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 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 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)
 
-1 a
a b 0
1 a b
 
=a3+b3
SOLUTION:  L.H.S= 
 
-1 a
a b 0
1 a b
 

Expand By R1

 =  b
 
b 0
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)  
 
b+c a+b
b c+a b+c
c a+b c+a
 
= a3+b3+c3-3abc

SOLUTION:

 L.H.S= 
 
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
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
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)
 
a a2
1 b b2
1 c c2
 
Taking (a+b+c) from C1
 =(a+b+c)
 
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

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