Ex 3.2, 5 - Chapter 3 Class 12 Matrices (Term 1)
Last updated at May 29, 2018 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 (Term 1)
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)]
