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 1
 

By
(-1/4)R3→ R3

hence the rank of given matrix is 3.



(ii) 
 
1 -4 -7
2 -5 1
-2 3
3 -7 4
 
SOLUTION: 
 
1 -4 -7
2 -5 1
-2 3
3 -7 4
 
= ∼R∼
 
1 -4 -7
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

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