The equation, $\left[ \begin{matrix} 1 & x & y \end{matrix} \right] \left[ \begin{matrix} 1 & 3 & 1 \ 0 & 2 & -1 \ 0 & 0 & 1 \end{matrix} \right] \left[ \begin{matrix} 1 \ x \ y \end{matrix} \right] =\left[ 0 \right] $ has for (i) $y=0$, (p) rational roots (ii) $y=-1$ (q) irrational roots (r) integral roots

Multiplying the matrices gives the quadratic equation x^2 + 3xy + y^2 - x = 0. For y = 0, the equation becomes x^2 - x = 0, yielding x = 0 and x = 1, which are both rational roots. For y = -1, the equation becomes x^2 - 4x + 1 = 0; using the quadratic formula gives x = 2 + sqrt(3) and x = 2 - sqrt(3), which are irrational roots. This matches case (i) with (p) and case (ii) with (q).

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