Symbolic transformations

Task number: 2604

Show without calculating the determinants that:

\( \det \begin{pmatrix} 0 & x & y & z \\ x & 0 & z & y \\ y & z & 0 & x \\ z & y & x & 0 \\ \end{pmatrix} = \det \begin{pmatrix} 0 & 1 & 1 & 1 \\ 1 & 0 & z^2 & y^2 \\ 1 & z^2 & 0 & x^2 \\ 1 & y^2 & x^2 & 0 \\ \end{pmatrix} \)

  • Resolution

    Assume first that \(x,y,z\ne 0\), and 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 \(xyz\).

    \( \begin{vmatrix} 0 & x & y & z \\ x & 0 & z & y \\ y & z & 0 & x \\ z & y & x & 0 \\ \end{vmatrix} =xyz \begin{vmatrix} 0 & 1 & 1 & 1 \\ x & 0 & \frac{z}{y} & \frac{y}{z} \\ y & \frac{z}{x} & 0 & \frac{x}{z} \\ z & \frac{y}{x} & \frac{x}{y} & 0 \\ \end{vmatrix} =\frac{1}{xyz} \begin{vmatrix} 0 & 1 & 1 & 1 \\ xyz & 0 & z^2 & y^2 \\ yxz & z^2 & 0 & x^2 \\ zxy & y^2 & x^2 & 0 \\ \end{vmatrix} = \begin{vmatrix} 0 & 1 & 1 & 1 \\ 1 & 0 & z^2 & y^2 \\ 1 & z^2 & 0 & x^2 \\ 1 & y^2 & x^2 & 0 \\ \end{vmatrix} \).

    Every term is a rational function in variables \(x,y\) a \(z\), both outer determinants are polynomials. When some variable equals \(0\), the polynomials (i.e. the outer determinants) are equal also in the nonconnectivity points of the rational functions (i.e. the inner determinants), since the polynomials are continuous and have the same limits.

    The equality holds for any values of \(x,y\) and \(z\).

Difficulty level: Easy task (using definitions and simple reasoning)
Proving or derivation task
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