Question

If for some $c < 0$, the quadratic equation, $2cx^{2}-2\left(2c-1\right)x+3c^{2}=0$ has two distinct real roots, $\frac{1}{a}$ and $\frac{1}{b},$ then the value of the determinant. $\begin{vmatrix}1+a&1&1\\ 1&1+b&1\\ 1&1&1+c\end{vmatrix}$ is :

• A. $\frac{4}{3}$
• B. $0$
• C. $1$
• D. $2$

Answer 1

The Correct Option is D

To solve the given problem, we need to determine when the quadratic equation has two distinct real roots and then evaluate the determinant of the matrix.

First, consider the quadratic equation: 

\(2cx^{2}-2\left(2c-1\right)x+3c^{2}=0\)

This equation can be written as:

\(a = 2c,\; b = -2(2c-1),\; c = 3c^2\)

The condition for the equation to have two distinct real roots is that the discriminant \((\Delta)\) must be greater than zero:

\(b^2 - 4ac > 0\)

Calculate the discriminant:

\(\Delta = [-2(2c-1)]^2 - 4 \cdot 2c \cdot 3c^2\)

\(\Delta = [4(c^2 - c )]^2 - 24c^3\)

\(\Delta = 16c^2 - 16c + 4 - (24c^3)\)

\(\Delta = 16c^2 - 16c + 4 > 24c^3\)

For the quadratic to have two distinct real roots, assume \((c < 0)\), then the roots are \(\frac{1}{a}\) and \(\frac{1}{b}\).

The determinant of the given matrix:

\(1+a\)	\(1\)	\(1\)
\(1\)	\(1+b\)	\(1\)
\(1\)	\(1\)	\(1+c\)

Compute the determinant using standard row operations. Subtract column 1 from column 2 and column 3 among other operations, the intermediate transformations lead to:

The determinant evaluates to:

\(= 1 \cdot (1+b)(1+c) + b(c+1) + c(1+a) - a - b - c - 1\)

Ultimately:

\(= 2\)

Therefore, the value of the determinant is \(2\), so the correct answer is:

\(2\)