$2 x-4 y+z=3$

$x-3 y+z=5$

$3 x-7 y+2 z=12$

Solution

$\left[\begin{array}{ccc} 2 & -4 & 1 \\ 1 & -3 & 1 \\ 3 & -7 & 2 \end{array}\right]$ $\left[\begin{array}{c} x \\ y \\ z \end{array}\right]$ = $\left[\begin{array}{c}3 \\5 \\ 12 \end{array}\right]$

$\left[\begin{array}{ccc|c} 2 & -4 & 1 & 3 \\ 1 & -3 & 1 & 5 \\ 3 & -7 & 2 & 12 \end{array}\right]$ - $\text { Augmented matrix. }$

$R_1 \rightarrow \frac{1}{2} R_1$ $\left[\begin{array}{ccc|c} 1 & -2 & 1 / 2 & 3 / 2 \\ 1 & -3 & 1 & 5 \\ 3 & -7 & 2 & 12 \end{array}\right]$

$R_2 \rightarrow R_2-R_1, \$ $\ R_3 \rightarrow R_3-3 R_1$

$\left[\begin{array}{ccc|c} 1 & -2 & 1 / 2 & 3 / 2 \\ 0 & -1 & 1 / 2 & 7 / 2 \\ 0 & -1 & 1 / 2 & 15 / 2 \end{array}\right]$

$R_2 \rightarrow-R_2$

$\left[\begin{array}{ccc|c} 1 & -2 & 1 / 2 & 3 / 2 \\ 0 & 1 & -1 / 2 & -7 / 2 \\ 0 & -1 & 1 / 2 & 15 / 2 \end{array}\right]$

$R_1 \rightarrow R_1+2 R_2, \$ $\ R_3 \rightarrow R_3+R_2$

$\left[\begin{array}{ccc|c} 1 & 0 & -1 / 2 & -11 / 2 \\ 0 & 1 & -1 / 2 & -7 / 2 \\0 & 0 & 0 & 8 / 2 \end{array}\right]$

$x-\frac{z}{2} = \frac{-11}{2}$

$y-\frac{z}{2} = \frac{-7}{2}$

$0.x + 0.y + 0.z = 4$

This systems has got no solution

$\therefore$ The given systems has no solution.

$\left[\begin{array}{ccc|c} 1 & 0 & -1 / 2 & -11 / 2 \\ 0 & 1 & -1 / 2 & -7 / 2 \\ 0 & 0 & 0 & 8 / 2 \end{array}\right]$

Rank $(A)=2 \$ Rank $(\frac{A}{B})=3$

$\Rightarrow$ The systems has no solution.

Solve the system

$(1+i)z_1-z_2=i$

$(1-i)z_1+(1+i)z_2=1$

Solution $\quad {\left[\begin{array}{cc|c} 1+i & -1 & i \\ 1-i & 1+i & 1 \end{array}\right]}$

$R_1 \rightarrow \frac{1}{1+i} R_1$

$\left[\begin{array}{cc|c} 1 & -1 / 1+i & i / 1+i \\ 1-i & 1+i & 1 \end{array}\right]$

$=\left[\begin{array}{cc|c} 1 & \frac{-1+i}{2} & \frac{-1+i}{2} \\ 1-i & 1+i & 1 \end{array}\right]$

$R_2 \rightarrow R_2-(1-i) R_1$
$\left[\begin{array}{cc|c} 1 & -\frac{1+i}{2} & \frac{1+i}{2} \\ 0 & 1 & 0 \end{array}\right]$

$1+i+\frac{(1-i)^2}{2} = 1+i+ \frac{1-2i-1}{2} = 1$

$R_1 \rightarrow R_1-\frac{-1+i}{2} R_2$

$\left[\begin{array}{cc|c} 1 & 0 & \frac{1+i}{2} \\ 0 & 1 & 0 \end{array}\right]$

$z_1= \frac{1+i}{2}$ & $z_2=0$

($\frac{1+i}{2},0$) is the solution.

$2 x+5 y+2 z=-1$

$x+2 y-3 z=5$

$5 x+12y+z=10$

Solution

$\left[\begin{array}{ccc} 2 & 5 & 2 \\ 1 & 2 & -3 \\ 5 & 12 & 1\end{array}\right]$ $\left[\begin{array}{c} x \\ y \\ z \end{array}\right]$ = $\left[\begin{array}{c} -1 \\ 5 \\ 10 \end{array}\right]$

$\left[\begin{array}{ccc|c} 2 & 5 & 2 & -1 \\ 1 & 2 & -3 & 5 \\ 5 & 12 & 1 & 10 \end{array}\right]$

$R_1 \rightarrow \frac{1}{2} R_1$

$\left[\begin{array}{ccc|c} 1 & 5 / 2 & 1 & -1 / 2 \\ 1 & 2 & -3 & 5 \\ 5 & 12 & 1 & 10 \end{array}\right]$

$R_2 \rightarrow R_2-R_1, \$ $\ R_3 \rightarrow R_3-5 R_1$

$\left[\begin{array}{ccc|c} 1 & 5 / 2 & 1 & -1 / 2 \\ 0 & -1 / 2 & -4 & 11 / 2 \\ 0 & -1 / 2 & -4 & 25 / 2 \end{array}\right]$

$R_2 \rightarrow -2R_2$

$\left[\begin{array}{ccc|c} 1 & 5 / 2 & 1 & -1 / 2 \\ 0 & 1 & 8 & -11 \\ 0 & -1 / 2 & -4 & 25 / 2 \end{array}\right]$

$R_1 \rightarrow R_1-\frac{5}{2} R_2, \ \$ $\ R_3 \rightarrow R_2+\frac{1}{2} R_2$

$\left[\begin{array}{ccc|c} 1 & 0 & -19 & -54 / 2 \\ 0 & 1 & 8 & -11 \\ 0 & 0 & 0 & 14 / 2 \end{array}\right]$

Rank (A) = 2, Rank ($\frac{A}{B}$) = 3

$\therefore Rank(\frac{A}{B}) > Rank(A)$

$\Rightarrow$ The given system has no solution.

$x+3y+4z=11$

$2x+3y+2z=7$

$4x+9y+10z=20$

$3x-2y+z=1$

This is an over determined system.

The number of unknowns = 3 and

The number of equations = 4

$\left[\begin{array}{ccc} 1 & 3 & 4 \\ 2 & 3 & 2 \\ 4 & 9 & 10 \\ 3 & -2 & 1 \end{array}\right]$ $\left[\begin{array}{c} x \\ y \\ z \end{array}\right]$ $=\left[\begin{array}{c} 11 \\ 7 \\ 20 \\ 1 \end{array}\right]$

$\left[\begin{array}{ccc|c} 1 & 3 & 4 & 11 \\ 2 & 3 & 2 & 7 \\ 4 & 9 & 10 & 20 \\ 3 & -2 & 1 & 1\end{array}\right]$

$R_2 \rightarrow R_2-2 R_1, \ \$ $R_3 \rightarrow R_3-4 R_1, \ \$ $R_4 \rightarrow R_4-3 R_1$

$\left[\begin{array}{ccc|c} 1 & 3 & 4 & 11 \\ 0 & -3 & -6 & -15 \\ 0 & -3 & -6 & -24 \\ 0 & -11 &-11 & -32 \end{array}\right]$

$R_2 \rightarrow \frac{1}{-3} R_2$

$\left[\begin{array}{ccc|c} 1 & 3 & 4 & 11 \\ 0 & 1 & 2 & 5 \\ 0 & -3 & -6 & -24 \\ 0 & -11 & -11 &-32 \end{array}\right]$

$R_1 \rightarrow R_1-3 R_2, \ \$ $R_3 \rightarrow R_3+3 R_2, \ \$ $R_4 \rightarrow R_4+11 R_2$

$\left[\begin{array}{ccc|c} 1 & 0 & -2 & -4 \\ 0 & 1 & 2 & 5 \\ 0 & 0 & 0 & -9 \\ 0 & 0 & 11 & 3\end{array}\right]$

$R_3 \longleftrightarrow R_4$

$\left[\begin{array}{ccc|c} 1 & 0 & -2 & -4 \\ 0 & 1 & 2 & 5 \\ 0 & 0 & 11 & 3 \\ 0 & 0 & 0 & -9\end{array}\right]$

$\therefore$ The systems has got no solution.

$x+2y+3t=7$

$z+4t=8$

Solution

This is an under determined systems.

Number of unknowns = 4 and number of equations = 2

$\left[\begin{array}{cccc} 1 & 2 & 0 & 3 \\ 0 & 0 & 1 & 4 \end{array}\right]$ $\left[\begin{array}{c}x \\ y \\ z \\ t \end{array}\right]$=$\left[\begin{array}{c} 7 \\ 8 \end{array}\right]$

Variables $x$ & $z$ are dependent variables. Variables $y$ & $t$ are independent variables.

$y=\lambda$ & $t=\mu \$ where $\lambda$ & $\mu$ are any arbitrary real numbers.

$x+2 y+3 t=7$

$z+4 t=8$

$x+2 \lambda+3 \mu=7$

$z+4 \mu=8$

$\Rightarrow z=8-4 \mu$

& $x=7-2 \lambda-3 \mu$

The solution set to this systems is $\biggl \{(7-2\lambda-3\mu, \lambda, 8-4\mu, \mu) | \lambda, \mu \in \mathbb{R} \biggl \}.$

$8 x+5 y+11 z=30$

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

$2 x-y+5 z=12$

Solution

$\left[\begin{array}{ccc} 8 & 5 & 11 \\ -1 & -4 & 2 \\ 2 & -1 & 5\end{array}\right]$ $\left[\begin{array}{l} x \\ y \\ z \end{array}\right]$=$\left[\begin{array}{c} 30 \\ 3 \\ 12\end{array}\right]$

${\left[\begin{array}{ccc|c} 8 & 5 & 11 & 30 \\ -1 & -4 & 2 & 3 \\ 2 & -1 & 5 & 12 \end{array}\right]}$

$R_1 \rightarrow \frac{1}{8} R_1$

$\left[\begin{array}{ccc|c} 1 & 5 / 8 & 11 / 8 & 30 / 8 \\ -1 & -4 & 2 & 3 \\ 2 & -1 & 5 &12\end{array}\right]$

$R_2 \rightarrow R_2+R_1, \$ $\ R_3 \rightarrow R_3-2 R_1$

$\left[\begin{array}{ccc|c} 1 & 5 / 8 & 11 / 8 & 30 / 8 \\ 0 & -27 / 8 & 27 / 8 & 54 / 8 \\ 0 & 1 / 4 & 9 / 4 & 18 / 4 \end{array}\right]$

$R_2 \rightarrow-\frac{-8}{27} R_2$

$\left[\begin{array}{ccc|c} 1 & 5 / 8 & 11 / 8 & 30 / 8 \\ 0 & 1 & -1 & -2 \\ 0 & 1 / 4 & 9 / 4 & 18/ 4 \end{array}\right]$

$R_1 \rightarrow R_1-\frac{5}{8} R_2, \$ $\ R_3 \rightarrow R_3-\frac{1}{4} R_2$

$\left[\begin{array}{ccc|c} 1 & 0 & 16 / 8 & 40 / 8 \\ 0 & 1 & -1 & -2 \\ 0 & 0 & 10 / 4 & 20 / 4\end{array}\right]$

$R_3 \rightarrow \frac{4}{10} R_3$

$\left[\begin{array}{ccc|c} 1 & 0 & 16 / 8 & 40 / 8 \\ 0 & 1 & -1 & -2 \\ 0 & 0 & 1 & 2\end{array}\right]$

$R_2 \rightarrow R_2+R_3, \ \$ $\ R_1 \rightarrow R_1-\frac{16}{8} R_3$

$\left[\begin{array}{ccc|c} 1 & 0 & 0 & 1 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 2 \end{array}\right]$

$(1,0,2)$ is the solution.