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Ex 4.4
Ex 4.4, 2
Ex 4.4, 3 Important
Ex 4.4, 4 Important
Ex 4.4, 5
Ex 4.4, 6 Important
Ex 4.4, 7
Ex 4.4, 8
Ex 4.4, 9
Ex 4.4, 10 Important
Ex 4.4, 11 Important
Ex 4.4, 12
Ex 4.4, 13
Ex 4.4, 14 Important You are here
Ex 4.4, 15 Important
Ex 4.4, 16
Ex 4.4, 17 (MCQ) Important
Ex 4.4, 18 (MCQ) Important
Last updated at June 11, 2023 by Teachoo
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