Example.1
Express D = 
as the product of two
determinants and evaluate it.
Solution
D =
= 
´
= 2 (a – b) (b – c)
(c – a) (x – y) (y – z) (z – x).
Example.2
Let a, b, c be real numbers with a2 + b2
+ c2 = 1. Show that the equation .
represents a straight
line.
Solution
C1 → aC1
+ bC2 + cC3
applying C2 → C2
– bC1 and C3 → C3 – cC1
applying R3 → R3
+ xR1 + yR2, we get
&⇒ (x2 + y2 +1)(aby + a2x
+ ac) = 0
&⇒ ax + by + c = 0
Example.3
Prove that 
Solution
Let D
= 
Multiplying C1,
by a Then D = 
Applying C1
→ C1 + bC2
+ cC3, we get
D = 
Taking (a2
+ b2 +c2) common from C1,
\ D
= 
Applying C2
→ C2 – bC1
and C3 → C3 - cC1,
Then D = 
Multiplying R1
by x, 
D
= 
Applying R1
→ R1 + yR2
+ zR3 ,
D
= 
Expanding along R1
\
D = 
= 
= 
= 
= 
= 
Hence D =
Example.4
Prove that 
=
2 sin
.
Solution
Do C1 → C1+C2+C3,
followed by R2 →
R2 – R1 and R3 → R3 – R1. Then expand along C1,
so that the determinant is.
= 
= 
= 
´ [
]
This can be simplified using cosq + cos2q+cos3q
= 2 cos 2q cosq + cos 2q= cos2q (2cos q+1),
cosq
– cos 2q = 
cosq
– cos 3q = 
and cos 3q – cos 2q = – 2 sin 
.
The determinant can now be written cos 2q (2 cosq +1) 
= 4 sin2 
Further simplification follows after noting
that
= 
= 
and 
After substituting these results, the
determinant equals
= 
Using the identity (a–b) (a2 + ab
+ b2) = a3–b3. This proves the results.
Example.5
If numbers n , r
are two different positive integers such that n≥ r+2 and it is given that D( n, r) = 
,
then show that D(n , r) = 
D
( n-1, r-1) .
Solution
We know mCk =
m-1Ck-1
Now D (n, r) = 
=
D(n –1, r –1)
