Ex 3.2

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)]

Made by

#### Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.