Ex 3.2, 4 - Chapter 3 Class 12 Matrices
Last updated at April 16, 2024 by Teachoo
Addition/ subtraction of matrices
Last updated at April 16, 2024 by Teachoo
Ex 3.2, 4 If A = [■8(1&2&−3@5&0&2@1&−1&1)], B = [■8(3&−1&2@4&2&5@2&0&3)] and , C = [■8(4&1&2@0&3&2@1&−2&3)] then compute (A+B) and (B – C) . Also, verify that A + (B – C) = (A + B) – C Calculating A + B A + B = [■8(1&2&−3@5&0&2@1&−1&1)]+ [■8(3&−1&2@4&2&5@2&0&3)] = [■8(1+3&2−1&−3+2@5+4&0+2&2+5@1+2&−1+0&1+3)] = [■8(𝟒&𝟏&−𝟏@𝟗&𝟐&𝟕@𝟑&−𝟏&𝟒)] Calculating B – C B – C = [■8(3&−1&2@4&2&5@2&0&3)] – [■8(4&1&2@0&3&2@1&−2&3)] = [■8(3−4&−1−1&2−2@4−0&2−3&5−2@2−1&0−(−2)&3−3)] = [■8(−𝟏&−𝟐&𝟎@𝟒&−𝟏&𝟑@𝟏&𝟐&𝟎)] We need to verify A + (B – C) = (A + B) – C Solving L.H.S A + (B – C) = [■8(1&2&−3@5&0&2@1&−1&1)]+ [■8(−1&−2&0@4&−1&3@1&2&0)] = [■8(1−1&2−2&−3+0@5+4&0−1&2+3@1+1&−1+2&1+0)] = [■8(𝟎&𝟎&−𝟑@𝟗&−𝟏&𝟓@𝟐&𝟏&𝟏)] Solving R.H.S (A + B) – C = [■8(4&1&−1@9&2&7@3&−1&4)]− [■8(4&1&2@0&3&2@1&−2&3)] = [■8(4−4&1−1&−1−2@9−0&2−3&7−2@3−1&−1+2&4−3)] = [■8(𝟎&𝟎&−𝟑@𝟗&−𝟏&𝟓@𝟐&𝟏&𝟏)] = L.H.S Hence L.H.S = R.H.S Hence proved