91Ó°ÊÓ

Consider the given system of equations,

$\left\{\begin{array}{l}2y+3z=-1\\ 5x+3y=-6\\ 7x+z=1\end{array}\right.$

The augmented matrix for the given system of equations is

$\left[\begin{array}{cccr}0& 2& 3& -1\\ 5& 3& 0& -6\\ 7& 0& 1& 1\end{array}\right]$

Interchanging rows$R_1\mathrm{and}{R}_2$,

$\left[\begin{array}{cccr}5& 3& 0& -6\\ 0& 2& 3& -1\\ 7& 0& 1& 1\end{array}\right]$

Apply $\frac{R_1}{5}→R_1$and $R_3-7×R_1→R_3$,

localid="1644431494537" $\left[\begin{array}{cccr}1& \frac{3}{5}& 0& -\frac{6}{5}\\ 0& 2& 3& -1\\ 0& -\frac{21}{5}& 1& \frac{47}{5}\end{array}\right]$

Apply $\frac{R_2}{2}→R_2$and $R_3+\frac{21}{5}×R_2→R_3$,

localid="1644431562941" $\left[\begin{array}{cccr}1& \frac{3}{5}& 0& -\frac{6}{5}\\ 0& 1& \frac{3}{2}& -\frac{1}{2}\\ 0& 0& \frac{73}{10}& \frac{73}{10}\end{array}\right]$

Apply $R_3×\frac{10}{73}→R_3$,

$\left[\begin{array}{cccr}1& \frac{3}{5}& 0& -\frac{6}{5}\\ 0& 1& \frac{3}{2}& -\frac{1}{2}\\ 0& 0& 1& 1\end{array}\right]$

Now, the matrix is in row-echelon form.

Writing the corresponding system of equations,

$$\begin{gathered} x+\frac{3}{5}y=-\frac{6}{5}......\left(i\right) \\ y+\frac{3}{2}z=-\frac{1}{2}......\left(ii\right) \\ z=1......\left(iii\right) \end{gathered}$$

Substitute $z=1$in equation (ii),

$$\begin{gathered} y+\frac{3}{2}×1=-\frac{1}{2} \\ y=-\frac{3}{2}-\frac{1}{2} \\ y=-\frac{4}{2} \\ y=-2 \end{gathered}$$

Substitute $y=-2$in equation (i),

$$\begin{gathered} x+\frac{3}{5}×\left(-2\right)=-\frac{6}{5} \\ x-\frac{6}{5}=-\frac{6}{5} \\ x=0 \end{gathered}$$

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

$$\begin{gathered} 0+\frac{3}{5}\left(-2\right)=-\frac{6}{5} \\ 0-\frac{6}{5}=-\frac{6}{5} \\ -\frac{6}{5}=-\frac{6}{5} \end{gathered}$$

This is true.

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

role="math" localid="1644591135020" $$\begin{gathered} -2+\frac{3}{2}\left(1\right)=-\frac{1}{2} \\ -2+\frac{3}{2}=-\frac{1}{2} \\ -\frac{1}{2}=-\frac{1}{2} \end{gathered}$$

This is also true.