Using properties of determinants, prove the following : |(1,1+p,1+p+q)(2,3+2p,1+3p+2q)(3,6+3p,1+6p+3q)|=1 ​

Question

Using properties of determinants, prove the following :

\(\begin{vmatrix} 1 & 1+p & 1+p+q \\[0.3em] 2 & 3+2p &1+3p+2q \\[0.3em] 3 & 6+3p & 1+6p+3q \end{vmatrix}=1\)

Answer 1

Let |A| = \(\begin{vmatrix} 1 & 1+p & 1+p+q \\[0.3em] 2 & 3+2p &1+3p+2q \\[0.3em] 3 & 6+3p & 1+6p+3q \end{vmatrix}\) 

Using the transformation R₂ \(\longrightarrow\) R₂ - 2R₁, 

R₃\(\longrightarrow\) R₃ - 3R₁

Using R₃ \(\longrightarrow\) R₃ - 3R₂

Expanding along column C₁,we get

|A| = 1.