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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

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 You are here

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, 17 If A = [■8(3&−2@4&−2)] and I= [■8(1&0@0&1)] , find k so that A2 = kA – 2I Finding A2 A2 = A × A = [■8(3&−2@4&−2)][■8(3&−2@4&−2)] = [■8(3(3)+(−2)(4)&3(−2)+(−2)(−2)@4(3)+(−2)(4)&4(−2)+(−2)(−2))] = [■8(9−8&−6+4@12−8&−8+4)] = [■8(𝟏&−𝟐@𝟒&−𝟒)] ∴ A2 = [■8(1&−2@4&−4)] Now , given that A2 = kA – 2I Putting values [■8(1&−2@4&−4)] = k [■8(3&−2@4&−2)] − 2 [■8(1&0@0&1)] [■8(1&−2@4&−4)] = [■8(3k&−2k@4k&−2k)] − [■8(1×2&0×2@0×2&1×2)] [■8(1&−2@4&−4)] = [■8(3k&−2k@4k&−2k)] − [■8(2&0@0&2)] [■8(1&−2@4&−4)] = [■8(3k−2&−2k−0@4k−0&−2k−2)] [■8(𝟏&−𝟐@𝟒&−𝟒)] = [■8(𝟑𝐤−𝟐&−𝟐𝐤@𝟒𝐤&−𝟐𝐤−𝟐)] Since matrices are equal. Comparing its corresponding elements. 1 = 3k – 2 1 + 2 = 3k 3 = 3k 3/3 = k 1 = k k = 1 Thus, k = 1