Fsc Part 1 Mathematics (Complete Solution)

# Q5 Without expansion verify that

(i)

 α β+γ 1 β γ+α 1 γ α+β 1

=0

Solution:

L.H.S=

 α β+γ 1 β γ+α 1 γ α+β 1

=

 α+β+γ β+γ 1 α+β+γ γ+α 1 α+β+γ α+β 1

C1+C2

Taking common (α+β+γ) from C1

= (α+β+γ)

 1 β+γ 1 1 γ+α 1 1 α+β 1

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

= (α+β+γ)(0)

=0

(ii)

 1 2 3x 2 3 6x 3 5 9x

=0

Solution:

L.H.S=

 1 2 3x 2 3 6x 3 5 9x

=3x

 1 2 1 2 3 1 3 6 1

Taking 3x common by C3

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

= (3x)(0)

=0

=R.H.S

(iii)

 1 a2 a/bc 1 b2 b/ca 1 c2 c/ab

= 0

SOLUTION:

L.H.S=

 1 a2 a/bc 1 b2 b/ca 1 c2 c/ab

=
 1 abc

 1 a2 abc x a/bc 1 b2 abc x b/ca 1 c2 abc x c/ab

Multiplying C3 by
abc and dividing
determinant by
abc.
=
 1 abc

 1 a2 a2 1 b2 b2 1 c2 c2

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

=
 1 abc
(0)

=0

=R.H.S

(iv)

 a-b b-c c-a b-c c-a a-b c-a a-b b-c

=0

Solution:

L.H.S=

 a-b b-c c-a b-c c-a a-b c-a a-b b-c

=

 a-b b-c c-a b-c c-a a-b c-a a-b b-c

C1+(C2+C3)
=

 0 b-c c-a 0 c-a a-b 0 a-b b-c

All elements of C1 are zero, so the value of the determinant is zero

=0

=R.H.S

(v)

 bc ca ab 1/a 1/b 1/c a b c

= 0

SOLUTION:

L.H.S=

 bc ca ab 1/a 1/b 1/c a b c

=
 1 abc

 bc ca ab abc x 1/a abc x 1/b abc x 1/c a b c

Multiplying R2 by
abc and dividing
determinant by
abc.
=
 1 abc

 bc ca ab bc ca ab a b c

R1 and R2 are identical , so the value of the determinant is zero.

=
 1 abc
(0)

=0

=R.H.S

(vi)

 mn l l2 nl m m2 lm n n2

=

 1 l2 l3 1 m2 m3 1 n2 n3

SOLUTION:

L.H.S=

 mn l l2 nl m m2 lm n n2

=
 1 lmn

 lmn l2 l3 lmn m2 m3 lmn n2 n3

Multiplying R1 by l
R2 by m, R3 by n
and dividing
determinant by
lmn.
=
 lmn lmn

 1 l2 l3 1 m2 m3 1 n2 n3

Taking lmn common from C1
=

 1 l2 l3 1 m2 m3 1 n2 n3

=R.H.S

(vii)

 2a 2b 2c a+b 2b b+c a+c b+c 2c

=0

Solution:

L.H.S=

 2a 2b 2c a+b 2b b+c a+c b+c 2c

=2

 a b c a+b 2b b+c a+c b+c 2c

taking 2 common by R1
= 2

 a b c b b b c c c

R2-R1
R3-R1
=2bc

 a b c 1 1 1 1 1 1

taking  common b by R2
, c by R3

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

=0

=R.H.S

(viii)

 7 2 6 6 3 2 -3 5 1

=

 7 2 7 6 3 5 -3 5 -3

+

 7 2 -1 6 3 -3 -3 5 4

SOLUTION:

R.H.S=

 7 2 7 6 3 5 -3 5 -3

+

 7 2 -1 6 3 -3 -3 5 4

 7 2 7+(-1) 6 3 5+(-3) -3 5 -3+4

 7 2 6 6 3 2 -3 5 1

=L.H.S

(ix)

 -a 0 c 0 a -b b -c 0

= 0

SOLUTION:

L.H.S=

 -a 0 c 0 a -b b -c 0

=
 1 abc

 -ac 0 ac 0 ab -ab bc -bc 0

Multiplying C1 by c
C2 by b, C3 by a
and dividing
determinant by
abc.
=
 1 abc

 0 0 ac 0 ab -ab 0 -bc 0

C1+(C2+C3)

All entries of C1 are zero, so the value of determinant is equal to 0.

=
 1 abc
(0)

=0

=R.H.S