91Ó°ÊÓ

Consider the given system of equations,

$\left\{\begin{array}{l}x-y+2z=-4\\ 2x+y+3z=2\\ -3x+3y-6z=12\end{array}\right.$

The augmented matrix for the given system of equations is

$\left[\begin{array}{cccr}1& -1& 2& -4\\ 2& 1& 3& 2\\ -3& 3& -6& 12\end{array}\right]$

Apply $R_2-2×R_1→R_2$and $R_3→R_3+3×R_1$,

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

Apply $R_2→\frac{R_2}{3}$,

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

Now, the matrix is in row-echelon form.

Writing the corresponding system of equations,

$$\begin{gathered} x-y+2z=-4......\left(i\right) \\ y-\frac{1}{3}z=\frac{10}{3}......\left(ii\right) \\ 0=0......\left(iii\right) \end{gathered}$$

As equation (iii) is a true statement.

Therefore, it is a dependent system.

Hence, the system of linear equations have infinitely many solution.

Solve for $y$in terms of $z$in equation (i),

$$\begin{gathered} -3y+z=-10 \\ -3y=-10-z \\ y=\frac{10+z}{3} \end{gathered}$$

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

$$\begin{gathered} x-\frac{10-z}{3}+2z=-4 \\ \frac{3x-10-z+6z}{3}=-4 \\ 3x-10+5z=-12 \\ x=-\frac{2}{3}-\frac{5}{3}z \end{gathered}$$

Here, $z$is any real number.

Substitute the values in equation (i),

$$\begin{gathered} \left(-\frac{2}{3}-\frac{5}{3}z\right)-\left(\frac{10+z}{3}\right)+2z=-4 \\ -\frac{2}{3}-\frac{5}{3}z-\frac{10-z}{3}+2z=-4 \\ -2-5z-10-z+6z=-12 \\ -12=-12 \end{gathered}$$

This is true.