Fsc Part 1 Mathematics (Complete Solution)

# Q13 Find the inverse

Find the inverse of   A=

 2 1 0 1 1 0 2 -4 5

and show that A-1A=I3

SOLUTION:

lAl=

 2 1 0 1 1 0 2 -4 5

=2(-1+12)-1(1-6)+0(-4+2)

=22+5

=27

Cofactors of A =

 A11 A12 A13 A21 A22 A23 A31 A32 A33

Aij=(-1)i+jMij

A11  =  (-1)1+1

 -1 3 -4 1

=(-1)2 (-1+12) =11
A12  =  (-1)1+2

 1 3 2 1

=(-1)3 (1-6) =-(-5)=5
A13  =  (-1)1+3

 1 -1 2 -4

=(-1)4 (-4+2) =1(-2)=-2
A21  =  (-1)2+1

 1 0 -4 1

=(-1)3 (1+0) =-1
A22  =  (-1)2+2

 2 0 2 1

=(-1)4 (2-0) =2
A23  =  (-1)2+3

 2 1 2 -4

=(-1)5 (-8-2) =10
A31  =  (-1)3+1

 1 0 -1 3

=(-1)4 (3+0) =3
A32  =  (-1)3+2

 2 0 1 3

=(-1)5 (6-0) =-6
A33  =  (-1)3+3

 2 1 1 -1

=(-1)6 (-2-1) =-3
Cofactors of A =

 11 5 -2 -1 2 10 3 -6 -3

 11 -1 3 5 2 -6 -2 10 -3

SO,  A-1=
=
 1 27

 11 -1 3 5 2 -6 -2 10 -3

=

 11/27 -1/27 3/27 5/27 2/27 -6/27 -2/27 10/27 -3/27

A-1A=

 11/27 -1/27 3/27 5/27 2/27 -6/27 -2/27 10/27 -3/27

 2 1 0 1 1 0 2 -4 5

=

 (22-1+6)/27 (11+1-12)/27 (0-3+3)/27 (1=+2-12)/27 (5-2+24)/27 (0+6-6)/27 (-4+10-6)/27 (-2-10+12)/27 (0+30-3)/27

=

 27/27 0/27 0/27 0/27 27/27 0/27 0/27 0/27 27/27

=

 1 0 0 0 1 0 0 0 1

=I3

Hence A-1A=I3

SO PROVED