Express D = as the product of two determinants and evaluate it.

D =

= ´

= 2 (a – b) (b – c) (c – a) (x – y) (y – z) (z – x).

Let a, b, c be real numbers with a^{2} + b^{2}
+ c^{2} = 1. Show that the equation .

represents a straight line.

C_{1} → aC_{1}
+ bC_{2} + cC_{3} _{ }

applying C_{2} → C_{2}
– bC_{1} and C_{3} → C_{3} – cC_{1}

applying R_{3} → R_{3}
+ xR_{1} + yR_{2}, we get

&⇒ (x^{2} + y^{2} +1)(aby + a^{2}x
+ ac) = 0

&⇒ ax + by + c = 0

Prove that

Let D =

Multiplying C_{1},
by a Then D =

Applying C_{1}
→ C_{1} + bC_{2}
+ cC_{3}, we get

D =

Taking (a^{2}
+ b^{2} +c^{2}) common from C_{1},

\ D =

Applying C_{2}
→ C_{2} – bC_{1}
and C_{3} → C_{3} - cC_{1},

Then D =

Multiplying R_{1}
by x, D
=

Applying R_{1}
→ R_{1} + yR_{2}
+ zR_{3} ,

D =

Expanding along R_{1}

\ D =

=

=

=

=

=

Hence D =

Prove that = 2 sin.

Do C_{1 }→ C_{1}+C_{2}+C_{3},
followed by R_{2 }→
R_{2} – R_{1} and R_{3} → R_{3} – R_{1}. Then expand along C_{1},
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 sin^{2}

Further simplification follows after noting that

=

= and

After substituting these results, the determinant equals

=

Using the identity (a–b) (a^{2 }+ ab
+ b^{2}) = a^{3}–b^{3}. This proves the results.

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) .

We know ^{m}C_{k}^{ }=
^{m-1}C_{k-1}

Now D (n, r) =

=D(n –1, r –1)