Using properties of determinants prove that: |(a,a+2b,a+2b+3c)(3a,4a+6b,5a+7b+9c)(6a,9a+12b,11a+15b+18c)|

Question

Using properties of determinants prove that:

\(\begin{vmatrix} a & a+2b & a+2b+3c \\[0.3em] 3a & 4a+6b &5a+7b+9c \\[0.3em] 6a & 9a+12b & 11a+15b+18c \end{vmatrix}\) = -a³

Answer 1

\(\begin{vmatrix} a & a+2b & a+2b+3c \\[0.3em] 3a & 4a+6b &5a+7b+9c \\[0.3em] 6a & 9a+12b & 11a+15b+18c \end{vmatrix}\)

= (1/6)[0 + 0 + 6a{a(a + b) - a(2a + b)[expansion by first column] 

= - a³