Question

Let \(a - 2b + c = 1\).
If \(f(x) = \begin{vmatrix} x + a & x + 2 & x + 1 \\ x + b & x + 3 & x + 2 \\ x + c & x + 4 & x + 3 \end{vmatrix}\), then:

• A. \(f(50) = 1\)
• B. \(f(-50) = -1\)
• C. \(f(50) = -501\)
• D. \(f(-50) = 501\)

Hint:

Apply \(R_1 \to R_1 - 2R_2 + R_3\); the first row collapses to (a - 2b + c, 0, 0) = (1, 0, 0), so f is constant.

Answer 1

The Correct Option is A

Apply \(R_1 \to R_1 - 2R_2 + R_3\) (this does not change the determinant).
First entry: \((x + a) - 2(x + b) + (x + c) = a - 2b + c = 1\).
Second entry: \((x + 2) - 2(x + 3) + (x + 4) = 0\).
Third entry: \((x + 1) - 2(x + 2) + (x + 3) = 0\).
Expanding along the new first row: \(f(x) = 1 \cdot \left[(x + 3)^2 - (x + 2)(x + 4)\right] = (x^2 + 6x + 9) - (x^2 + 6x + 8) = 1\).
So \(f(x) = 1\) for all x, hence \(f(50) = 1\).
Answer: (1) \(f(50) = 1\)