Question

If \( x, y, z \) \(\text{ are the three cube roots of 27, then the determinant of the matrix}\) \[ \begin{pmatrix} x & y & z \\ y & z & x \\ z & x & y \end{pmatrix} \] \(\text{is:}\)

• A. 0
• B. 1
• C. \( (x)(y)(x - z)(y - z) \)
• D. \( 0.02 \)

Hint:

When dealing with cube roots of unity, the rows of the matrix will often be linearly dependent, leading to a determinant of zero.

Answer 1

The Correct Option is A

We are given that \( x, y, z \) are the cube roots of 27. The cube roots of 27 are: \[ x = 3, \, y = 3 \omega, \, z = 3 \omega^2 \] where \( \omega \) is a primitive cube root of unity, satisfying \( \omega^3 = 1 \) and \( 1 + \omega + \omega^2 = 0 \). We are asked to find the determinant of the matrix: \[ \begin{pmatrix} x & y & z \\ y & z & x \\ z & x & y \end{pmatrix} \]

Step 1: Substitute the values of \( x, y, z \).
Substituting \( x = 3 \), \( y = 3\omega \), and \( z = 3\omega^2 \), we get: \[ \begin{pmatrix} 3 & 3\omega & 3\omega^2 \\ 3\omega & 3\omega^2 & 3 \\ 3\omega^2 & 3 & 3\omega \end{pmatrix} \]

Step 2: Factor out the common factor.
We can factor out 3 from each row: \[ 3^3 \times \begin{pmatrix} 1 & \omega & \omega^2 \\ \omega & \omega^2 & 1 \\ \omega^2 & 1 & \omega \end{pmatrix} \] The determinant of the remaining matrix is zero because the rows are linearly dependent (since \( \omega \) is a cube root of unity, and the rows are permutations of each other). Thus, the determinant is zero. Thus, the correct answer is \( 0 \), corresponding to option (a).