Cube Roots Unity Determinant Evaluation
This solution details the calculation of the determinant involving p, q, and r, where these represent the cube roots of unity.
Cube Roots of Unity Properties
The cube roots of unity are solutions to the equation \(x^3=1\). These are \(1, \omega, \omega^2\), satisfying:
- \(\omega^3 = 1\)
- \(1 + \omega + \omega^2 = 0\)
Step 1: Assign Values and Compute Squares
To evaluate the determinant, let's assign values to p, q, and r. We choose:
\(p = 1\), \(q = \omega\), \(r = \omega^2\)
Calculate the squares:
- \(p^2 = 1^2 = 1\)
- \(q^2 = \omega^2\)
- \(r^2 = (\omega^2)^2 = \omega^4 = \omega^3 \cdot \omega = 1 \cdot \omega = \omega\)
Step 2: Substitute into Determinant
The given determinant is:
\( \Delta = \begin{vmatrix} p^2+q^2 & r^2 & r^2 \\ p^2 & q^2+r^2 & p^2 \\ q^2 & q^2 & r^2+p^2 \end{vmatrix} \)Substitute the computed squares:
\( \Delta = \begin{vmatrix} 1+\omega^2 & \omega & \omega \\ 1 & \omega^2+\omega & 1 \\ \omega^2 & \omega^2 & \omega+1 \end{vmatrix} \)Step 3: Simplify Using Properties
Apply the property \(1 + \omega + \omega^2 = 0\) to simplify terms:
- \(1+\omega^2 = -\omega\)
- \(\omega^2+\omega = -1\)
- \(\omega+1 = -\omega^2\)
The determinant becomes:
\( \Delta = \begin{vmatrix} -\omega & \omega & \omega \\ 1 & -1 & 1 \\ \omega^2 & \omega^2 & -\omega^2 \end{vmatrix} \)Step 4: Evaluate Determinant
Factor \(\omega\) from the first row (R1) and \(\omega^2\) from the third row (R3):
\( \Delta = (\omega) (\omega^2) \begin{vmatrix} -1 & 1 & 1 \\ 1 & -1 & 1 \\ 1 & 1 & -1 \end{vmatrix} \)Using \(\omega^3 = 1\):
\( \Delta = 1 \cdot \begin{vmatrix} -1 & 1 & 1 \\ 1 & -1 & 1 \\ 1 & 1 & -1 \end{vmatrix} \)Calculate the determinant value:
\(\Delta = 1 \times [ -1((-1)(-1) - (1)(1)) - 1((1)(-1) - (1)(1)) + 1((1)(1) - (-1)(1)) ]\)
\(\Delta = [ -1(1 - 1) - 1(-1 - 1) + 1(1 + 1) ]\)
\(\Delta = [ -1(0) - 1(-2) + 1(2) ]\)
\(\Delta = 0 + 2 + 2 = 4\)
Conclusion
The value of the determinant is 4.
