Question

\[ D = \begin{vmatrix} -\frac{bc}{a^2} & \frac{c}{a} & \frac{b}{a} \\ \frac{c}{b} & -\frac{ac}{b^2} & \frac{a}{b} \\ \frac{b}{c} & \frac{a}{c} & -\frac{ab}{c^2} \end{vmatrix} \]

• A. \( 0 \)
• B. \( 4 \)
• C. \( -1 \)
• D. \( \frac{a^2 + b^2 + c^2}{a^2 b^2 c^2} \)

Hint:

To evaluate determinants, expand along a row or column with the most zeros or simplest terms. Compute 2×2 determinants carefully to simplify calculations.

Answer 1

The Correct Option is B

We need to evaluate the determinant:

\[ D = \begin{vmatrix} -\frac{bc}{a^2} & \frac{c}{a} & \frac{b}{a} \\ \frac{c}{b} & -\frac{ac}{b^2} & \frac{a}{b} \\ \frac{b}{c} & \frac{a}{c} & -\frac{ab}{c^2} \end{vmatrix}. \]

Step 1: Expand the Determinant Along the First Row

Using the determinant expansion along the first row:

\[ D = -\frac{bc}{a^2} \begin{vmatrix} - \frac{ac}{b^2} & \frac{a}{b} \\ \frac{a}{c} & -\frac{ab}{c^2} \end{vmatrix} - \frac{c}{a} \begin{vmatrix} \frac{c}{b} & \frac{a}{b} \\ \frac{b}{c} & -\frac{ab}{c^2} \end{vmatrix} + \frac{b}{a} \begin{vmatrix} \frac{c}{b} & -\frac{ac}{b^2} \\ \frac{b}{c} & \frac{a}{c} \end{vmatrix}. \]

Step 2: Compute the 2×2 Determinants

Each determinant is evaluated as follows:

\[ \begin{vmatrix} - \frac{ac}{b^2} & \frac{a}{b} \\ \frac{a}{c} & -\frac{ab}{c^2} \end{vmatrix} = \left(-\frac{ac}{b^2} \times -\frac{ab}{c^2} \right) - \left( \frac{a}{b} \times \frac{a}{c} \right) = \frac{a^2 b}{b^2 c^2} - \frac{a^2}{bc}. \]

\[ \begin{vmatrix} \frac{c}{b} & \frac{a}{b} \\ \frac{b}{c} & -\frac{ab}{c^2} \end{vmatrix} = \left(\frac{c}{b} \times -\frac{ab}{c^2} \right) - \left( \frac{a}{b} \times \frac{b}{c} \right) = -\frac{ac}{b c^2} - \frac{a}{c b}. \]

\[ \begin{vmatrix} \frac{c}{b} & -\frac{ac}{b^2} \\ \frac{b}{c} & \frac{a}{c} \end{vmatrix} = \left(\frac{c}{b} \times \frac{a}{c} \right) - \left( -\frac{ac}{b^2} \times \frac{b}{c} \right) = \frac{a}{b} + \frac{a}{b}. \]

Step 3: Compute the Final Value

After simplifications, we obtain:

\[ D = 4. \]