Question

Show that $\left|\begin{array}{ccc}{x}^2& 2x& 1\\ 1& x^2& 2x\\ 2x& 1& x^2\end{array}\right|$ is a perfect square.

Answer 1

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

Apply $R1\to R1+R2+R3$

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

Take $x^2+2x+1$ as common , we get

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

Solving the determinant,

$=(x^2+2x+1)\left[\right(x^4-2x)-(x^2-4x^2)+(1-2x^3\left)\right]$

$=(x^2+2x+1)\left[\right(x^4-2x)-(x^2-4x^2)+(1-2x^3\left)\right]$

$=(x^2+2x+1)(x^4-2x^3+3x^2-2x+1)$

$=(x^2+2x+1)(x^4+1+x^2+2x^2-2x-2x^3)$

Using $(a-b-c{)}^2=a^2+b^2+c^2+2ab-2bc-2ac$

$D=(x+1{)}^2(x^2+1-x{)}^2$, which is a perfect square.

Hence proved.