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CBSE Class 12 Maths 2010 Solved Paper

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Question : 26
Total: 29
Using properties of determinants show the following:
|
(b+c)2abca
ab(a+c)2bc
acbc(a+b)2
|
= 2 abc (a+b+c)3
Solution:  
Consider,
Δ =
|
(b+c)2abca
ab(a+c)2bc
acbc(a+b)2
|
= 2 abc (a+b+c)3
By performing R1 → aR1,R2 → bR2,R3 → cR3 and dividing the determinant by abc, we get
Δ =
1
abc
|
a(b+c)2a2ba2c
ab2b(a+c)2b2c
ac2bc2c(a+b)2
|

Now, taking a, b, c common from C1,C2 and C3
Δ =
abc
abc
|
(b+c)2a2a2
b2(a+c)2b2
c2c2(a+b)2
|

⇒ Δ =
|
(b+c)2a2a2
b2(c+a)2b2
c2c2(a+b)
|

Applying C1 → C1–C2,C2 → C2–C3
Δ = (a+b+c)2
|
b+c−a0a2
b−c−ac+a−bb2
0c−a−b(a+b)2
|

Applying R3 → R3−(R1+R2)
Δ = (a+b+c)2
|
b+c−a0a2
b−c−ac+a−bb2
2a−2b−2a2ab
|

Applying C1 → C1+C2
Δ = (a+b+c)2
|
b+c−a0a2
0c+a−bb2
−2b−2a2ab
|

Applying C3 → C3+bC2
Δ = (a+b+c)2
|
b+c−a0a2
0c+a−bbc+ab
−2b−2a0
|

Applying C1 → aC1 and C2 → bC2
Δ =
(a+b+c)2
ab
|
ab+ac−a20a2
0bc+ab−b2bc+ab
−2ab−2ab0
|

Applying C1 → C1–C2
Δ =
(a+b+c)2
ab
|
ab+ac−a20a2
−bc−ab+b2bc+ab−b2bc+ab
0−2ab0
|

Expanding along R3
=
(a+b+c)2
ab
(2ab(ab2c+a2b2+abc2+a2bc−a2bc−a3b+a2bc+a3b−a2b2))

= 2 (a+b+c)2 (ab2c+abc2+a2bc)
= 2 (a+b+c)3 abc = R.H.S.
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