Question

Using the properties of determinants, show that:

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

Answer 1

Consider, $LHS=\left|\begin{array}{ccc}1& x& x^2\\ x^2& 1& x\\ x& x^2& 1\end{array}\right|$

$C_1\to C_1+C_2+C_3$

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

Taking $(1+x+x^2)$ common from $C_1$

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

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

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

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

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

Expanding along the first column ,

$=(1+x+x^2)(x-1{)}^2\left|\begin{array}{cc}1& x\\ -(1+x)& 1\end{array}\right|$

$=(1+x+x^2)(x-1{)}^2(1+x+x^2)$

$=(x^3-1{)}^2$

$=(1-x^3{)}^2$