Q.Test for consistency and if possible, solve the following systems of equations by rank method.

We test consistency by row-reducing $[A|B]$ to echelon form and comparing $\rho(A)$ with $\rho([A|B])$: if the two ranks are equal to the number of unknowns ($n=3$) the solution is unique, if they are equal but less than 3 there are infinitely many solutions, and if they differ the system is inconsistent.

Step 1. Part (i): write $[A|B]$ and reduce.

Step 2. Part (i): read off the ranks and solve. Three non-zero rows, so $\rho(A)=\rho([A|B])=3=n$ — consistent with a unique solution. Back-substituting: $-7z=-7\Rightarrow z=1$; $3y=3\Rightarrow y=1$; $x-y+2z=2\Rightarrow x=2+1-2=1$. So $x=1,\ y=1,\ z=1$.

Step 3. Part (ii): write $[A|B]$ and reduce.

Step 4. Part (ii): read off the ranks and solve. Only two non-zero rows, so $\rho(A)=\rho([A|B])=2<3=n$ — consistent with infinitely many solutions (one free parameter). Put $z=t$: from row 2, $10y-5z=-1\Rightarrow y=\dfrac{z}{2}-\dfrac1{10}=\dfrac t2-\dfrac1{10}$; from row 1, $x=1+3y-2z=1+3\!\left(\dfrac t2-\dfrac1{10}\right)-2t=\dfrac{7}{10}-\dfrac t2$. So $x=\dfrac{7}{10}-\dfrac t2,\ y=\dfrac t2-\dfrac1{10},\ z=t,\ t\in\mathbb R$.

Step 5. Part (iii): write $[A|B]$ and reduce.

Step 6. Part (iii): read off the ranks. The last row reads $0=-2$, which is impossible, so $\rho(A)=2$ (only the first two rows are independent) while $\rho([A|B])=3$ (the third row is non-zero purely because of its constant term). Since $\rho(A)\ne\rho([A|B])$, the system is inconsistent — no solution.

Step 7. Part (iv): write $[A|B]$ and reduce.

(Rows 2 and 3 were already exact multiples of row 1: $R_2=3R_1$ and $R_3=2R_1$.)

Step 8. Part (iv): read off the ranks and solve. Only one non-zero row, so $\rho(A)=\rho([A|B])=1<3=n$ — consistent with infinitely many solutions (two free parameters). Put $y=s,\ z=t$: from $2x-y+z=2$, $x=1+\dfrac s2-\dfrac t2$. So $x=1+\dfrac s2-\dfrac t2,\ y=s,\ z=t,\ s,t\in\mathbb R$.

Rank method — row-reduce $[A|B]$ and compare $\rho(A)$ with $\rho([A|B])$

• Concluding 'no solution' whenever the ranks are unequal without checking which one is bigger — inconsistency needs $\rho([A|B])>\rho(A)$ specifically.
• Stopping at row-echelon form without correctly reading off the free parameter(s) once $\rho(A)<n$.
• Missing that a row like $R_2=3R_1$ makes an equation entirely redundant, as in part (iv).