Example 17 - Chapter 3 Class 12 Matrices
Last updated at May 29, 2018 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)] 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
Examples
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Example 6
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Example 8
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Example 12
Example 13
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Example 15
Example 16
Example 17 You are here
Example 18 Important
Example 19 Important
Example 20
Example 21
Example 22 Important
Example 23
Example 24
Example 25
Example 26
Example 27 Important
Example 28 Important
About the Author
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.