Question

The value of \(\begin {vmatrix} \frac{√3}{2}&\frac{1}{2} \\\frac{√3}{2} &\frac{1}{2}\end {vmatrix}\)

• A. 0
• B. 1
• C. \(\frac{\sqrt{3}}{2}\)
• D. \(\frac{1}{2}\)

Answer 1

The Correct Option is A

To find the value of the determinant \(\begin{vmatrix} \frac{\sqrt{3}}{2} & \frac{1}{2} \\ \frac{\sqrt{3}}{2} & \frac{1}{2} \end{vmatrix}\), we use the formula for a 2x2 matrix determinant:

\(\text{determinant} = ad - bc\)

For our matrix, let:

• \(a = \frac{\sqrt{3}}{2}\)
• \(b = \frac{1}{2}\)
• \(c = \frac{\sqrt{3}}{2}\)
• \(d = \frac{1}{2}\)

Substitute these values into the determinant formula:

\(\text{determinant} = \left(\frac{\sqrt{3}}{2}\right)\left(\frac{1}{2}\right) - \left(\frac{1}{2}\right)\left(\frac{\sqrt{3}}{2}\right)\)

Calculating each term:

\(\frac{\sqrt{3}}{2} \times \frac{1}{2} = \frac{\sqrt{3}}{4}\) and \(\frac{1}{2} \times \frac{\sqrt{3}}{2} = \frac{\sqrt{3}}{4}\)

Now, evaluate the determinant:

\(\frac{\sqrt{3}}{4} - \frac{\sqrt{3}}{4} = 0\)

Therefore, the value of the determinant is 0.