Question

Using properties of determinants, prove that

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

Answer 1

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

Apply $R_1\to R_1-R_2,R_2\to R_2-R_3$

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

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

Take $(a-1)$ common from first and second row

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

Now expand along third column

$=(a-1{)}^2[1(a+1-2)]=(a-1{)}^3$

Hence proved.