Ex 3.2, 17 - Find k so that A2 = kA - 2I, if A = [3 -2 4 -2]

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

Ex 3.2, 17 - Chapter 3 Class 12 Matrices