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Ex 3.2, 5 - Chapter 3 Class 12 Matrices
Last updated at June 7, 2023 by Teachoo
Ex 3.2
Ex 3.2, 1
Ex 3.2, 2 (i)
Ex 3.2, 2 (ii) Important
Ex 3.2, 2 (iii)
Ex 3.2, 2 (iv)
Ex 3.2, 3 (i)
Ex 3.2, 3 (ii) Important
Ex 3.2, 3 (iii)
Ex 3.2, 3 (iv) Important
Ex 3.2, 3 (v)
Ex 3.2, 3 (vi) Important
Ex 3.2, 4
Ex 3.2, 5 You are here
Ex 3.2, 6
Ex 3.2, 7 (i)
Ex 3.2, 7 (ii) Important
Ex 3.2, 8
Ex 3.2, 9
Ex 3.2, 10
Ex 3.2, 11
Ex 3.2, 12 Important
Ex 3.2, 13 Important
Ex 3.2, 14
Ex 3.2, 15
Ex 3.2, 16 Important
Ex 3.2, 17 Important
Ex 3.2, 18
Ex 3.2, 19 Important
Ex 3.2, 20 Important
Ex 3.2, 21 (MCQ) Important
Ex 3.2, 22 (MCQ) Important
Chapter 3 Class 12 Matrices
Serial order wise
Transcript
Ex 3.2, 5 If A = [■8(2/3&1&5/3@1/3&2/3&4/3@7/3&2&2/3)] and B = [■8(2/5&3/5&1@1/5&2/5&4/5@7/5&6/5&2/5)] ,then compute 3A – 5B 3A – 5B = 3 [■8(2/3&1&5/3@1/3&2/3&4/3@7/3&2&2/3)] - 5 [■8(2/5&3/5&1@1/5&2/5&4/5@7/5&6/5&2/5)] = [■8(2/3 ×3&1×3&5/3×3@1/3×3&2/3×3&4/3×3@7/3×3&2×3&2/3×3)] – [■8(2/5×5&3/5×5&1×5@1/5×5&2/5×5&4/5×5@7/5×5&6/5×5&2/5×5 )] = [■8(𝟐&𝟑&𝟓@𝟏&𝟐&𝟒@𝟕&𝟔&𝟐)] – [■8(𝟐&𝟑&𝟓@𝟏&𝟐&𝟒@𝟕&𝟔&𝟐)] = [■8(2−2&3−3&5−5@1−1&2−2&4−4@7−7&6−6&2−2)] = [■8(𝟎&𝟎&𝟎@𝟎&𝟎&𝟎@𝟎&𝟎&𝟎)] Hence, 3A – 5B = [■8(0&0&0@0&0&0@0&0&0)]
