Question

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

Answer 1

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

$R_1\to R_1-R_2$

$R_2\to R_2-R_3$

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

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

Taking out common $a-1$ from both $R_1$ and $R_2$

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

Expanding along $C_3$

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