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CBSE Class 12 Maths 2010 Solved Paper

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Question : 26 of 29
 
Marks: +1, -0
Using properties of determinants show the following:
|(b+c)2abcaab(a+c)2bcacbc(a+b)2|
= 2 abc (a+b+c)3
Solution:
Consider,
Δ =
|(b+c)2abcaab(a+c)2bcacbc(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 =
1abc|a(b+c)2a2ba2cab2b(a+c)2b2cac2bc2c(a+b)2|

Now, taking a, b, c common from C1,C2 and C3
Δ =
abcabc|(b+c)2a2a2b2(a+c)2b2c2c2(a+b)2|

⇒ Δ =
|(b+c)2a2a2b2(c+a)2b2c2c2(a+b)|

Applying C1 → C1–C2,C2 → C2–C3
Δ = (a+b+c)2
|b+c−a0a2b−c−ac+a−bb20c−a−b(a+b)2|

Applying R3 → R3−(R1+R2)
Δ = (a+b+c)2
|b+c−a0a2b−c−ac+a−bb22a−2b−2a2ab|

Applying C1 → C1+C2
Δ = (a+b+c)2
|b+c−a0a20c+a−bb2−2b−2a2ab|

Applying C3 → C3+bC2
Δ = (a+b+c)2
|b+c−a0a20c+a−bbc+ab−2b−2a0|

Applying C1 → aC1 and C2 → bC2
Δ = (a+b+c)2ab
|ab+ac−a20a20bc+ab−b2bc+ab−2ab−2ab0|

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

Expanding along R3
= (a+b+c)2ab
(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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