Using properties of determinants prove that:|(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 that:

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

|(1,1+p,1+p+q)(2,3+2p,1+3p+2q)(3,6+3p, 1+6p+3q)| = 1

Answer 1

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

= (1/2)[0 + 3(1 + q) - (1 + 6p + 3q) + p(6 + 3p - 3p)] 

[expansion by first row] 

= (1/2)(3 + 3q - 1 - 6p - 3q + 6p) =1