Fsc Part 1 Mathematics (Complete Solution)

# Q3 Show that

(i)

 a11 a12 a13+a13 a21 a22 a23+a23 a31 a32 a33+a33

=

 a11 a12 a13 a21 a22 a23 a31 a32 a33

+

 a11 a12 a13 a21 a22 a23 a31 a32 a33

SOLUTION:

R.H.S=

 a11 a12 a13 a21 a22 a23 a31 a32 a33

+

 a11 a12 a13 a21 a22 a23 a31 a32 a33

By using Property 6 of determinants we know that if any row (column) of a determinant consists of two terms, it can be written as the sum of two determinants.So,

 a11 a12 a13+a13 a21 a22 a23+a23 a31 a32 a33+a33

=L.H.S

hence

 a11 a12 a13+a13 a21 a22 a23+a23 a31 a32 a33+a33

=

 a11 a12 a13 a21 a22 a23 a31 a32 a33

+

 a11 a12 a13 a21 a22 a23 a31 a32 a33

So L.H.S=R.H.S proved

(ii)

 2 3 0 3 9 6 2 15 1

= 9

 2 1 0 1 1 2 2 5 1

SOLUTION:

L.H.S=

 2 3 0 3 9 6 2 15 1

taking 3 common from R2

= 3

 2 3 0 1 3 2 2 15 1

Taking 3 common from C2

= (3)(3)

 2 1 0 1 1 2 2 5 1

= 9

 2 1 0 1 1 2 2 5 1

=R.H.S

Hence Proved

(iii)

 a+l a a a a+l a a a a+l

= l2(3a+l)

SOLUTION:

L.H.S=

 a+l a a a a+l a a a a+l

=

 3a+l 3a+l 3a+l a a+l a a a a+l

R1+(R2+R3)
= (3a+l))

 1 1 1 a a+l a a a a+l

Taking common (3a+l) from R1
= (3a+l)

 1 0 0 a l 0 a 0 l

C2-C1
C3-C1

Expand by R1

= (3a+l)[ 1

 l 0 0 l

-0+0]

=(3a+l)[1(l2-0)]

=(3a+l)l2

=R.H.S

Hence Proved

(iv)

 1 1 1 x y z yz zx xy

=

 1 1 1 x y z x2 y2 z2

SOLUTION:

L.H.S=

 1 1 1 x y z yz zx xy

=
 1 xyz

 x y z x2 y2 z2 xyz xyz xyz

Multiplying C1 by x,
C2 by y , C3 by z and
divided determinant
by xyz
=
 xyz xyz

 x y z x2 y2 z2 1 1 1

taking xyz common from R3
= (-1)

 x y z 1 1 1 x2 y2 z2

Interchanging R1 and R3
=(-1)(-1)

 1 1 1 x y z x2 y2 z2

Interchanging R1 and R2
=

 1 1 1 x y z x2 y2 z2

=R.H.S

(v)

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

= 4abc

SOLUTION:

L.H.S=

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

=

 0 -2c -2b b c+a b c c a+b

R1-(R2+R3)
= -2

 0 c b b c+a b c c a+b

= -2

 0 c b b a 0 c 0 a

R2-R1
R3-R1

Expand by R1

=  -2( 0 - c

 b 0 c a

+ b

 b a c 0

)

=-2[ -c (ab-0)+b(0-ac) ]

=-2[ -abc-abc ]

=-2(-2abc)

=4abc

=R.H.S

Hence Proved

(vi)

 b -1 a a b 0 1 a b

=a3+b3
SOLUTION:  L.H.S=

 b -1 a a b 0 1 a b

Expand By R1

=  b

 b 0 a b

- (-1)

 a 0 1 b

+ a

 a b 1 a

=b(b2-0)+1(ab-0)+a(a2-b)

=b3+ab+a3-ab

=a3+b3

=R.H.S

(vii)

 rcosΦ 1 -sinΦ 0 1 0 rsinΦ 0 cosΦ

= r2
SOLUTION:  L.H.S=

 rcosΦ 1 -sinΦ 0 1 0 rsinΦ 0 cosΦ

Expand By R1

=  rcosΦ

 r 0 0 cosΦ

-1

 0 0 rsinΦ cosΦ

+(-sinΦ )

 0 1 rsinΦ 0

=rcosΦ (rcosΦ )-0-sinΦ (0-r2sinΦ )

=r2cos2Φ +r2sin2Φ

=r2(cos2Φ +sin2Φ)

=r2(1)

=r2

=R.H.S

(viii)

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

= a3+b3+c3-3abc

SOLUTION:

L.H.S=

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

=

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

C1+C2
= (a+b+c)

 1 b+c a+b 1 c+a b+c 1 a+b c+a

Taking (a+b+c) common from C1
= (a+b+c)

 1 b+c a+b 0 a-b c-a 0 a-c c-b

R2-R1
R3-R1

Expand by C1

=  (a+b+c)( 1

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

-0+0)

=(a+b+c)[ (a-b)(c-b)-(a-c)(c-a)]

=(a+b+c)[ac-ab-bc+b2-(ac-a2-c2+ac)]

=(a+b+c)(a2+b2+c2-ab-bc-ac)

= a3+b3+c3-3abc

=R.H.S

Hence Proved

(ix)

 a+λ b c a b+λ c a b c+λ

= λ2(a+b+c+λ)

SOLUTION:

L.H.S=

 a+λ b c a b+λ c a b c+λ

=

 a+b+c+λ b c a+b+c+λ b+λ c a+b+c+λ b c+λ

C1+(C2+C3)
= (a+b+c+λ)

 1 b c 1 b+λ c 1 b c+λ

Taking (a+b+c+λ) common from C1
= (a+b+c+λ)

 1 b c 0 λ 0 0 0 λ

R2-R1
R3-R1

Expand by C1

=  (a+b+c+λ)( 1

 λ 0 0 λ

-0+0)

=(a+b+c+λ)[ 1(λ2-0)]

=(a+b+c+λ)λ2

2(a+b+c+λ)

=R.H.S

(x)

 1 1 1 a b c a2 b2 c2

= (a-b)(b-c)(c-a)

SOLUTION:

L.H.S=

 1 1 1 a b c a2 b2 c2

=

 0 0 1 a-b b-c c a2-b2 b2-c2 c2

C1-C2
C2-C3
=(a-b)(b-c)

 0 0 1 a-b b-c c (a-b)(a+b) (b+c)(b-c) c2

= (a-b)(b-c)(0-0+1

 1 1 a+b b+c

)

=(a-b)(b-c)[ 1(b+c-a-b) ]

=(a-b)(b-c)(c-a)

=R.H.S

Hence Proved

(xi)

 b+c a a2 c+a b b2 a+b c c2

= (a+b+c)(a-b)(b-c)(c-a)

SOLUTION:

L.H.S=

 b+c a a2 c+a b b2 a+b c c2

=

 a+b+c a a2 a+b+c b b2 a+b+c c c2

C1+C2
=(a+b+c)

 1 a a2 1 b b2 1 c c2

Taking (a+b+c) from C1
=(a+b+c)

 0 a-b a2-b2 0 b-c b2-c2 1 c c2

R1-R2
R2-R3

Expand by C1

= (a+b+c)( 0-0+1

 a-b (a-b)(a+b) b-c (b-c)(b+c)

)
= (a+b+c)( a-b)(b-c)

 1 (a+b) 1 (b+c)

=(a+b+c)(a-b)(b-c)(b+c-a-b)

=(a+b+c)(a-b)(b-c)(c-a)

=R.H.S

Hence Proved