Ex 3.2, 15 - Chapter 3 Class 12 Matrices
Last updated at April 16, 2024 by Teachoo
Ex 3.2
Ex 3.2, 1
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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)]
