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If p, q, r are the cube roots of unity, then what is \(\begin{vmatrix} p^2+q^2 & r^2 & r^2 \\ p^2 & q^2+r^2 & p^2 \\ q^2 & q^2 & r^2+p^2 \end{vmatrix}\) equal to ?

This question was previously asked in
NDA 2 2026 GAT Question Paper (13-Sep-2026)
The correct answer is
4

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.

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