Question

Check the equality of the following determinant:
$\left|\begin{array}{ccc}x-2& 2x-3& 3x-4\\ x-4& 2x-9& 3x-16\\ x-8& 2x-27& 3x-64\end{array}\right|=0$

Answer 1

$\left|\begin{array}{ccc}x-2& 2x-3& 3x-4\\ x-4& 2x-9& 3x-16\\ x-8& 2x-27& 3x-64\end{array}\right|=0\left|\begin{array}{ccc}x& 2x& 3x\\ x& 2x& 3x\\ x& 2x& 3x\end{array}\right|-\left|\begin{array}{ccc}2& 3& 4\\ 4& 9& 16\\ 8& 27& 64\end{array}\right|=0\left|\begin{array}{ccc}x& 2x& 3x\\ x& 2x& 3x\\ x& 2x& 3x\end{array}\right|-\left|\begin{array}{ccc}2& 3& 4\\ 2^2& 3^2& 4^2\\ 2^3& 3^3& 4^3\end{array}\right|=0\left|\begin{array}{ccc}x& 2x& 3x\\ x& 2x& 3x\\ x& 2x& 3x\end{array}\right|-2\times 3\times 4\left|\begin{array}{ccc}1& 3& 4\\ 2& 3& 4\\ 2^2& 3^2& 4^2\end{array}\right|=0\Rightarrow 24[1(48-36)-1(32-16)+1(18-12)]24[12-16+6]=24[12-10]=48\left|\begin{array}{ccc}x& 2x& 3x\\ x& 2x& 3x\\ x& 2x& 3x\end{array}\right|-48=0x^3\left|\begin{array}{ccc}1& 2& 3\\ 1& 2& 3\\ 1& 2& 3\end{array}\right|-48=0\Rightarrow x^3\left(0\right)-48=0$

Hence it is not true for any values of $x$