Question

Prove that :

$\left|\begin{array}{ccc}1& 1+p& 1+p+q\\ 2& 3+2p& 4+3p+2q\\ 3& 6+3p& 10+6p+3q\end{array}\right|=1$

Answer 1

$$\begin{gathered} \mathrm{Let}\mathrm{LHS}=\mathrm{\Delta }=\left|11+p1+p+q \\ 23+2p4+3p+2q \\ 36+3p10+6p+3q\right| \\ =\left|111+\mathrm{p} \\ 234+3\mathrm{p} \\ 3610+6\mathrm{p}\right|+\left|1\mathrm{p}\mathrm{q} \\ 22\mathrm{p}2\mathrm{q} \\ 33\mathrm{p}3\mathrm{q}\right| \\ =\left|111 \\ 234 \\ 3610\right|+\left|11\mathrm{p} \\ 233\mathrm{p} \\ 366\mathrm{p}\right|+\left(\mathrm{pq}\right)\left|111 \\ 222 \\ 333\right|\left[\mathrm{Taking}\mathrm{out}pq\mathit{}\mathrm{common}\mathrm{from}\mathrm{last}\mathrm{determinant}\right] \\ =\left|111 \\ 234 \\ 3610\right|+\left(\mathrm{p}\right)\left|111 \\ 233 \\ 366\right|+0\left[\mathrm{Taking}\mathrm{out}p\mathrm{common}\mathrm{from}\mathrm{second}\mathrm{determinant}\right] \\ =\left|111 \\ 234 \\ 3610\right|+0\left[\because \mathrm{Value}\mathrm{of}\mathrm{determinant}\mathrm{with}\mathrm{two}\mathrm{identical}\mathrm{columns}\mathrm{is}\mathrm{zero}\right] \\ =\left|100 \\ 212 \\ 337\right|\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\left[\mathrm{Applying}{\mathrm{C}}_2\to {\mathrm{C}}_2-{\mathrm{C}}_1\mathrm{and}{\mathrm{C}}_3\to {\mathrm{C}}_3-\mathrm{C}1\right] \\ =\left\{1\times \left|\begin{array}{cc}1& 2\\ 3& 7\end{array}\right|\right\}\left[\mathrm{Expanding}\mathrm{along}{\mathrm{R}}_1\right] \\ =7-6 \\ =1 \\ =\mathrm{RHS} \end{gathered}$$