91Ó°ÊÓ

Consider the given system of equations,

$\left\{\begin{array}{l}-x-3y+2z=14\\ -x+2y-3z=-4\\ 3x+y-2z=6\end{array}\right.$

The augmented matrix for the given system of equations is

$\left[\begin{array}{cccr}-1& -3& 2& 14\\ -1& 2& 3& -4\\ 3& 1& -2& 6\end{array}\right]$

Apply $\frac{R_1}{-1}→R_1$and $R_2+R_1→R_2$,

$\left[\begin{array}{cccr}1& 3& -2& -14\\ 0& 5& -5& -18\\ 3& 1& -2& 6\end{array}\right]$

Apply $R_3-3×R_1→R_3$and $\frac{R_2}{5}→R_2$,

$\left[\begin{array}{cccr}1& 3& -2& -14\\ 0& 1& -1& -\frac{18}{5}\\ 0& -8& 4& 48\end{array}\right]$

Apply $R_3+8×R_{2→}{R}_3$,

$\left[\begin{array}{cccr}1& 3& -2& -14\\ 0& 1& -1& -\frac{18}{5}\\ 0& 0& -4& \frac{96}{5}\end{array}\right]$

Apply $\frac{R_3}{4}→R_3$,

$\left[\begin{array}{cccr}1& 3& -2& -14\\ 0& 1& -1& -\frac{18}{5}\\ 0& 0& 1& -\frac{24}{5}\end{array}\right]$

Now, the matrix is in row-echelon form.

Writing the corresponding system of equations,

$$\begin{gathered} x+3y-2z=-14......\left(i\right) \\ y-z=-\frac{18}{5}......\left(ii\right) \\ z=-\frac{24}{5}......\left(iii\right) \end{gathered}$$

Substitute $z=-\frac{24}{5}$in equation (ii),

$$\begin{gathered} y-\left(-\frac{24}{5}\right)=-\frac{18}{5} \\ y=-\frac{18}{5}-\frac{24}{5} \\ y=-\frac{42}{5} \end{gathered}$$

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

$$\begin{gathered} x+3\left(-\frac{42}{5}\right)-2\left(-\frac{24}{5}\right)=-14 \\ x-\frac{126}{5}+\frac{48}{5}=-14 \\ x=-14+\frac{78}{5} \\ x=\frac{8}{5} \end{gathered}$$

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

$$\begin{gathered} \frac{8}{5}+3\left(-\frac{42}{5}\right)-2\left(-\frac{24}{5}\right)=-14 \\ \frac{8}{5}-\frac{126}{5}+\frac{48}{5}=-14 \\ -\frac{70}{5}=-14 \\ -14=-14 \end{gathered}$$

This is true.

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

$$\begin{gathered} -\frac{42}{5}-\left(-\frac{24}{5}\right)=-\frac{18}{5} \\ -\frac{42}{5}+\frac{24}{5}=-\frac{18}{5} \\ -\frac{18}{5}=-\frac{18}{5} \end{gathered}$$

This is also true.