Let A = \(\left( \begin{matrix} 1 & 4 \\ 2 & 8 \end{matrix} \right) \)
Order of A is 2 \(\times\) 2 [Since minimum of (2,2) is 2]
Consider the second order minor \(\left| \begin{matrix} 1 & 4 \\ 2 & 8 \end{matrix} \right| \)
= 8-8
= 0
Since the second order minor vanishes \(\rho (A)\neq 2\)
Consider a first order minor \(\left[ 1 \right] \neq 0\)
There is a minor of order 1, which is not zero
\(\therefore \rho \left( A \right) =1\)

Let \(A=\left( \begin{matrix} i & -1 \\ 3 & -6 \end{matrix} \right) \)
Order of A is 2 \(\times\) 2
\(\therefore \rho (A)\le 2\) [Since minimum of (2, 2) is 2]
 Consider the second order minor
\(\left| \begin{matrix} 1 & -1 \\ 3 & -6 \end{matrix} \right| =-6-(-3)\)
= -6 + 3 = -3
\(\neq 0\)
There is a minor of order 2, which is not zero
\(\therefore \rho (A)=2\)

Given A =\(\left( \begin{matrix} 1 & -3 \\ 9 & 1 \end{matrix}\begin{matrix} 4 & 7 \\ 2 & 0 \end{matrix} \right) \)
\(\left( \begin{matrix} 1 & -3 \\ 0 & 28 \end{matrix}\begin{matrix} 4 & 0 \\ -34 & -63 \end{matrix} \right) { R }_{ 2 }\rightarrow { R }_{ 2 }-9{ R }_{ 1 }\)
\(-\left( \begin{matrix} 1 & -3 \\ 0 & 0 \end{matrix}\begin{matrix} 4 & 0 \\ \frac { 10 }{ 3 } & -63 \end{matrix} \right) { R }_{ 2 }\rightarrow { R }_{ 2 }+\frac { 28 }{ 3 } .{ R }_{ 1 }\)
The last equivalent matrix is in echelon form and there are 2 non-zero rows
\(\therefore \rho (A)=2\)

Let \(A=\left( \begin{matrix} 5 & 6 \\ 7 & 8 \end{matrix} \right) \)
Order of A is 2 \(\times\) 2
\(\therefore \rho (A)\le 2\)  [Since minimum of (2, 2) is 2]
Consider the second order minor
\(\left| \begin{matrix} 5 & 6 \\ 7 & 8 \end{matrix} \right| =40-42\)
= \(-2\neq 0\)
There is a minor of order 2, which is not zero
\(\therefore \rho (A)=2\)

Let A =\(\begin{pmatrix} -5 & -7 \\ 5 & 7 \end{pmatrix}\)
Order of A is 2 \(\times\) 2 
∴\(\rho \)(A)\(\le \)2
Consider the second order minor \(\begin{vmatrix} -5 & -7 \\ 5 & 7 \end{vmatrix}\)=0
Since the second order minor vanishes,\(\rho (A)\neq 2\)
Consider a first order minor \(\left| -5 \right| \neq 0\)
There is a minor of order 1, which is not zero 
\(\therefore \rho (A)=1\)

Let A =\(\begin{pmatrix} 1 & 5 \\ 3 & 9 \end{pmatrix}\)
Order of A is 2 \(\times\) 2
∴ \(\rho \) (A) \(\le \) 2
Consider the second order minor
\(\begin{vmatrix} 1 & 5 \\ 3 & 9 \end{vmatrix}=-6\neq 0\)
There is a minor of order 2, which is not zero.
∴ \(\rho \) (A) \(\le \) 2