Question

If \(\begin{vmatrix} a-b-c & 2a & 2a \\ 2b & b-c-a & 2b \\ 2c & 2c & c-a-b \end{vmatrix} = (a+b+c)(x + a + b + c)^2\), \(x \ne 0\), then x is equal to :

• A. \(-2(a+b+c)\)
• B. \(2(a+b+c)\)
• C. abc
• D. \(-(a+b+c)\)

Hint:

The determinant equals (a+b+c)^3; so (x + a + b + c)^2 = (a+b+c)^2, and x is non-zero.

Answer 1

The Correct Option is A

Apply \(R_1 \to R_1 + R_2 + R_3\). The new first row entries are: \((a - b - c) + 2b + 2c = a + b + c\); \(2a + (b - c - a) + 2c = a + b + c\); \(2a + 2b + (c - a - b) = a + b + c\).
So \(\Delta = (a+b+c)\begin{vmatrix} 1 & 1 & 1 \\ 2b & b-c-a & 2b \\ 2c & 2c & c-a-b \end{vmatrix}\).
Apply \(C_2 \to C_2 - C_1\) and \(C_3 \to C_3 - C_1\): the first row becomes (1, 0, 0); the second row becomes \((2b, -(a+b+c), 0)\); the third row becomes \((2c, 0, -(a+b+c))\).
The determinant is therefore \((a+b+c)^2\), and \(\Delta = (a+b+c)^3\).
Given \(\Delta = (a+b+c)(x+a+b+c)^2\), we get \((x + a+b+c)^2 = (a+b+c)^2\), so \(x + a + b + c = \pm(a+b+c)\). Thus x = 0 or \(x = -2(a+b+c)\). Since \(x \ne 0\), \(x = -2(a+b+c)\).
Answer: (1) \(-2(a+b+c)\)