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Solved Subjective Question on Determinant Set 1

Posted on - 04-01-2017

Math

IIT JEE

Example.1

If a, b and g are such that a + b + g = 0, then prove that

.

Solution

Operating C2 → C2 – cos g C1, C3 → C3 – cosb C1,

D =

D =

=

Example.2

Solution

Since are coplanar there must exist k1, k2 , k3 not all zero (say k1 > 0) such that

Now by operating C1→ ,

Example.3

Prove that = –64(a -b) (a -g) (a -d) (b -g) (b -d) (g- d).

Solution

Let (b + g -a - d)2 = A, (g + a -b -d)2 = B and (a +b -g -d)2 = C

then A – B = (2g - 2d) (2b -2a) = - 4(a - b) (g - d)

and D = = ( A – B) ( B –C) ( C – A)

= - 64 ( a - b ) ( a - g) ( a - d) ( b- g) ( b - d) ( g - d) .

Example.4

In D ABC, prove that .

Solution

D =

=

=

=

=

= 0 [Q C2 and C3 are identical when (a2 – b2) and (a2– c2) are taken common]

Second Method

D =

= = 0

Example.5

If x, y, z are not all zero and if ax+by+cz= 0, bz+cy+az = 0, cx+ay+bz =0 prove that x:y:z = 1:1:1 or 1: w :w2 or 1: w2:w where w is the complex cube roots of unity.

Solution

For non-trival solution = 0

= ( a+ b+ c) (a2 +b2 +c2 – ab – bc – ca) =0 . . . . (1).

= ( a+ b+c ) ( a+ wb +w2c)( a+ w2b + wc) = 0 . . . . (2).

Now, by using cramer rule find x, y, z and using ( a+b+c) = 0 ,

( a+ wb +w2c)=0 and ( a+ w2b + wc) = 0, prove the desired result.

 
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