Question

Show that Δ = ${\displaystyle \left|\begin{array}{ccc}x& \text{p}& \text{q}\\ \text{p}& x& \text{q}\\ \text{q}& \text{q}& x\end{array}\right|=(x-\text{p})(x^2+\text{p}x-2{\text{q}}^2)}$ 

Answer 1

Applying C₁ → C₁ – C₂, we have

Δ = ${\displaystyle \left|\begin{array}{ccc}x-\text{p}& \text{p}& \text{q}\\ \text{p}-x& x& \text{q}\\ 0& \text{q}& x\end{array}\right|}$

= ${\displaystyle (x-\text{p})\left|\begin{array}{ccc}1& \text{p}& \text{q}\\ -1& x& \text{q}\\ 0& \text{q}& x\end{array}\right|}$

= ${\displaystyle (x-\text{p})\left|\begin{array}{ccc}0& \text{p}+x& 2\text{p}\\ -1& x& \text{q}\\ 0& \text{q}& x\end{array}\right|}$  Applying R₁ → R₁ – R₂ 

Expanding along C₁, we have

Δ = ${\displaystyle (x-\text{p})(\text{p}x+x^2-2{\text{q}}^2)}$

= ${\displaystyle (x-\text{p})(x^2+\text{p}x-2{\text{q}}^2)}$