(i) Let A = \(\begin{pmatrix} 5 &6 \\[0.3em] 7 & 8 \end{pmatrix}\)

order of A is 2 x 2.

ρ(A) ≤ 2 

Consider the second order minor 

\(\begin{vmatrix} 5 &6 \\[0.3em] 7 & 8 \end{vmatrix}\)= 40 – 42 = -2 ≠ 0 

There is a minor of order 2, which is not zero. 

ρ(A) = 2

(ii) Let A = \(\begin{pmatrix} 1 &-1 \\[0.3em] 3 & -6 \end{pmatrix}\)

Order of A is 2 × 2 

ρ(A) ≤ 2 

Consider the second order minor 

\(\begin{vmatrix} 1 & -1 \\[0.3em] 3 & -6 \end{vmatrix}\)= -6 + 3 = -3 ≠ 0

ρ(A) = 2

 (iii) Let A = \(\begin{pmatrix} 1 &4 \\[0.3em] 2 & 8 \end{pmatrix}\)

Since A is of order 2 × 2, ρ(A) ≤ 2 

Now \(\begin{vmatrix} 1 & 4 \\[0.3em] 2 & 8 \end{vmatrix}\) = 8 – 8 = 0 

Since second order minor vanishes ρ(A) ≠ 2 But first order minors, |1|, |4|, |2|, |8| are non zero. ρ(A) = 1

(iv) Let A =\( \begin{pmatrix} 2 & -1 & 1 \\ 3 & 1 & -5 \\ 1 & 1 & 1 \end{pmatrix} \)

Oradar of A is 3 x 3

ρ(A) ≤ 3 

Consider the third order minor

\( \begin{pmatrix} 2 & -1 & 1 \\ 3 & 1 & -5 \\ 1 & 1 & 1 \end{pmatrix} \)= 2(1 + 5) + 1 (3 + 5) + 1(3 – 1) 

= 2(6) + 8 + 2 

= 22 ≠ 0 

There is a minor of order 3, which is non zero. ρ(A) = 3