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Ex 3.2, 5 - Chapter 3 Class 12 Matrices (Term 1)
Last updated at March 30, 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/[email protected]/3&2/3&4/[email protected]/3&2&2/3)] and B = [ 8(2/5&3/5&[email protected]/5&2/5&4/[email protected]/5&6/5&2/5)] ,then compute 3A 5B 3A 5B = 3 [ 8(2/3&1&5/[email protected]/3&2/3&4/[email protected]/3&2&2/3)] - 5 [ 8(2/5&3/5&[email protected]/5&2/5&4/[email protected]/5&6/5&2/5)] = [ 8(2/3 3&1 3&5/3 [email protected]/3 3&2/3 3&4/3 [email protected]/3 3&2 3&2/3 3)] [ 8(2/5 5&3/5 5&1 [email protected]/5 5&2/5 5&4/5 [email protected]/5 5&6/5 5&2/5 5 )] = [ 8(2&3&[email protected]&2&[email protected]&6&2)] [ 8(2&3&[email protected]&2&[email protected]&6&2)] = [ 8(2 2&3 3&5 [email protected] 1&2 2&4 [email protected] 7&6 6&2 2)] = [ 8(0&0&[email protected]&0&[email protected]&0&0)] Hence, 3A 5B = [ 8(0&0&[email protected]&0&[email protected]&0&0)]
