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

Let
$△=\left|\begin{array}{ccc}{a}^2+2a& 2a+1& 1\\ 2a+1& a+2& 1\\ 3& 3& 1\end{array}\right|\text{Operating}{R}_1\to R_1-R_2,\text{we have}=\left|\begin{array}{ccc}{a}^2-1& a-1& 0\\ 2a+1& a+2& 1\\ 3& 3& 1\end{array}\right|\text{Operating}{R}_2\to R_2-R_3,\text{we have}=\left|\begin{array}{ccc}{a}^2-1& a-1& 0\\ 2a-2& a-1& 0\\ 3& 3& 1\end{array}\right|=\left|\begin{array}{ccc}(a+1)(a-1)& a-1& 0\\ 2(a-1)& a-1& 0\\ 3& 3& 1\end{array}\right|\text{Taking~}(a-1)\text{as common from}{R}_1\text{and}{R}_2\text{both, we have}={(a-1)}^2\left|\begin{array}{ccc}a+1& 1& 0\\ 2& 1& 0\\ 3& 3& 1\end{array}\right|\text{Expanding along}{C}_3\text{, we have}={(a-1)}^2[1(a+1)-2]={(a-1)}^2(a-1)={(a-1)}^3$