Question

If x, y, z are different and \(A=\begin{vmatrix} x & x^2 & 1+x^3 \\ y & y^2 & 1+y^3 \\ z & z^2 & 1+z^3 \end{vmatrix}=0\), then the value of xyz is :

• A. 1
• B. 0
• C. -1
• D. 2

Answer 1

The Correct Option is C

To solve the problem, we need to evaluate the determinant \(A\) of the given \(3 \times 3\) matrix:

\[\begin{vmatrix} x & x^2 & 1+x^3 \\ y & y^2 & 1+y^3 \\ z & z^2 & 1+z^3 \end{vmatrix}=0\]

We are asked to find the value of \(xyz\) given that this determinant is zero. The matrix represents a system where the columns are polynomials in \(x\), \(y\), and \(z\) respectively. Each row can be treated as a polynomial function of a variable. If the determinant of a matrix composed of polynomials is zero, it implies that the rows (or columns) of the matrix are linearly dependent, meaning one row can be expressed as a combination of others.

For the polynomial vector formed by rows being dependent, a Vandermonde matrix comes in play:

\[\begin{vmatrix} 1 & x & x^2 \\ 1 & y & y^2 \\ 1 & z & z^2 \end{vmatrix} \times \begin{vmatrix} x & x^2 & x^3+1 \\ y & y^2 & y^3+1 \\ z & z^2 & z^3+1 \end{vmatrix} = 0\]

The simplified difference for a determinant with added terms like \(x^3+1\) compared to a normal polynomial matrix is zero when expanded due to symmetric polynomial properties.

Given the condition that the determinant is zero and \(x\), \(y\), \(z\) are distinct, the expression:

\[\text{Determinant} = c_{1}c_{2}c_{3} \left( (x-y)(y-z)(z-x) \right)^2 = 0\]

Since \(x\), \(y\), and \(z\) are distinct, \((x-y)(y-z)(z-x)\neq0\). Therefore, for the determinant to be zero, at least one of \(c_1, c_2, c_3\) put (as expressions with xyz factors) must equate to zero which implies \(xyz=-1\) due to symmetry complexes in its multi-linear form when crossed with polynomial roots.

Therefore, the value of \(xyz\) is:

-1