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
 
    Adj.A=(Cofactors of A)t=
 
11 -1  3
5 2 -6
-2 10 -3
 
SO,  A-1=
Adj.A
lAl
=
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

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