If p , q , r are the cube roots of unity, then what is vmatrix...

Let $a = p^2$, $b = q^2$, and $c = r^2$. The given determinant can be written as:

$\Delta = \begin{vmatrix} a+b \text{ and } c \text{ and } c \\ a \text{ and } b+c \text{ and } a \\ b \text{ and } b \text{ and } c+a \end{vmatrix}$

Applying the row operation $R_1 \rightarrow R_1 - R_2 - R_3$, we get:

$\Delta = \begin{vmatrix} 0 \text{ and } -2b \text{ and } -2a \\ a \text{ and } b+c \text{ and } a \\ b \text{ and } b \text{ and } c+a \end{vmatrix}$

Expanding the determinant along the first row:

$\Delta = 0 - (-2b) \begin{vmatrix} a \text{ and } a \\ b \text{ and } c+a \end{vmatrix} + (-2a) \begin{vmatrix} a \text{ and } b+c \\ b \text{ and } b \end{vmatrix}$

$\Delta = 2b(ac + a^2 - ab) - 2a(ab - b^2 - bc)$

$\Delta = 2abc + 2a^2b - 2ab^2 - 2a^2b + 2ab^2 + 2abc$

$\Delta = 4abc$

Substituting back $a = p^2$, $b = q^2$, and $c = r^2$:

$\Delta = 4(p^2)(q^2)(r^2) = 4(pqr)^2$

Since $p, q, r$ are the cube roots of unity (1, \omega, \omega^2$), their product is$pqr = 1 \cdot \omega \cdot \omega^2 = \omega^3 = 1$.<br><br>\Delta = 4(1)^2 = 4$

Answer: $4$