

Ex 3.4
Ex 3.4, 2 Deleted for CBSE Board 2022 Exams
Ex 3.4, 3 Deleted for CBSE Board 2022 Exams
Ex 3.4, 4 Deleted for CBSE Board 2022 Exams
Ex 3.4, 5 Deleted for CBSE Board 2022 Exams You are here
Ex 3.4, 6 Deleted for CBSE Board 2022 Exams
Ex 3.4, 7 Deleted for CBSE Board 2022 Exams
Ex 3.4, 8 Important Deleted for CBSE Board 2022 Exams
Ex 3.4, 9 Deleted for CBSE Board 2022 Exams
Ex 3.4, 10 Deleted for CBSE Board 2022 Exams
Ex 3.4, 11 Deleted for CBSE Board 2022 Exams
Ex 3.4, 12 Deleted for CBSE Board 2022 Exams
Ex 3.4, 13 Deleted for CBSE Board 2022 Exams
Ex 3.4, 14 Deleted for CBSE Board 2022 Exams
Ex 3.4, 15 Important Deleted for CBSE Board 2022 Exams
Ex 3.4, 16 Deleted for CBSE Board 2022 Exams
Ex 3.4, 17 Important Deleted for CBSE Board 2022 Exams
Ex 3.4, 18 (MCQ) Deleted for CBSE Board 2022 Exams
Last updated at Dec. 8, 2016 by Teachoo
Ex3.4, 5 Find the inverse of each of the matrices, if it exists. [ 8(2&1@7&4)] Let A = [ 8(2&1@7&4)] We know that A = IA [ 8(2&1@7&4)] = [ 8(1&0@0" " &1)] A R1 R1 1/7 R2 [ 8( / ( )&1 1/7(4)@7&4)] = [ 8(1 1/7(0)&0 1/7(1)@0" " &1)] A [ 8( &1 4/7@7&4)] = [ 8(1 0&( 1)/7@0" " &1)] A [ 8( &3/7@7&4)] = [ 8(1&( 1)/7@0" " &1)] A R2 R2 7R1 [ 8(1&3/7@ ( )&4 7(3/7) )] = [ 8(1&( 1)/7@0 7(1)&1 7(( 1)/7) )] A [ 8(1&3/7@ &4 3)] = [ 8(1&( 1)/7@0 7&1+1)] A [ 8(1&3/7@ &1)] = [ 8(1&( 1)/7@ 7&2)] A R1 R1 3/7R2 [ 8(1 3/7(0)& / / ( )@0&1)] = [ 8(1 3/7( 7)&( 1)/7 3/7(2)@ 7&2)] A [ 8(1 0& / / @0&1)] = [ 8(1+3&( 1)/7 6/7@ 7&2)] A [ 8(1& @0&1)] = [ 8(4& 1@ 7" " &2)] A I = [ 8(4& 1@ 7" " &2)] A This is similar to I = A-1A Thus, A-1 = [ 8(4& 1@ 7" " &2)] A