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)
 
2 3x
2 3 6x
3 5 9x
 
=0

Solution:

L.H.S= 
 
2 3x
2 3 6x
3 5 9x
 
      =3x 
 
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)
      = 
 
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
 
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
 
by adding C3
 
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

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