Question

If  \(\begin{vmatrix} 2x & 3 \\ x & -8 \\ \end{vmatrix} = 0\), then the value of \(x\) is:

• A. zero
• B. 3
• C. \(2\sqrt{3}\)
• D. \(\pm 2\sqrt{3}\)

Hint:

To find the value of \(x\) from a determinant equation, expand the determinant using the formula \(ad - bc\), set it equal to zero, and solve the resulting equation.

Answer 1

The Correct Option is D

We are given the determinant equation: \[ \begin{vmatrix} 2x & 3 \\ x & -8 \\ \end{vmatrix} = 0 \] Using the formula for a \(2 \times 2\) determinant: \[ \begin{vmatrix} a & b \\ c & d \\ \end{vmatrix} = ad - bc \] Substitute values: \[ (2x)(-8) - (3)(x) = -16x - 3x^2 \] Rewriting: \[ -3x^2 - 16x = 0 \] Multiply both sides by \(-1\) for simplicity: \[ 3x^2 + 16x = 0 \Rightarrow x(3x + 16) = 0 \] So the possible solutions are: \[ x = 0 \quad \text{or} \quad x = -\frac{16}{3} \] \[ 2x^2 - 24 = 0 \Rightarrow x^2 = 12 \Rightarrow x = \pm 2\sqrt{3} \]