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Ex 3.2, 15 - Chapter 3 Class 12 Matrices
Last updated at June 7, 2023 by Teachoo
Ex 3.2
Ex 3.2, 1
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 You are here
Ex 3.2, 16 Important
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
Chapter 3 Class 12 Matrices
Serial order wise
Transcript
Ex 3.2, 15 Find A2 – 5A + 6I if A = [■8(2&0&1@2&1&3@1&−1&0)] Finding A2 A2 = AA = [■8(2&0&1@2&1&3@1&−1&0)] [■8(2&0&1@2&1&3@1&−1&0)] = [■8(2(2)+0(2)+1(1)&2(0)+0(1)+1(−1)&2(1)+0(3)+1(0)@2(2)+1(2)+3(1)&2(0)+1(1)+3(−1)&2(1)+1(3)+3(0)@1(2)+−1(2)+0(1)&1(0)+−1(1)+0(−1)&1(1)+−1(3)+0(0))] = [■8(4+0+1&0+0−1&2+0+0@4+2+3&0+1−3&2+3+0@2−2+0&0−1+0&1−3+0)] = [■8(𝟓&−𝟏&𝟐@𝟗&−𝟐&𝟓@𝟎&−𝟏&−𝟐)] Now calculating A2 – 5A + 6I = [■8(5&−1&2@9&−2&5@0&−1&−2)] – 5 [■8(2&0&1@2&1&3@1&−1&0)]+ 6 [■8(1&0&0@0&1&0@0&0&1)] = [■8(5&−1&2@9&−2&5@0&−1&−2)] – [■8(2×5&0×5&1×5@2×5&1×5&3×5@1×5&−1×5&0×5)] + [■8(1×6&0×6&0×6@0×6&1×6&0×6@0×6&0×6&1×6)] = [■8(𝟓&−𝟏&𝟐@𝟗&−𝟐&𝟓@𝟎&−𝟏&−𝟐)] - [■8(𝟏𝟎&𝟎&𝟓@𝟏𝟎&𝟓&𝟏𝟓@𝟓&−𝟓&𝟎)] + [■8(𝟔&𝟎&𝟎@𝟎&𝟔&𝟎@𝟎&𝟎&𝟔)] = [■8(5−10+6&−1−0+0&2−5+0@9−10+0&−2−5+6&5−15+0@0−5+0&−1+5+0&−2−0+6)] = [■8(𝟏&−𝟏&−𝟑@−𝟏&−𝟏&−𝟏𝟎@−𝟓&𝟒&𝟒)]
