Question

Using properties of determinants,prove that:
$\begin{vmatrix}1& 1+p& 1+p+q\\\ 2& 3+2p& 4+3p+2q\\\ 3& 6+3p& 10+6p+3q\end{vmatrix}=1$

Answer 1

$\triangle=\begin{vmatrix}1& 1+p& 1+p+q\\\ 2& 3+2p& 4+3p+2q\\\ 3& 6+3p& 10+6p+3q\end{vmatrix}$
Applying $R_2→R_2-2R_1$ and R3→R3-3R1$R_3→R_3-3R_1$,we have
$\triangle=\begin{vmatrix}1& 1+p& 1+p+q\\\ 0& 1& 2+p\\\ 0& 3& 7+3p\end{vmatrix}$
Applying $R_3→R_3-3R_2$ we have:
$Δ=\begin{vmatrix}1& 1+p& 1+p+q\\\ 0& 1& 2+p\\\ 0& 0& 1\end{vmatrix}$
Expanding along $C_1$,we have:
$Δ=1\begin{vmatrix}1& 2+p\\\ 0& 1\end{vmatrix}=1(1-0)=1$
Hence,the given result is proved.