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

Toolbox:
• If each element of a row(or a column) of a determinant is multiplied by a constant k,then its value gets multiplied by k.
• By this property we can take out any common factor from any one row or any column of a given determinant.
Let $\Delta = \begin{vmatrix} 1 & 1+p& 1+p+q\\2 & 3+2p & 4+3p+2q\\3&6+3p&10+6p+3q\end{vmatrix}$

Apply $R_2\rightarrow R_2-2R_1$ and $R_1\rightarrow R_3-3R_1$ we have,

$\Delta = \begin{vmatrix} 1 & 1+p& 1+p+q\\0 & 1 & 2+p\\0&3&7+3p\end{vmatrix}$

Apply $R_3\rightarrow R_3-3R_2$ we get,

$\Delta = \begin{vmatrix} 1 & 1+p& 1+p+q\\0 & 1 & 2+p\\0&0&1\end{vmatrix}$

Now expanding along $C_1$ we get,

$\Delta=1(1-0)=1.$

Hence proved.