Question

Using properties of determinants prove the following questions.

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

Answer 1

$LHS=\left|\begin{array}{ccc}1& 1+p& 1+p+q\\ 2& 3+2p& 4+3p+2q\\ 3& 6+3p& 10+6p+3q\end{array}\right|$
$=\left|\begin{array}{ccc}1& 1+p& 1+p+1\\ 0& 1& 2+p\\ 0& 3& 7+3p\end{array}\right|$
(using $R_2\to R_2-2R_1$ and$R_3\to R_3-3R_1$)

$=\left|\begin{array}{ccc}0& 1+p& 1+p+q\\ 0& 1& 2+p\\ 0& 0& 1\end{array}\right|$
(using $R_3\to R_3-3R_2$)
Expanding along $C_1$, we get =$1(1\times 1-0)=1=RHS$