\(\left[ \begin{matrix} 0 \\ \begin{matrix} 0 \\ 8 \end{matrix} \end{matrix}\begin{matrix} 1 \\ \begin{matrix} 2 \\ 1 \end{matrix} \end{matrix}\begin{matrix} 2 \\ \begin{matrix} 4 \\ 0 \end{matrix} \end{matrix}\begin{matrix} 1 \\ \begin{matrix} 3 \\ 2 \end{matrix} \end{matrix} \right] \)
Let A =\(\left[ \begin{matrix} 0 \\ \begin{matrix} 0 \\ 8 \end{matrix} \end{matrix}\begin{matrix} 1 \\ \begin{matrix} 2 \\ 1 \end{matrix} \end{matrix}\begin{matrix} 2 \\ \begin{matrix} 4 \\ 0 \end{matrix} \end{matrix}\begin{matrix} 1 \\ \begin{matrix} 3 \\ 2 \end{matrix} \end{matrix} \right] \)
A is a matrix of order 3 \(\times\) 4
∴ \(\rho \)(A) ≤ min(3, 4) = 3 
The highest order of minor of A is 3
It is \(\left| \begin{matrix} 0 & 1 & 2 \\ 0 & 2 & 4 \\ 8 & 1 & 0 \end{matrix} \right| \) = 0+0-8(4-4) = 0
[Expanded along C₁]
Also, \(\left| \begin{matrix} 0 & 2 & 1 \\ 0 & 4 & 3 \\ 8 & 0 & 2 \end{matrix} \right| =0+0-8\left| \begin{matrix} 2 & 1 \\ 4 & 3 \end{matrix} \right| \)
[Expanded along C₁]
= -8(6-4) = -8(2) = -16 ≠ 0
∴ \(\rho \)(A) = 3

\(\left[ \begin{matrix} 1 & -2 & 3 \\ 2 & 4 & -6 \\ 5 & 1 & -1 \end{matrix} \right] \)
Let A =\(\left[ \begin{matrix} 1 & -2 & 3 \\ 2 & 4 & -6 \\ 5 & 1 & -1 \end{matrix} \right] \)
A is a matrix of order 3 \(\times\) 3
∴ \(\rho \)(A) ≤ min(3, 3) = 2
The highest order of minor of A is 3
It is \(\left| \begin{matrix} 1 & -2 & 3 \\ 2 & 4 & -6 \\ 5 & 1 & -1 \end{matrix} \right| =1\left| \begin{matrix} 4 & -6 \\ 1 & -1 \end{matrix} \right| +2\left| \begin{matrix} 2 & -6 \\ 5 & -1 \end{matrix} \right| +3\left| \begin{matrix} 2 & 4 \\ 5 & 1 \end{matrix} \right| \)
[Expanded along R₁]
= 1(-4+6)+2(-2+30)+3(2-20)
= 1(2)+2(28)+3(-18)
= 2+56-54 = 58-54 = 4 ≠ 0
∴ \(\rho \)(A) = 3

Let A = \(\left[ \begin{matrix} 1 & 2 & 3 \\ 2 & 1 & 4 \\ 3 & 0 & 5 \end{matrix} \right] \). Applying elementary row operations, we get
A \(\overset { \begin{matrix} { R }_{ 2 }\longrightarrow { R }_{ 2 }-2{ R }_{ 1 } \\ { R }_{ 3 }\longrightarrow { R }_{ 3 }-3{ R }_{ 1 } \end{matrix} }{ \longrightarrow } \left[ \begin{matrix} 1 & 2 & 3 \\ 0 & -3 & -2 \\ 0 & -6 & -4 \end{matrix} \right] \overset { { R }_{ 3 }\longrightarrow { R }_{ 3 }-2{ R }_{ 2 } }{ \longrightarrow } \left[ \begin{matrix} 1 & 2 & 3 \\ 0 & -3 & -2 \\ 0 & 0 & 0 \end{matrix} \right] \).
The last equivalent matrix is in row-echelon form. It has two non-zero rows. So, ρ(A) = 2.

\(\left[ \begin{matrix} 1 \\ 3 \end{matrix}\begin{matrix} -2 \\ -6 \end{matrix}\begin{matrix} -1 \\ -3 \end{matrix}\begin{matrix} 0 \\ 1 \end{matrix} \right] \)
Let A =\(\left[ \begin{matrix} 1 \\ 3 \end{matrix}\begin{matrix} -2 \\ -6 \end{matrix}\begin{matrix} -1 \\ -3 \end{matrix}\begin{matrix} 0 \\ 1 \end{matrix} \right] \)
A is a matrix of order (2 \(\times\) 4)
∴ \(\rho \)(A) ≤ min(2, 4) = 2
The highest order of minor of A is 2
It is \(\left| \begin{matrix} 1 & -2 \\ 3 & -6 \end{matrix} \right| \) = -6 + 6 = 0
Also, \(\left| \begin{matrix} -1 & 0 \\ -3 & 1 \end{matrix} \right| \) = -1 + 0 = -1 ≠ 0
∴ \(\rho \)(A) = 2

\(\left[ \begin{matrix} 2 & -4 \\ -1 & 2 \end{matrix} \right] \)
Let A =\(\left[ \begin{matrix} 2 & -4 \\ -1 & 2 \end{matrix} \right] \)
A is a matrix of order 2 \(\times\) 2
∴ \(\rho \)(A) ≤ min(2,2) = 2
The highest order of minor of A is 2
it is \(\left| \begin{matrix} 2 & -1 \\ -1 & 2 \end{matrix} \right| \)= 4 - 4 = 0
So, \(\rho \)(A)<2
Next consider the minor of order 1 |2| = 2 ≠ 0
∴ \(\rho \)(A) = 1