Evaluate the following determinant:

Solution:

(ii) To Prove: $\left|\begin{array}{ccc}a^{2}+2 a & 2 a+1 & 1 \\ 2 a+1 & a+2 & 1 \\ 3 & 3 & 1\end{array}\right|=(a-1)^{3}$

LHS $=\left|\begin{array}{ccc}a^{2}+2 a & 2 a+1 & 1 \\ 2 a+1 & a+2 & 1 \\ 3 & 3 & 1\end{array}\right|$

Applying $R_{1} \rightarrow R_{1}-R_{2}$

$=\left|\begin{array}{ccc}a^{2}+2 a-2 a-1 & 2 a+1-a-2 & 1-1 \\ 2 a+1 & a+2 & 1 \\ 3 & 3 & 1\end{array}\right|$

$=\left|\begin{array}{ccc}a^{2}-1 & a-1 & 0 \\ 2 a+1 & a+2 & 1 \\ 3 & 3 & 1\end{array}\right|$

Taking $(a-1)$ common from $R_{1}$

$=(a-1)\left|\begin{array}{ccc}a+1 & 1 & 0 \\ 2 a+1 & a+2 & 1 \\ 3 & 3 & 1\end{array}\right|$

Applying $R_{2} \rightarrow R_{2}-R_{3}$

$=(a-1)\left|\begin{array}{ccc}a+1 & 1 & 0 \\ 2 a+1-3 & a+2-3 & 1-1 \\ 3 & 3 & 1\end{array}\right|$

$=(a-1)\left|\begin{array}{ccc}a+1 & 1 & 0 \\ 2 a-2 & a-1 & 0 \\ 3 & 3 & 1\end{array}\right|$

Taking $(a-1)$ common from $R_{2}$

$=(a-1)^{2}\left|\begin{array}{ccc}a+1 & 1 & 0 \\ 2 & 1 & 0 \\ 3 & 3 & 1\end{array}\right|$

Expanding through $C_{3}$

$=(a-1)^{2}[1(1(a+1)-2)]$

$=(a-1)^{2}[1(a+1-2)]$

$=(a-1)^{2}(a-1)$

$=(a-1)^{3}=\mathrm{RHS}$

Hence, $\left|\begin{array}{ccc}a^{2}+2 a & 2 a+1 & 1 \\ 2 a+1 & a+2 & 1 \\ 3 & 3 & 1\end{array}\right|=(a-1)^{3}$.