Fsc Part 1 Mathematics (Complete Solution)

# Q10 Find the rank of the following matrices.

(i)

 1 -1 2 -1 2 -6 5 1 3 5 4 -3

SOLUTION:

 1 -1 2 -1 2 -6 5 1 3 5 4 -3

=∼R∼

 1 -1 2 1 0 -4 1 -1 0 8 -2 6

By
R2-2R1→ R3
R3-3R1→ R3
=∼R∼

 1 -1 2 1 0 1 -1/4 1/4 0 4 -1 -3

By
(-1/4)R2→ R3 ,(1/2)R3→ R3
=∼R∼

 1 0 7/4 5/4 0 1 -1/4 1/4 0 0 0 -4

By
R1+R2→ R1
R3-4R2→ R3
= ∼R∼

 1 0 7/4 5/4 0 1 -1/4 1/4 0 0 0 1

By
(-1/4)R3→ R3

hence the rank of given matrix is 3.

(ii)

 1 -4 -7 2 -5 1 1 -2 3 3 -7 4

SOLUTION:

 1 -4 -7 2 -5 1 1 -2 3 3 -7 4

= ∼R∼

 1 -4 -7 0 3 15 0 2 10 0 5 25

R2-2R1→ R2
R3-R1→ R3
R4-3R1→ R4
=∼R∼

 1 -4 -7 0 1 5 0 2 10 0 1 5

R2 -R3 → R3
(1/5)R3→ R3
=∼R∼

 1 0 13 0 1 5 0 0 0 0 0 0

By
R1+4R2→ R1
R3-2R2→ R3
R1-R2→ R4

hence the rank of given matrix is 2.

(iii)

 3 -1 3 0 -1 1 2 -1 -3 -2 2 3 4 2 5 2 5 -2 -3 3

SOLUTION:

 3 -1 3 0 -1 1 2 -1 -3 -2 2 3 4 2 5 2 5 -2 -3 3

=R∼

 1 -1 3 0 -1 3 2 -1 -3 -2 2 3 4 2 5 2 5 -2 -3 3

By
R1↔R2
=∼R∼

 1 2 -1 -3 -2 0 -7 6 9 5 0 -1 6 8 9 0 1 0 3 7

By
R2-3R1 → R2
R3-2R1 → R3
R4-2R1 → R4
=∼R∼

 1 2 -1 -3 -2 0 1 0 3 7 0 -1 6 8 9 0 -7 6 9 5

By
R2↔ R4
= ∼R∼

 1 0 -1 -9 -16 0 1 0 3 7 0 0 6 11 16 0 0 6 30 54

R1-2R2 → R1
R3+R2 → R3
R4+7R2 → R4
= ∼R∼

 1 0 -1 -9 -16 0 1 0 3 7 0 0 6 11 16 0 0 1 5 9

(1/6)R4 → R4
= ∼R∼

 1 0 -1 -9 -16 0 1 0 3 7 0 0 1 5 9 0 0 6 11 16

R3↔R4
= ∼R∼

 1 0 0 -4 -7 0 1 0 3 7 0 0 1 11 9 0 0 0 -19 -38

R1+R3 → R1
R4-6R1 → R4
= ∼R∼

 1 0 0 -4 -7 0 1 0 3 7 0 0 1 5 9 0 0 0 1 2

(-/19)R4 → R4
= ∼R∼

 1 0 0 0 1 0 1 0 0 1 0 0 1 0 -1 0 0 0 1 2

R1+4R4 → R1
R2-3R4 → R2
R3+5R4 → R3

Hence the tank of given matrix is 4