Q.Test the consistency of the system $x+y+z=6$, $x-y+z=2$, $x+2y-z=2$ using the rank method, and solve it if consistent.

Comparing $\rho(A)$ and $\rho([A|B])$ against the number of unknowns tells us whether — and how — the system can be solved.

Elimination gives $y=2$, then $x+z=4$ and $x-z=-2$, so $x=1,z=3$.

Checking consistency

The coefficient matrix $A=\begin{pmatrix}1&1&1\\1&-1&1\\1&2&-1\end{pmatrix}$ has $\det A=1(1-2)-1(-1-1)+1(2+1)=-1+2+3=4\neq0$, so $\rho(A)=3$; since $\rho(A)$ cannot exceed $\rho([A|B])$ and the augmented matrix has the same 3 independent rows, $\rho([A|B])=3$ too. With $\rho(A)=\rho([A|B])=3=n$, the system has a UNIQUE solution.

Solving

Subtracting equation 2 from equation 1: $(x+y+z)-(x-y+z)=6-2 \Rightarrow 2y=4 \Rightarrow y=2$.

From equation 1: $x+2+z=6 \Rightarrow x+z=4$.

From equation 3: $x+2(2)-z=2 \Rightarrow x-z=-2$.

Adding $x+z=4$ and $x-z=-2$: $2x=2 \Rightarrow x=1$, so $z=4-1=3$.

Check (independent recomputation, substituting into ALL three original equations): $1+2+3=6$ ✓; $1-2+3=2$ ✓; $1+2(2)-3=1+4-3=2$ ✓ — all three are satisfied.

Solving the system first and only checking consistency as an afterthought — the rank test is meant to confirm a unique solution EXISTS before spending effort solving, and is essential when a quick manual solve isn't obvious.