![Ex 3.2, 17 - Find k so that A2 = kA - 2I, if A = [3 -2 4 -2]](https://d1avenlh0i1xmr.cloudfront.net/45271a36-43cc-4b92-9292-8fda58fff6c9/slide58.jpg)
Ex 3.2, 17 - Chapter 3 Class 12 Matrices (Term 1)
Last updated at Jan. 17, 2020 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
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
Transcript
Ex 3.2, 17 If A = [■8(3&−[email protected]&−2)] and I= [■8(1&[email protected]&1)] , find k so that A2 = kA – 2I Finding A2 A2 = A × A = [■8(3&−[email protected]&−2)][■8(3&−[email protected]&−2)] = [■8(3(3)+(−2)(4)&3(−2)+(−2)(−2)@4(3)+(−2)(4)&4(−2)+(−2)(−2))] = [■8(9−8&−[email protected]−8&−8+4)] = [■8(1&−[email protected]&−4)] ∴ A2 = [■8(1&−[email protected]&−4)] Now , given that A2 = kA – 2I Putting values [■8(1&−[email protected]&−4)] = k [■8(3&−[email protected]&−2)] − 2 [■8(1&[email protected]&1)] [■8(1&−[email protected]&−4)] = [■8(3k&−[email protected]&−2k)] − [■8(1×2&0×[email protected]×2&1×2)] [■8(1&−[email protected]&−4)] = [■8(3k&−[email protected]&−2k)] − [■8(2&[email protected]&2)] [■8(1&−[email protected]&−4)] = [■8(3k−2&−2k−[email protected]−0&−2k−2)] [■8(1&−[email protected]&−4)] = [■8(3k−2&−[email protected]&−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
