Learn in your speed, with individual attention - Teachoo Maths 1-on-1 Class

Ex 3.2

Ex 3.2, 1

Ex 3.2, 2 (i)

Ex 3.2, 2 (ii) Important

Ex 3.2, 2 (iii)

Ex 3.2, 2 (iv)

Ex 3.2, 3 (i)

Ex 3.2, 3 (ii) Important

Ex 3.2, 3 (iii)

Ex 3.2, 3 (iv) Important

Ex 3.2, 3 (v)

Ex 3.2, 3 (vi) Important

Ex 3.2, 4

Ex 3.2, 5

Ex 3.2, 6

Ex 3.2, 7 (i)

Ex 3.2, 7 (ii) Important You are here

Ex 3.2, 8

Ex 3.2, 9

Ex 3.2, 10

Ex 3.2, 11

Ex 3.2, 12 Important

Ex 3.2, 13 Important

Ex 3.2, 14

Ex 3.2, 15

Ex 3.2, 16 Important

Ex 3.2, 17 Important

Ex 3.2, 18

Ex 3.2, 19 Important

Ex 3.2, 20 Important

Ex 3.2, 21 (MCQ) Important

Ex 3.2, 22 (MCQ) Important

Chapter 3 Class 12 Matrices

Serial order wise

Last updated at June 7, 2023 by Teachoo

Ex 3.2, 7 Find X and Y, if (ii) 2X + 3Y = [■8(2&3@4&0)] and 3X + 2Y = [■8(2&−2@−1&5)] Given 2X + 3Y = [■8(𝟐&𝟑@𝟒&𝟎)] Multiplying by 3 3 × (2X+ 3Y) = 3 [■8(2&3@4&0)] 6X + 9Y = [■8(2 × 3&3 × 3@4 × 3&0 × 3)] 6X + 9Y = [■8(6&9@12&0)] Given 3X + 2Y = [■8(𝟐&−𝟐@−𝟏&𝟓)] Multiplying by 2 2 × (3X + 2Y) = 2 × [■8(2&−2@−1&5)] 6X + 4Y = [■8(2 ×2&−2 ×2@−1 ×2&5 ×2)] 6X + 4Y = [■8(4&−4@−2&10)] Subtracting (1) from (2), (6X + 9Y) – (6X + 4Y) = [■8(6&9@12&0)] – [■8(4&−4@−2&10)] 6X + 9Y – 6X – 4Y = [■8(6−4&9−(−4)@12−(−2)&0−10)] 9Y – 4Y + 6X – 6X = [■8(2&9+4@12+2&−10)] 5Y + 0 = [■8(𝟐&𝟏𝟑@𝟏𝟒&−𝟏𝟎)] Y = 1/5 [■8(2&13@14&−10)] Y = [■8(𝟐/𝟓&𝟏𝟑/𝟓@𝟏𝟒/𝟓&−𝟏𝟎/𝟓)] = [■8(𝟐/𝟓&𝟏𝟑/𝟓@𝟏𝟒/𝟓&−𝟐)] Putting value of Y in (1) 6X + 9Y = [■8(6&9@12&0)] 6X + 9 [■8(2/5& 13/5@14/5&−2)] = [■8(6&9@12&0)] 6X + [■8(9 × 2/5&9 ×13/5@9 ×14/5&9 ×−2)] = [■8(6&9@12&0)] 6X + [■8(18/5&117/5@126/5&−18)] = [■8(6&9@12&0)] 6X = [■8(𝟔&𝟗@𝟏𝟐&𝟎)] – [■8(𝟏𝟖/𝟓&𝟏𝟏𝟕/𝟓@𝟏𝟐𝟔/𝟓&−𝟏𝟖)] 6X = [■8(6−18/5&9−117/5@12−126/5&0−(−18))] 6X = [■8((6 × 5 − 18)/5&(9 × 5 − 117)/5@ (12 × 5 − 126)/5&18)] 6X = [■8((30 − 18)/5&(45 − 117)/5@ (60 − 126)/5&18)] 6X = [■8(𝟏𝟐/𝟓&(−𝟕𝟐)/𝟓@ (−𝟔𝟔)/𝟓&𝟏𝟖)] X = 1/6 [■8(12/5&(−72)/5@ (−66)/5&18)] X = [■8(1/6 × 12/5&1/6 ×(−72)/5@1/6 ×(−66)/5&1/6 ×18)] X = [■8(𝟐/𝟓&(−𝟏𝟐)/𝟓@(−𝟏𝟏)/𝟓&𝟑)] Thus, X = [■8(𝟐/𝟓& (−𝟏𝟐)/𝟓@ (−𝟏𝟏)/𝟓&𝟑)] , Y = [■8(𝟐/𝟓&𝟏𝟑/𝟓@𝟏𝟒/𝟓&−𝟐)]