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# Examine the consistency of the following system of equation. If it is consistent than solve the same. $x+y-z=1\;,2x+2y-2z=2\;,-3x-3y+3z=-3$

Toolbox:
Step 1
The matrix equation is $\begin{bmatrix} 1 & 1 & -1 \\ 2 & 2 & -2 \\ -3 & -3 & 3 \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix}1 \\ 2 \\ -3 \end{bmatrix}$
$AX = B$
The angmented matrix
$[A,B] = \begin{bmatrix} 1 & 1 & -1 & 1 \\ 2 & 2 & -2 & 2 \\ -3 & -3 & 3 & 3 \end{bmatrix} \sim \begin{bmatrix} 1 & 1 & -1 & 1 \\ 1 & 1 & -1 & 1 \\ 1 & 1 & -1 & 1 \end{bmatrix} R_2 \rightarrow \large\frac{1}{2} R_2$
$R_3 \rightarrow \large\frac{1}{3}R_3$
$\sim \begin{bmatrix} 1 & 1 & -1 & 1 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{bmatrix} R_2 \rightarrow R_2-R_1$
$R_3 \rightarrow R_3-R_1$
Step 2
The last equivalent matrix is in echelon form and it can be seen that $\rho(A)= \rho(A,B)=1$ [ one nonzero row]
$\therefore$ the equations are consistent with infinitely many solutions.
Step 3
The equations reduce to $x+y-z=1$
Let $x = s, y=t, s,t \in R.$
Then $z = s+t-1$
$\therefore$ the solutions are given by $(x,y,z) = (s,t,s+t-1)s,t \in R$