Example 17 - Calculate AC, BC, (A + B)C. Also, verify - Examples

  1. Chapter 3 Class 12 Matrices
  2. Serial order wise

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)] 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 20@16+12+0 )] = [ (10@20@28)] = AC + BC L.H.S. = R.H.S. Hence proved

About the Author

Davneet Singh's photo - Teacher, Computer Engineer, Marketer
Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 8 years. He provides courses for Maths and Science at Teachoo. You can check his NCERT Solutions from Class 6 to 12.