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Question:

Using properties of determinants, prove that \begin{vmatrix} a^2+2a & 2a+1 & 1 \ 2a+1 & a+2 & 1 \ 3 & 3 & 1 \end{vmatrix} = (a-1)^3

Solution:

\begin{vmatrix} a^2+2a & 2a+1 & 1 \ 2a+1 & a+2 & 1 \ 3 & 3 & 1 \end{vmatrix}
Apply R1→R1-R2, R2→R2-R3 ⇒ \begin{vmatrix} a^2-1 & a-1 & 0 \ 2(a-1) & a-1 & 0 \ 3 & 3 & 1 \end{vmatrix}
⇒ \begin{vmatrix} (a-1)(a+1) & a-1 & 0 \ 2(a-1) & a-1 & 0 \ 3 & 3 & 1 \end{vmatrix}
Take (a-1) common from first and second row = (a-1)^2 \begin{vmatrix} a+1 & 1 & 0 \ 2 & 1 & 0 \ 3 & 3 & 1 \end{vmatrix}
Now expand along third column = (a-1)^2[1(a+1-2)] = (a-1)^3
Hence proved.