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Chapter 3 Class 12 Matrices (Term 1)

Serial order wise

Last updated at Jan. 17, 2020 by Teachoo

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