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

  1. Chapter 3 Class 12 Matrices
  2. Serial order wise
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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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