Concept of determinant

Question

!!Delta(x) = |(x^2-5x+3, 2x-5,3), (3x^2+x+4, 6x+1,9), (7x^2-6x+9, 14x-6,21)| = Ax^3+Bx^2+Cx+D!!. Then,

(a) A=0

(b) B=0

(c) C=0

(d) None

Solution:

Method 1:

!!Ax^3+Bx^2+Cx+D = |(x^2-5x+3, 2x-5,3), (3x^2+x+4, 6x+1,9), (7x^2-6x+9, 14x-6,21)|!!

Differentiate with respect to !!x!! both the sides,

!!3Ax^2+2Bx+C=|(2x-5, 2x-5,3), (6x+1, 6x+1,9), (14x-6, 14x-6,21)|+ |(x^2-5x+3, 2,3), (3x^2+x+4, 6,9), (7x^2-6x+9, 14,21)|+|(x^2-5x+3, 2x-5,0), (3x^2+x+4, 6x+1,0), (7x^2-6x+9, 14x-6,0)| !!

first determinnt is zero because two columns are same and last determinant also zero because last column is zero.

!!3Ax^2+2Bx+C= 0+ (2xx3)|(x^2-5x+3, 1,1), (3x^2+x+4, 3,3), (7x^2-6x+9, 7,7)|+0!!

!!=>3Ax^2+2Bx+C= 0+0+0!!

Means !!A=0, B=0 and C=0!!

Methos 2: You can apply row and column opreations to solve also.

Therefore, option (A),(B) and (C) are correct.