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Solved Subjective Question on Determinant Set 2

Posted on - 04-01-2017

Math

IIT JEE

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)

 
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