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Learn in your speed, with individual attention - Teachoo Maths 1-on-1 Class


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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

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.