Ex 3.2

Chapter 3 Class 12 Matrices
Serial order wise

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