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Mathematics Question on Matrices

If p,q,rp, q, r are distinct, then the value of ∣pp21+p3 qq21+q3 rr21+r3 ∣\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)(1 + pqr)(q - p)(r - p)(r - q)

B

(1−pqr)(q+p)(r+p)(r−q)(1 - pqr)(q + p)(r + p)(r - q)

C

(1+pqr)(q−p)(r+p)(r−q)(1 + pqr)(q - p)(r + p)(r - q)

D

(1−pqr)(q+p)(r−p)(r+q)(1 - pqr)(q + p)(r - p)(r + q)

Answer

(1+pqr)(q−p)(r−p)(r−q)(1 + pqr)(q - p)(r - p)(r - q)

Explanation

Solution

To evaluate the determinant, expand and simplify:

The determinant is:

Δ=∣pp21+p3 qq21+q3 rr21+r3∣.\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:

Δ=∣pp(p−1)p3(p−1) qq(q−1)q3(q−1) rr(r−1)r3(r−1)∣.\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−1p - 1, q−1q - 1, and r−1r - 1 from the columns:

Δ=(p−1)(q−1)(r−1)∣pp21 qq21 rr21∣.\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:

∣pp21 qq21 rr21∣=(q−p)(r−p)(r−q).\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:

Δ=(1+pqr)(q−p)(r−p)(r−q).\Delta = (1 + pqr)(q - p)(r - p)(r - q).

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