Applying C3→ C3 – C2, we get,Applying C2 → C2 + C1, we get,Applying C2→ C2 – 5C1 and C3 →C3 – 5C1 we get,= 1[( – 7)( – 36) – ( – 20)( – 13)] = 252 – 260= – 8So, Δ = – 8

1 A 1 B 1 C 1 D 1 E 1 F 1 G 1 H 2 A 2 B 2 C 2 D 2 E 2 F 2 G 2 H 2 I 2 J 2 K 2 L 2 M 2 N 2 O 2 P 2 Q 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 A 52 B 52 C 52 D 52 E 52 F 52 G 52 H 52 I 53 54

RD Sharma - Volume 1

Exercise 6.1

1 A 1 B 1 C 1 D 1 E 1 F 1 G 2 3 4 5 6 7 8 9 10 A 10 B 10 C 10 D 10 E 10 F 11 12 A 12 B

Exercise 6.2

Exercise 6.3

1 A 1 B 1 C 1 D 2 A 2 B 2 C 2 D 3 4 5 6 7 8 9 10 11 12 A 12 B 13 A 13 B

Exercise 6.4

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32

Exercise 6.5

1 2 3 4 5

MCQ

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33

Very short answer

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56

Q. 1 E3.8( 5 Votes )

Evaluate the foll

Class 12th RD Sharma - Volume 1 6. Determinants

Answer :

Applying C₃→ C₃ – C₂, we get,

Applying C₂ → C₂ + C₁, we get,

Applying C₂→ C₂ – 5C₁ and C₃ →C₃ – 5C₁ we get,

= 1[( – 7)( – 36) – ( – 20)( – 13)] = 252 – 260

= – 8

So, Δ = – 8

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