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Ex 4.4, 14 For the matrix A = [■8(3&2@1&1)] , find the numbers a and b such that A2 + aA + bI = O. Finding A2 A2 = A.A = [■8(3&2@1&1)] [■8(3&2@1&1)] = [■8(3(3)+2(1)&3(2)+2(1)@1(3)+1(1)&1(2)+1(1))] = [■8(9+2&6+2@3+1&2+1)] = [■8(11&8@4&3)] Now, A2 + aA + bI = O Putting values [■8(11&8@4&3)] + a [■8(3&2@1&1)] + b [■8(1&0@0&1)] = O [■8(11&8@4&3)] + [■8(3a&2a@a&a)] + [■8(b&0@0&b)] = O [■8(11+3a+b&8+2a+0@4+a+0&3+a+b)] = O [■8(3a+b+11&2a+8@4+a&a+b+3)] = [■8(0&0@0&0)] Since the matrices are equal, Comparing corresponding elements 3a + b + 11 = 0 2a + 8 = 0 4 + a = 0 a + b + 3 = 0 Solving (3) a + 4 = 0 a = –4 Putting value a in (1) 11 + 3 a + b = 0 11 + 3 (–4) + b = 0 11 – 12 +b = 0 –1 + b = 0 b = 1 Hence, a = −4, b = 1

  1. Chapter 4 Class 12 Determinants
  2. Serial order wise

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Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science and Computer Science at Teachoo