Let \(A=\left[ \begin{matrix} 7 & -1 \\ 2 & 1 \end{matrix} \right] \)
The order of A is 2 x 2
\(\rho (A)\le min(2,2)\)
\(\left[ \begin{matrix} 7 & -1 \\ 2 & 1 \end{matrix} \right] =7-(-2)=7+29\neq 0\)
The highest order of non-vanishing minor of A is 2
\(\therefore \rho (A)=2\)

Let A = \(\left( \begin{matrix} 2 & -4 \\ -1 & 2 \end{matrix} \right) \)
The order of A is 2 \(\times\) 2
\(\rho (A)\le min(2,2)\)
\(\Rightarrow \rho (A)\le 2\)
\(\left| \begin{matrix} 2 & -4 \\ -1 & 2 \end{matrix} \right| =4-4=0\)
Since the second order minor vanishes \(\rho (A)\neq 2\)
 We have to try for atleast one non-zero first order minor.
ie. atleast one non-zero element of A.
This is possible because A has non-zero element
\(\therefore \rho (A)-1\)

\(\Delta =\left| \begin{matrix} 1 & 2 \\ 1 & 1 \end{matrix} \right| =1(1)-(1)(2)=1-2=-1\)
Since \(\Delta \neq O\)  Cramer's rule can be applied and the system is consistent with unique solution.
\(\Delta x=\left| \begin{matrix} 3 & 2 \\ 2 & 1 \end{matrix} \right| =3-4=-1\)
\(\Delta y=\left| \begin{matrix} 1 & 3 \\ 1 & 2 \end{matrix} \right| =2-3=-1\)
\(x=\cfrac { \Delta x }{ \Delta } =\cfrac { -1 }{ -1 } =1\)
\(y=\cfrac { \Delta x }{ \Delta } =\cfrac { -1 }{ -1 } =1\)
\(\therefore\) solution set is {1, 1}

The non-homogeneous equations are
x + 2y = 3, 2x + 4y = 6

Here \(\rho (A)=1\) and \(\rho \left( \left[ A,B \right] \right) =1\)
Since \(\rho (A)=\rho \left[ \left( A,B \right) \right] =1<\) Number of unknowns, the given system is consistent with infinitely many solutions.
To find the solution, let us rewrite the above echelon form into the matrix form, we get
\(\left( \begin{matrix} 1 & 2 \\ 0 & 0 \end{matrix} \right) \left( \begin{matrix} x \\ y \end{matrix} \right) =\left( \begin{matrix} 3 \\ 0 \end{matrix} \right) \)
\(\Rightarrow x+2y=3\)
let \(y=k,k\varepsilon R\)
\((1)\Rightarrow x+2k=3\Rightarrow x=3-2k\)
\(\therefore\) Solution set is \(\left\{ 3-2k,k \right\} ,k\epsilon R\)
For different values of k; we get infinite number of solutions

Given non-homogeneous equations are x + y + z = 6, x + 2y + 3z = 14, x + 4y + 7z = 30

Here \(\rho (A)=\rho (A,B) = 2\) 
\(\therefore\) The given system is consistent