Ex 3.2, 15 - Chapter 3 Class 12 Matrices
Last updated at Dec. 16, 2024 by Teachoo
Last updated at Dec. 16, 2024 by Teachoo
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)]
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 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
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