Question

The sum of all the roots of the equation $\begin{vmatrix} x & -3 & 2 \\ -1 & -2 & x-1 \\ 1 & x-2 & 3 \end{vmatrix} = 0$ is

• A. 13
• B. 3
• C. 2
• D. 7

Hint:

When a question asks for the sum or product of the roots of an equation derived from a determinant, you often don't need to find the full expanded polynomial. For the sum of roots of a cubic equation resulting from a $3 \times 3$ determinant, you only need the coefficients of the $x^3$ and $x^2$ terms. The $x^3$ term comes from the product of the diagonal entries containing $x$, and the $x^2$ term from products of one $x$ term with constants from the other rows/columns.

Answer 1

The Correct Option is B

Step 1: Expand the Determinant: \[ D(x) = x((-2)(3) - (x-1)(x-2)) - (-3)((-1)(3) - (x-1)(1)) + 2((-1)(x-2) - (-2)(1)) = 0 \] Simplify inner terms: Term 1: \( x [ -6 - (x^2 - 3x + 2) ] = x [ -x^2 + 3x - 8 ] = -x^3 + 3x^2 - 8x \) Term 2: \( 3 [ -3 - x + 1 ] = 3 [ -x - 2 ] = -3x - 6 \) Term 3: \( 2 [ -x + 2 + 2 ] = 2 [ -x + 4 ] = -2x + 8 \)
Step 2: Form the Polynomial Equation: Summing the terms: \[ (-x^3 + 3x^2 - 8x) + (-3x - 6) + (-2x + 8) = 0 \] \[ -x^3 + 3x^2 + (-8 - 3 - 2)x + (-6 + 8) = 0 \] \[ -x^3 + 3x^2 - 13x + 2 = 0 \] Multiply by -1: \[ x^3 - 3x^2 + 13x - 2 = 0 \]
Step 3: Find Sum of Roots: For a cubic equation \( ax^3 + bx^2 + cx + d = 0 \), the sum of the roots is given by \( -\frac{b}{a} \). Here, \( a = 1 \) and \( b = -3 \). \[ \text{Sum} = -\frac{-3}{1} = 3 \]