Question

Calculate the value of the following determinant:
$\left|\begin{array}{cccc}a& 1& 1& 1\\ 1& a& 1& 1\\ 1& 1& a& 1\\ 1& 1& 1& a\end{array}\right|$.

Answer 1

$Given\Delta =\left|\begin{array}{cccc}a& 1& 1& 1\\ 1& a& 1& 1\\ 1& 1& a& 1\\ 1& 1& 1& a\end{array}\right|$

$R_1=R_1+R_2+R_3+R_4$

$\Rightarrow \Delta =\left|\begin{array}{cccc}3+a& 3+a& 3+a& 3+a\\ 1& a& 1& 1\\ 1& 1& a& 1\\ 1& 1& 1& a\end{array}\right|$

$\Rightarrow \Delta =(3+a)\left|\begin{array}{cccc}1& 1& 1& 1\\ 1& a& 1& 1\\ 1& 1& a& 1\\ 1& 1& 1& a\end{array}\right|$

$R_1=R_1-R_2$

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

Expanding along $R_1$

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

$\Rightarrow \Delta =(3+a)(a-1)\left|\begin{array}{ccc}1& 1& 1\\ 1& a& 1\\ 1& 1& a\end{array}\right|R_1=R_1-R_2\Rightarrow \Delta =(3+a)(a-1)\left|\begin{array}{ccc}0& 1-a& 0\\ 1& a& 1\\ 1& 1& a\end{array}\right|$

Expanding along $R_1$

$\Rightarrow \Delta =(3+a)(a-1)\{0-(1-a\left)\right\{a-1\}+0\}$

$\Rightarrow \Delta =(3+a){(a-1)}^3$