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Solution


Q.

Let α be a root of the equation (a-c)x2+(b-a)x+(c-b)=0 where a,b,c are distinct real numbers such that the matrix [α2α1111abc] is singular. Then, the value of (a-c)2(b-a)(c-b)+(b-a)2(a-c)(c-b)+(c-b)2(a-c)(b-a) is           [2023]
1 6  
2 3  
3 9  
4 12  

Ans.

(2)

Given matrix is singular,  

∴    |α2α1111abc|=0

⇒α2(c-b)-α(c-a)+1(b-a)=0

It is singular if α=1.

⇒c-b-c+a+b-a=0

Now, (a-c)2(b-a)(c-b)+(b-a)2(a-c)(c-b)+(c-b)2(a-c)(b-a) 

=(a-c)3(a-c)(b-a)(c-b)+(b-a)3(a-c)(b-a)(c-b)+(c-b)3(a-c)(b-a)(c-b)

=(a-c)3+(b-a)3+(c-b)3(a-c)(b-a)(c-b)

=3(a-c)(b-a)(c-b)(a-c)(b-a)(c-b)=3                  (∵a-c+b-a+c-b=0)