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: Understanding the Concept
The problem asks for the sum of the roots of a polynomial equation that is defined by setting a determinant to zero. We first need to expand the determinant to get the polynomial equation. Then, we can use Vieta's formulas to find the sum of the roots without actually solving for the roots.
Step 2: Key Formula or Approach
For a polynomial equation of the form \(a_n x^n + a_{n-1} x^{n-1} + \dots + a_1 x + a_0 = 0\), the sum of the roots is given by \( -\frac{a_{n-1}}{a_n} \). We will expand the determinant to find the polynomial and identify the required coefficients.
Step 3: Detailed Explanation
We are given the equation:
\[ \begin{vmatrix} x & -3 & 2 \\ -1 & -2 & x-1 \\ 1 & x-2 & 3 \end{vmatrix} = 0 \] Let's expand the determinant along the first row:
\[ x \begin{vmatrix} -2 & x-1 \\ x-2 & 3 \end{vmatrix} - (-3) \begin{vmatrix} -1 & x-1 \\ 1 & 3 \end{vmatrix} + 2 \begin{vmatrix} -1 & -2 \\ 1 & x-2 \end{vmatrix} = 0 \] \[ x[(-2)(3) - (x-1)(x-2)] + 3[(-1)(3) - (x-1)(1)] + 2[(-1)(x-2) - (-2)(1)] = 0 \] \[ x[-6 - (x^2 - 3x + 2)] + 3[-3 - (x - 1)] + 2[-x + 2 + 2] = 0 \] \[ x[-6 - x^2 + 3x - 2] + 3[-3 - x + 1] + 2[-x + 4] = 0 \] \[ x[-x^2 + 3x - 8] + 3[-x - 2] + 2[-x + 4] = 0 \] \[ -x^3 + 3x^2 - 8x - 3x - 6 - 2x + 8 = 0 \] Combine like terms:
\[ -x^3 + 3x^2 + (-8-3-2)x + (-6+8) = 0 \] \[ -x^3 + 3x^2 - 13x + 2 = 0 \] To make the leading coefficient positive, we multiply by -1:
\[ x^3 - 3x^2 + 13x - 2 = 0 \] This is a cubic equation. For a cubic equation of the form \(ax^3 + bx^2 + cx + d = 0\), the sum of the roots (let's call them \(r_1, r_2, r_3\)) is given by \(r_1 + r_2 + r_3 = -\frac{b}{a}\).
In our equation, \(a = 1\), \(b = -3\), \(c = 13\), and \(d = -2\).
Sum of the roots = \( - \frac{-3}{1} = 3 \).
Step 4: Final Answer
The sum of all the roots of the equation is 3.