Fsc Part 1 Mathematics (Complete Solution)

Q2
Without expansion show that

(i)
 
7 8
3 4 5
2 3 4
 
= 0

SOLUTION:

L.H.S= 
 
6 7 8
3 4 5
2 3 4
 
=        
 
6 1 1
3 1 1
2 1 1
 
C2-C1
C3-C2

C2 and C3 are identical. so the value of the determinant is zero

=0

=R.H.S

Hence L.H.S=R.H.S



(ii)
 
3 -1
1 1 0
2 -3 5
 
= 0

SOLUTION:

L.H.S= 
 
2 3 -1
1 1 0
2 -3 5
 
=        
 
-1 3 -1
0 1 0
5 -3 5
 
C1-C2

C1 and C3 are identical. so the value of the determinant is zero

=0

=R.H.S

Hence L.H.S=R.H.S



(iii)
 
1 2 3
4 5 6
7 8 9
 
= 0

SOLUTION:

L.H.S= 
 
1 2 3
4 5 6
7 8 9
 
=        
 
1 2 3
3 3 3
3 3 3
 
R2-R1
R3-R2

R2 and R3 are identical. so the value of the determinant is zero

=0

=R.H.S

Hence L.H.S=R.H.S

Other Topics

;