# Ex 3.2, 4 - Chapter 3 Class 12 Matrices

Last updated at April 16, 2024 by Teachoo

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 You are here

Ex 3.2, 5

Ex 3.2, 6

Ex 3.2, 7 (i)

Ex 3.2, 7 (ii) Important

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 April 16, 2024 by Teachoo

Ex 3.2, 4 If A = [■8(1&2&−3@5&0&2@1&−1&1)], B = [■8(3&−1&2@4&2&5@2&0&3)] and , C = [■8(4&1&2@0&3&2@1&−2&3)] then compute (A+B) and (B – C) . Also, verify that A + (B – C) = (A + B) – C Calculating A + B A + B = [■8(1&2&−3@5&0&2@1&−1&1)]+ [■8(3&−1&2@4&2&5@2&0&3)] = [■8(1+3&2−1&−3+2@5+4&0+2&2+5@1+2&−1+0&1+3)] = [■8(𝟒&𝟏&−𝟏@𝟗&𝟐&𝟕@𝟑&−𝟏&𝟒)] Calculating B – C B – C = [■8(3&−1&2@4&2&5@2&0&3)] – [■8(4&1&2@0&3&2@1&−2&3)] = [■8(3−4&−1−1&2−2@4−0&2−3&5−2@2−1&0−(−2)&3−3)] = [■8(−𝟏&−𝟐&𝟎@𝟒&−𝟏&𝟑@𝟏&𝟐&𝟎)] We need to verify A + (B – C) = (A + B) – C Solving L.H.S A + (B – C) = [■8(1&2&−3@5&0&2@1&−1&1)]+ [■8(−1&−2&0@4&−1&3@1&2&0)] = [■8(1−1&2−2&−3+0@5+4&0−1&2+3@1+1&−1+2&1+0)] = [■8(𝟎&𝟎&−𝟑@𝟗&−𝟏&𝟓@𝟐&𝟏&𝟏)] Solving R.H.S (A + B) – C = [■8(4&1&−1@9&2&7@3&−1&4)]− [■8(4&1&2@0&3&2@1&−2&3)] = [■8(4−4&1−1&−1−2@9−0&2−3&7−2@3−1&−1+2&4−3)] = [■8(𝟎&𝟎&−𝟑@𝟗&−𝟏&𝟓@𝟐&𝟏&𝟏)] = L.H.S Hence L.H.S = R.H.S Hence proved