Find the rank of A = \(\begin{pmatrix} 1 & 2 & 3 \\ 2 & 3 & 4 \\ 3 &

1.	Find the rank of A = \(\begin{pmatrix} 1 & 2 & 3 \\ 2 & 3 & 4 \\ 3 & 5 & 7 \end{pmatrix}\)
Answer» Given A = \(\begin{pmatrix} 1 & 2 & 3 \\ 2 & 3 & 4 \\ 3 & 5 & 7 \end{pmatrix}\) Consider \(\begin{pmatrix} 1 & 2 & 3 \\ 2 & 3 & 4 \\ 3 & 5 & 7 \end{pmatrix}\) = 1((21 – 20) – 2(14 – 12) + 3(10 – 9)) = 1 – 4 + 3 = 0 Since third order minor equals zero, ρ(A) < 3 Consider = \(\begin{vmatrix} 1 & 2 \\ 2 & 3 \end{vmatrix}\) = 3 – 4 = -1 ≠ 0 There is a minor of order 2 which is not zero. Hence ρ(A) = 2

Find the rank of A = \(\begin{pmatrix} 1 & 2 & 3 \\ 2 & 3 & 4 \\ 3 & 5 & 7 \end{pmatrix}\)

Answer»

Given A = \(\begin{pmatrix} 1 & 2 & 3 \\ 2 & 3 & 4 \\ 3 & 5 & 7 \end{pmatrix}\)

Consider \(\begin{pmatrix} 1 & 2 & 3 \\ 2 & 3 & 4 \\ 3 & 5 & 7 \end{pmatrix}\)

= 1((21 – 20) – 2(14 – 12) + 3(10 – 9))

= 1 – 4 + 3

= 0

Since third order minor equals zero, ρ(A) < 3

Consider = \(\begin{vmatrix} 1 & 2 \\ 2 & 3 \end{vmatrix}\) = 3 – 4 = -1 ≠ 0

There is a minor of order 2 which is not zero. Hence ρ(A) = 2