# Ex 3.2, 17 - Chapter 3 Class 12 Matrices (Term 1)

Last updated at Jan. 17, 2020 by Teachoo

Last updated at Jan. 17, 2020 by Teachoo

Transcript

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

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 (Term 1)

Serial order wise

About the Author

Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 10 years. He provides courses for Maths and Science at Teachoo.