Question

If
\[ \begin{vmatrix} x^k & x^{k+2} & x^{k+3} \\ y^k & y^{k+2} & y^{k+3} \\ z^k & z^{k+2} & z^{k+3} \end{vmatrix} = (x - y)(y - z)(z - x)\left( \frac{1}{x} + \frac{1}{y} + \frac{1}{z} \right), \] then the value of \(k\) is:

• A. k =−3
• B. k = 3
• C. k = 1
• D. k =−1

Hint:

For determinant equations involving powers, check how the degree of each term affects the overall equation. Sometimes simplifying for specific values of k helps identify the correct answer.

Answer 1

The Correct Option is D

The determinant is:

\[\Delta= \begin{vmatrix} x^k & x^{k+2} & x^{k+3} \\ y^k & y^{k+2} & y^{k+3} \\ z^k & z^{k+2} & z^{k+3} \end{vmatrix}\]

Step 1: Factorization.

Factor out \(x^k y^k z^k\):

\[\Delta= x^k y^k z^k \begin{vmatrix} 1 & x^2 & x^3 \\ 1 & y^2 & y^3 \\ 1 & z^2 & z^3 \end{vmatrix}\]

Step 2: Determinant simplification.

Simplified determinant:

\[\begin{vmatrix} 1 & x^2 & x^3 \\ 1 & y^2 & y^3 \\ 1 & z^2 & z^3 \end{vmatrix} = (x-y)(y-z)(z-x)(xy+yz+zx).\]

Thus:

\[\Delta= x^k y^k z^k (x-y)(y-z)(z-x)(xy+yz+zx).\]

Step 3: Comparison.

Given determinant:

\[(x-y)(y-z)(z-x)\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\]

Rewrite:

\[\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{xy+yz+zx}{xyz}.\]

Full expression:

\[\Delta=(x-y)(y-z)(z-x)\cdot\frac{xy+yz+zx}{xyz}.\]

Step 4: Power Equating.

Compare \(x^k y^k z^k\) with \(\frac{1}{xyz}\)

\[x^k y^k z^k = \frac{1}{xyz}.\]

This implies k = -1

Therefore:

\[k - 1 - 1 = -1 \implies k = -1.\]

Conclusion: The value of k is:

\(\boxed{-1}\).