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.