Three possibilities that arise:

1) No solution.

2) Unique solution.

3) Multiple solution.

$-3 x-2 y+4 y=9$

$\quad$$\quad$$\quad$$3 y-2 z=5$

$\quad$$4 x-3 y+2 z=7$

${\left[\begin{array}{ccc}-3 & -2 & 4 \\ 0 & 3 & -2 \\ 4 & -3 & 2\end{array}\right]\left[\begin{array}{l}x \\ y \\ z\end{array}\right]=\left[\begin{array}{l}9 \\ 5 \\ 7\end{array}\right]}$

$x=3 \quad y=7 \quad z=8$

$(3,7,8)$

This is the solution of the given system.

$A x=B$ be the given system of linear equations

Let $\rho$ be the elementary row operations that one performed on $A$ & $B$. Let $\tilde{A}$ denote $\rho(A)$ & $\tilde{B}$ denote $\rho(B)$.

$\tilde{A} x=\tilde{B}$ - the newly obtained system.

$A$ is row equivalent to $\tilde{A}$ &

$B$ is now equivalent to $\tilde{B}$

i.e., $\vec{A}$ is obtained from $A$ just by applying the row elementary operations.

The systems,

$A x=B$ & $\tilde{A} x=\tilde{B}$

have the same set of solutions.

If $\rho$ is an elementary row operation then $\rho(A)=\rho(I) A$.

Let, $\rho_1, \rho_2, \ldots, \rho_s$ be a finite set of elementary row operations. Let, $A$ be an $n \times n$ matrix.

Then,

a) $(\rho_s \circ \rho_{s-1} \circ \rho_{s-3} \circ \cdots \circ \rho_2 \circ \rho_1)$ (A)

$=\left(\rho_s \circ \rho_{s-1} \cdot \cdots \circ \rho_2 \circ \rho_s\right)(I)(A)$

b) $\left(\rho_s \circ \rho_{s-1} \circ \cdots \circ \rho_2 \circ \rho_1\right)(A)$

$=\rho_s(I) \rho_{s-1}(I) \cdots \rho_2(I) \rho_1(I) A$

c) If $\rho$ is an elementary row operation & if $A$ & $B$ are any two matrices which can be multiplied, then

$\rho(A B) = \rho(A) \cdot B$

$\rho(A B) = \rho(I)(A B)$

$=(\rho(I) A)(B)$

$=\rho(A)(B) .$

$\therefore$ If $\rho_s, \rho_{s-1}, \ldots . \rho_2, \rho_1$ is a finite set of elementary row operations then

$\left(\rho_{s} \circ \rho_{s-1} \circ \rho_{s-2} \circ \cdots \circ \rho_2 \circ \rho_1\right)(A B)$

$=\left(\rho_s \cdot \rho_{s-1} \cdot \rho_{s-2} \circ \cdots \circ \rho_2 \circ \rho_1\right)(A) B .$

$A X=B$ is the given system.

Suppose, $x$ is the solution to the system, then $\rho(Ax)=\rho(B)$

where, $\rho$ is any elementary row operation.

$\rho(A) x=\rho(B)$

$\tilde{A} x=\tilde{D}$

$2 x-3 y=-21$

$3 x+2 y=1$

$8 x-5 y=-49$

This is an over determined system

${\left[\begin{array}{cc}2 & -3 \\ 3 & 2 \\ 8 & -5\end{array}\right]\left[\begin{array}{c}x \\ y\end{array}\right]=\left[\begin{array}{c}-21 \\ 1 \\ -49\end{array}\right] }$

${\left[\begin{array}{cc|c}2 & -3 & -21 \\ 3 & 2 & 1 \\ 8 & -5 & -49\end{array}\right]}$

$R_1 \rightarrow \frac{1}{2} R_1\left[\begin{array}{cc|c}1 & -\frac{3}{2} & -\frac{21}{2} \\ 3 & 2 & 1 \\ 8 & -5 & -49\end{array}\right]$

$\begin{aligned} & R_2 \rightarrow R_2-3 R_1 \\ & R_3 \rightarrow R_3-3 R_1\end{aligned}\left[\begin{array}{cc|c}1 & -\frac{3}{2} & -\frac{21}{2} \\ 0 & 2+\frac{9}{2} & 1+\frac{63}{2} \\ 0 & -5+\frac{24}{2} & -49+\frac{162}{2}\end{array}\right]$

$= \left[\begin{array}{cc|c}1 & -\frac{3}{2} & -\frac{21}{2} \\ 0 & \frac{12}{3} & \frac{65}{2} \\ 0 & \frac{14}{2} & 35\end{array}\right]$

$R_2 \rightarrow \frac{2}{13} R_2\left[\begin{array}{cc|c}1 & -\frac{3}{2} & -\frac{21}{2} \\ 0 & 1 & 5 \\ 0 & \frac{14}{2} & 35\end{array}\right]$

$\begin{aligned} & R_1 \rightarrow \frac{3}{2} R_2+R_1 \\ & R_3 \rightarrow \frac{-14}{2} R_2+R_3\end{aligned}\left[\begin{array}{ll|l}1 & 0 & \frac{15}{2}-\frac{21}{2} \\ 0 & 1 &\hspace {0.5cm} 5 \\ 0 & 0 & 35-35\end{array}\right]$

$=\left[\begin{array}{cc|c}1 & 0 & -3 \\ 0 & 1 & 5 \\ 0 & 0 & 0\end{array}\right]$

$x = -3, y = 5$

Solve the system:

$2 x-3 y+2 z=13$

$3 x+y-z=2$

$3 x-4 y-3z=1$

$\begin{aligned} & R_1 \rightarrow R_1+\frac{3}{2} R_2, \ & R_3 \rightarrow R_3-\frac{1}{2} R_2\end{aligned}$

$\left[\begin{array}{ccc|c}1 & 0 & 1 -\frac{12}{11} & \frac{13}{2}+-\frac{105}{11} \\ 0 & 1 & -\frac{8}{11} & -\frac{35}{11} \\ 0 & 0 & -6+\frac{4}{11} & -\frac{37}{2}+\frac{35}{22}\end{array}\right]$

${\left[\begin{array}{lll|l}1 & 0 & -\frac{1}{11} & \frac{38}{22} \\ 0 & 1 & -\frac{8}{11} & -\frac{35}{11} \\ 0 & 0 & -\frac{62}{11} & -\frac{372}{22}\end{array}\right]}$

$R_3 \rightarrow \frac{-11}{62} R_3\left[\begin{array}{ccc|c}1 & 0 & -\frac{1}{11} & \frac{38}{22} \\ 0 & 1 & -\frac{8}{11} & -3 \frac{5}{11} \\ 0 & 0 & 1 & 3\end{array}\right]$

$\begin{array}{l}R_1 \longrightarrow R_1+\frac{1}{11} R_3 ,\quad R_2 \longrightarrow R_2+\frac{8}{11} R_3\end{array}$

$\left[\begin{array}{lll|l}1 & 0 & 0 & \frac{38}{22}+\frac{3}{11} \\ 0 & 1 & 0 & -\frac{36}{11}+\frac{24}{11} \\ 0 & 0 & 1 & 3\end{array}\right]$

$= {\left[\begin{array}{lll|l}1 & 0 & 0 & 2 \\ 0 & 1 & 0 & -1 \\ 0 & 0 & 1 & 3\end{array}\right]}$

$(2,-1,3)$