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Ex 3.2, 16 - Prove A3 - 6A2 + 7A + 2I = O, given A = [1 0

Ex 3.2, 16 - Chapter 3 Class 12 Matrices - Part 2
Ex 3.2, 16 - Chapter 3 Class 12 Matrices - Part 3
Ex 3.2, 16 - Chapter 3 Class 12 Matrices - Part 4

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Ex 3.2, 16 If A = [■8(1&0&2@0&2&1@2&0&3)] , prove that A3 – 6A2 + 7A + 2I = O Finding A2 A2 = A × A = [■8(1&0&2@0&2&1@2&0&3)] [■8(1&0&2@0&2&1@2&0&3)] = [■8(1(1)+0 (0)+2(2)&1(0)+0(2)+2(0)&1(2)+0(1)+2(3)@0(1)+2(0)+1(2)&0(0)+2(2)+1(0)&0(2)+2(1)+1(3)@2(1)+0(0)+3(2)&2(0)+0(2)+3(0)&2(2)+0(1)+3(3))] = [■8(1+0+4&0+0+0&2+0+6@0+0+2&0+4+0&0+2+3@2+0+6&0+0+0&4+0+9)] = [■8(5&0&8@2&4&5@8&0&13)] Finding A3 A3 = A2. A = [■8(5&0&8@2&4&5@8&0&13)] [■8(1&0&2@0&2&1@2&0&3)] = [■8(5(1)+0 (0)+8(2)&5(0)+0(2)+8(0)&5(2)+0(1)+8(3)@2(1)+4(0)+5(2)&2(0)+4(2)+5(0)&2(2)+4(1)+5(3)@8(1)+0(0)+13(2)&8(0)+0(2)+13(0)&8(2)+0(1)+13(3))] = [■8(5+0+16&0+0+0&10+0+24@2+0+10&0+8+0&4+4+15@8+0+26&0+0+0&16+0+39)] = [■8(21&0&34@12&8&23@34&0&55)] Now calculating A3 - 6A2 +7A + 2I Putting values = [■8(21&0&34@12&8&23@34&0&55)] – 6 [■8(5&0&8@2&4&5@8&0&13)] + 7 [■8(1&0&2@0&2&1@2&0&3)] + 2 [■8(1&0&0@0&1&0@0&0&1)] = [■8(21&0&34@12&8&23@34&0&55)] – [■8(6(5)&0(5)&8(6)@2(6)&4(6)&5(6)@8(6)&0(6)&13(6))] + [■8(1(7)&0(7)&2(7)@0(7)&2(7)&1(7)@2(7)&0(7)&3(7))] + [■8(2(1)&2(0)&2(0)@2(0)&1(2)&0(2)@2(0)&0(2)&1(2))] = [■8(21−30+7+2&0−0+0+0&34−48+14+0@12−12+0+0&8−24+14+2&23−30+7+0@34−48+14+0&0+0+0+0&55−78+21+2)] = [■8(−30+30&0&−48+48@12−12&24−24&30−30@48−48&0&78−78)] = [■8(0&0&0@0&0&0@0&0&0)] = O Thus, A3 – 6A2 + 7A + 2I = O Hence proved

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Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.