Symbolic transformations
Task number: 2604
Show without calculating the determinants that for \(x,y,z\ne 0\) it holds:
\( \det \begin{pmatrix} 0 & x & y & z \\ x & 0 & z & y \\ y & z & 0 & x \\ z & y & x & 0 \\ \end{pmatrix} = \det \begin{pmatrix} 0 & 1 & 1 & 1 \\ 1 & 0 & z^2 & y^2 \\ 1 & z^2 & 0 & x^2 \\ 1 & y^2 & x^2 & 0 \\ \end{pmatrix} \)
Solution
Do the following transformations: divide the 2nd–4th column by \(x\), \(y\), \(z\) then multiply the 2nd–4th row by \(yz\), \(xz\) and \(xy\), and finally divide the first column by \(xyz\).
\( \begin{vmatrix} 0 & x & y & z \\ x & 0 & z & y \\ y & z & 0 & x \\ z & y & x & 0 \\ \end{vmatrix} =xyz \begin{vmatrix} 0 & 1 & 1 & 1 \\ x & 0 & \frac{z}{y} & \frac{y}{z} \\ y & \frac{z}{x} & 0 & \frac{x}{z} \\ z & \frac{y}{x} & \frac{x}{y} & 0 \\ \end{vmatrix} =\frac{1}{xyz} \begin{vmatrix} 0 & 1 & 1 & 1 \\ xyz & 0 & z^2 & y^2 \\ yxz & z^2 & 0 & x^2 \\ zxy & y^2 & x^2 & 0 \\ \end{vmatrix} = \begin{vmatrix} 0 & 1 & 1 & 1 \\ 1 & 0 & z^2 & y^2 \\ 1 & z^2 & 0 & x^2 \\ 1 & y^2 & x^2 & 0 \\ \end{vmatrix} \).
Comment
Each term in the equality is a rational function in the variables \(x,y\) and \(z\), and both extreme determinants are indeed polynomials.
Argument on \(\mathbb R\): If any of the variables is equal to \(0\), the polynomials (i.e., the extreme determinants) must also be equal at the points of discontinuity of the rational functions (i.e., the internal determinants), because the polynomials are continuous and must take on the same values in the limit.
One can also use the argument that two polynomials coincide if they have the same values for sufficiently many choices of variables.
Over other fields, it is probably easier to analyze the cases directly. For example, for \(x=0\) we have \( \begin{vmatrix} 0 & 0 & y & z \\ 0 & 0 & z & y \\ y & z & 0 & 0 \\ z & y & 0 & 0 \\ \end{vmatrix} = \begin{vmatrix} 0 & 1 & 1 & 1 \\ 1 & 0 & z^2 & y^2 \\ 1 & z^2 & 0 & 0 \\ 1 & y^2 & 0 & 0 \\ \end{vmatrix} \), which follows from the determinant of block matrices and the identity \( \begin{vmatrix} y & z \\ z & y \\ \end{vmatrix} = yz \begin{vmatrix} 1 & 1 \\ \frac{z}{y} & \frac{y}{z} \\ \end{vmatrix} = \begin{vmatrix} 1 & 1 \\ z^2 & y^2 \\ \end{vmatrix} \)
The equality of determinants in fact holds for any values of the variables \(x,y\) and \(z\) and over any field.
