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Ex 3.2
Ex 3.2, 2 (i)
Ex 3.2, 2 (ii) Important
Ex 3.2, 2 (iii)
Ex 3.2, 2 (iv)
Ex 3.2, 3 (i)
Ex 3.2, 3 (ii) Important
Ex 3.2, 3 (iii)
Ex 3.2, 3 (iv) Important
Ex 3.2, 3 (v)
Ex 3.2, 3 (vi) Important
Ex 3.2, 4
Ex 3.2, 5
Ex 3.2, 6
Ex 3.2, 7 (i)
Ex 3.2, 7 (ii) Important
Ex 3.2, 8
Ex 3.2, 9
Ex 3.2, 10
Ex 3.2, 11
Ex 3.2, 12 Important
Ex 3.2, 13 Important
Ex 3.2, 14
Ex 3.2, 15
Ex 3.2, 16 Important You are here
Ex 3.2, 17 Important
Ex 3.2, 18
Ex 3.2, 19 Important
Ex 3.2, 20 Important
Ex 3.2, 21 (MCQ) Important
Ex 3.2, 22 (MCQ) Important
Last updated at March 22, 2023 by Teachoo
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