Question:
If \(\Delta = \begin{vmatrix} x - 2 & 2x - 3 & 3x - 4 \\ 2x - 3 & 3x - 4 & 4x - 5 \\ 3x - 5 & 5x - 8 & 10x - 17 \end{vmatrix} = Ax^3 + Bx^2 + Cx + D\), then \(B + C\) is equal to :
If \(\Delta = \begin{vmatrix} x - 2 & 2x - 3 & 3x - 4 \\ 2x - 3 & 3x - 4 & 4x - 5 \\ 3x - 5 & 5x - 8 & 10x - 17 \end{vmatrix} = Ax^3 + Bx^2 + Cx + D\), then \(B + C\) is equal to :
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Apply \(R_2 \to R_2 - R_1\) and \(R_3 \to R_3 - R_2\) (old \(R_2\)) to factor the determinant, then expand and read off the coefficients of \(x^2\) and \(x\).
Updated On: Nov 5, 2026
- -1
- -3
- 9
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The Correct Option is B
Solution and Explanation
\(R_2 \to R_2 - R_1\): \((x - 1,\ x - 1,\ x - 1)\).
\(R_3 \to R_3 - R_2^{\text{old}}\): \((x - 2,\ 2x - 4,\ 6x - 12) = (x - 2)(1, 2, 6)\).
So \(\Delta = (x - 1)(x - 2)\begin{vmatrix} x - 2 & 2x - 3 & 3x - 4 \\ 1 & 1 & 1 \\ 1 & 2 & 6 \end{vmatrix}\).
The inner determinant \(= (x - 2)(6 - 2) - (2x - 3)(6 - 1) + (3x - 4)(2 - 1) = 4x - 8 - 10x + 15 + 3x - 4 = -3x + 3 = -3(x - 1)\).
\(\Delta = -3(x - 1)^2(x - 2) = -3(x^3 - 4x^2 + 5x - 2) = -3x^3 + 12x^2 - 15x + 6\).
So \(B = 12\), \(C = -15\), \(B + C = -3\).
Answer: (2) -3
\(R_3 \to R_3 - R_2^{\text{old}}\): \((x - 2,\ 2x - 4,\ 6x - 12) = (x - 2)(1, 2, 6)\).
So \(\Delta = (x - 1)(x - 2)\begin{vmatrix} x - 2 & 2x - 3 & 3x - 4 \\ 1 & 1 & 1 \\ 1 & 2 & 6 \end{vmatrix}\).
The inner determinant \(= (x - 2)(6 - 2) - (2x - 3)(6 - 1) + (3x - 4)(2 - 1) = 4x - 8 - 10x + 15 + 3x - 4 = -3x + 3 = -3(x - 1)\).
\(\Delta = -3(x - 1)^2(x - 2) = -3(x^3 - 4x^2 + 5x - 2) = -3x^3 + 12x^2 - 15x + 6\).
So \(B = 12\), \(C = -15\), \(B + C = -3\).
Answer: (2) -3
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