Question

Show that $\left|\begin{array}{ccc}x-3& x-4& x-\alpha \\ x-2& x-3& x-\beta \\ x-1& x-2& x-\gamma \end{array}\right|=0,\mathrm{where}\alpha ,\beta ,\gamma \mathrm{are}\mathrm{in}\mathrm{AP}.$

Answer 1

Given:α, β, γ areinA.P.

Now,
$2\beta =\alpha +\gamma$

$$\begin{gathered} ∆=\left|\begin{array}{ccc}x-3& x-4& x-\alpha \\ x-2& x-3& x-\beta \\ x-1& x-2& x-\gamma \end{array}\right| \\ ∆=\frac{1}{2}\left|\begin{array}{ccc}x-3& x-4& x-\alpha \\ 2x-4& 2x-6& 2x-2\beta \\ x-1& x-2& x-\gamma \end{array}\right|\left[\mathrm{Applying}{R}_2\to 2R_2\right] \\ ∆=\frac{1}{2}\left|\begin{array}{ccc}x-3& x-4& x-\alpha \\ 0& 0& -2\beta +\alpha +\gamma \\ x-1& x-2& x-\gamma \end{array}\right|\left[\because 2\beta =\alpha +\gamma \right]\left[\mathrm{Applying}{R}_2\to R_2-\left(R_1+R_3\right)\right] \\ ∆=\frac{1}{2}\left|\begin{array}{ccc}x-3& x-4& x-\alpha \\ 0& 0& 0\\ x-1& x-2& x-\gamma \end{array}\right| \\ ∆=0 \end{gathered}$$