Ex 3.2

Chapter 3 Class 12 Matrices
Serial order wise

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Ex 3.2, 4 If A = [■8(1&2&−[email protected]&0&[email protected]&−1&1)], B = [■8(3&−1&[email protected]&2&[email protected]&0&3)] and , C = [■8(4&1&[email protected]&3&[email protected]&−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&−[email protected]&0&[email protected]&−1&1)]+ [■8(3&−1&[email protected]&2&[email protected]&0&3)] = [■8(1+3&2−1&−[email protected]+4&0+2&[email protected]+2&−1+0&1+3)] = [■8(𝟒&𝟏&−𝟏@𝟗&𝟐&𝟕@𝟑&−𝟏&𝟒)] Calculating B – C B – C = [■8(3&−1&[email protected]&2&[email protected]&0&3)] – [■8(4&1&[email protected]&3&[email protected]&−2&3)] = [■8(3−4&−1−1&2−[email protected]−0&2−3&5−[email protected]−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&−[email protected]&0&[email protected]&−1&1)]+ [■8(−1&−2&[email protected]&−1&[email protected]&2&0)] = [■8(1−1&2−2&−[email protected]+4&0−1&[email protected]+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&−[email protected]&2&[email protected]&−1&4)]− [■8(4&1&[email protected]&3&[email protected]&−2&3)] = [■8(4−4&1−1&−1−[email protected]−0&2−3&7−[email protected]−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