# Example 17 - Chapter 3 Class 12 Matrices (Term 1)

Last updated at Jan. 17, 2020 by Teachoo

Last updated at Jan. 17, 2020 by Teachoo

Transcript

Example 17 If A = [■8(0&6&7@−6&0&8@7&−8&0)], B = [■8(0&1&1@1&0&2@1&2&0)] , C = [█(2@−2@3)] Calculate AC, BC and (A + B)C. Also, verify that (A + B)C = AC + BC Calculating AC AC = [■8(0&6&7@−6&0&8@7&8&0)]_(3 × 3) [█(2@−2@3)]_(3 × 1) = [█(0(2)+6(−2)+7(3)@−6(2)+0(−2)+8(3)@7(2)+8(−2)+0(3))]_(3 × 1) = [█(0−12+21@−12+0+24@14+16+0)] = [█(9@12@30)] Calculating BC BC = [■8(0&1&1@1&0&2@1&2&0)]_(3 × 3) [█(2@−2@3)]_(3 × 1) = [█(0(2)+1(−2)+1(3)@1(2)+0(−2)+2(3)@1(2)+2(−2)+0(3))]_(3 × 1) = [█(0−2+3@2+0+6@2−4+0)] = [█(1@8@−2)] Calculating AC + BC AC + BC =[█(9@12@30)] + [█(1@8@−2)] = [█(9+1@12+8@30−2)] = [█(10@20@28)] Calculating (A + B) C First calculating A + B A + B = [■8(0&6&7@−6&0&8@7&−8&0)] +[■8(0&1&1@1&0&2@1&2&0)] = [■8(0+0&6+1&7+1@−6+1&0+0&8+2@7+1&−8+2&0+0)] = [■8(0&7&8@−5&0&10@8&−6&0)] Now, Calculating (A + B) C (A + B) C = [■8(0&7&8@−5&0&10@8&−6&0)]_(3 × 3) [█(2@−2@3)]_(3 × 1) = [█(0(2)+7(−2)+8(3)@−5(2)+0(−2)+10(3)@8(2)+(−6)(−2)+0(3))]_(3 × 1) = [█(0−14+24@−10+0+30@16+12+0 )] = [█(10@20@28)] = AC + BC ∴ L.H.S. = R.H.S. Hence proved

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Example 12

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Example 17 You are here

Example 18 Important

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Example 20 Important

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Example 22 Important

Example 23 Deleted for CBSE Board 2022 Exams

Example 24 Important Deleted for CBSE Board 2022 Exams

Example 25 Important Deleted for CBSE Board 2022 Exams

Example 26 Important

Example 27 Important

Example 28

Chapter 3 Class 12 Matrices (Term 1)

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About the Author

Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 10 years. He provides courses for Maths and Science at Teachoo.