Question

If \( p, q, r \) are distinct, then the value of \[ \begin{vmatrix} p & p^2 & 1 + p^3 \\ q & q^2 & 1 + q^3 \\ r & r^2 & 1 + r^3 \\ \end{vmatrix} \] is:

• A. \( (1 + pqr)(q - p)(r - p)(r - q) \)
• B. \( (1 - pqr)(q + p)(r + p)(r - q) \)
• C. \( (1 + pqr)(q - p)(r + p)(r - q) \)
• D. \( (1 - pqr)(q + p)(r - p)(r + q) \)

Answer 1

The Correct Option is A

To evaluate the determinant, expand and simplify:

The determinant is:

\[ \Delta = \begin{vmatrix} p & p^2 & 1 + p^3 \\ q & q^2 & 1 + q^3 \\ r & r^2 & 1 + r^3 \end{vmatrix}. \]

Using the property of determinants, subtract the first column from the second and the third column from the first, simplifying the matrix to:

\[ \Delta = \begin{vmatrix} p & p(p - 1) & p^3(p - 1) \\ q & q(q - 1) & q^3(q - 1) \\ r & r(r - 1) & r^3(r - 1) \end{vmatrix}. \]

Factor out \(p - 1\), \(q - 1\), and \(r - 1\) from the columns:

\[ \Delta = (p - 1)(q - 1)(r - 1) \begin{vmatrix} p & p^2 & 1 \\ q & q^2 & 1 \\ r & r^2 & 1 \end{vmatrix}. \]

The remaining determinant simplifies using standard properties of symmetric determinants:

\[ \begin{vmatrix} p & p^2 & 1 \\ q & q^2 & 1 \\ r & r^2 & 1 \end{vmatrix} = (q - p)(r - p)(r - q). \]

Thus, the overall value of the determinant becomes:

\[ \Delta = (1 + pqr)(q - p)(r - p)(r - q). \]

Hence, the correct answer is \((1 + pqr)(q - p)(r - p)(r - q)\).