Question

If the elements of matrix \( A \) are the reciprocals of the elements of the matrix \( \begin{pmatrix} 1 & \omega & \omega^2 \\ \omega & \omega^2 & 1 \\ \omega^2 & 1 & \omega \end{pmatrix} \), where \( \omega \) is a complex cube root of unity, then

• A. \( A^{-1} = I \)
• B. \( A^{-1} = A^2 \)
• C. \( A^{-1} = A \)
• D. \( A^{-1} \) does not exist

Hint:

If the sum of elements of each row (or column) of a matrix is zero, then its determinant is zero and the inverse does not exist.

Answer 1

The Correct Option is D

Step 1: Recall properties of cube roots of unity.
If \( \omega \) is a cube root of unity, then \[ \omega^3 = 1, \quad 1+\omega+\omega^2 = 0 \] Step 2: Observe the structure of the given matrix.
Each row of the matrix satisfies \[ 1 + \omega + \omega^2 = 0 \] Hence, the sum of elements of every row is zero.
Step 3: Deduce the determinant.
If the sum of elements of each row is zero, then the rows are linearly dependent. Therefore, \[ \det(A) = 0 \] Step 4: Use the invertibility condition.
A matrix is invertible if and only if its determinant is non-zero.
Step 5: Conclusion.
Since \( \det(A) = 0 \), the inverse of \( A \) does not exist.