frac{8}{2}=4 \\ 4=4 \end{gathered}$$

This is true.

Substitute the values $y,z$in equation (ii),

$$\begin{gathered} -1-\frac{1}{13}\left(-3\right)=-\frac{10}{13} \\ -1+\frac{3}{13}=-\frac{10}{13} \\ -\frac{10}{13}=-\frac{10}{13} \end{gathered}$$

This is also true.