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If $ax^2_1+by^2_1+cz^2_1=ax^2_2+by^2_2+cz^2_2=ax^2_3+by^2_3+cz^2_3=d$ and $ax_2x_3+by_2y_3+cz_2z_3=ax_3x_1+by_3y_1+cz_3z_1=ax_1x_2+by_1y_2+cz_1z_2=f$, then prove that $$\begin{vmatrix} x_1 &y_1&z_1\\ x_2&y_2&z_2\\ x_3&y_3&z_3\\ \end{vmatrix}=(d-f)\sqrt{\frac{d+2f}{abc}}$$ where $a,b,c\ne0$


$\begin{vmatrix} x_1 &y_1&z_1\\ x_2&y_2&z_2\\ x_3&y_3&z_3\\ \end{vmatrix}=x_1y_2z_3-x_1y_3z_2-y_1x_2z_3+y_1x_3z_2+z_1x_2y_3-z_1x_3y_2$
I cannot understand how I should use the given equation to convert it in terms of $a,b,c,d,f.$ Please help.

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  • $\begingroup$ Are you sure $\left(d-f\right)$ shouldn't be $\pm\left(d-f\right)$ ? Swapping $b, y_1, y_2, y_3$ with $c, z_1, z_2, z_3$ flips the sign of the left-hand side. $\endgroup$ – darij grinberg Aug 9 '18 at 17:02
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Assuming both sides have the same sign, which I'll leave to you to prove: \begin{align*}(\text{equality to prove})&\Longleftrightarrow \begin{vmatrix} x_1 &y_1&z_1\\ x_2&y_2&z_2\\ x_3&y_3&z_3\\ \end{vmatrix}^2=(d-f)^2\frac{d+2f}{abc}\\&\Longleftrightarrow \begin{vmatrix} x_1 &y_1&z_1\\ x_2&y_2&z_2\\ x_3&y_3&z_3\\ \end{vmatrix}\begin{vmatrix} ax_1 &by_1&cz_1\\ ax_2&by_2&cz_2\\ ax_3&by_3&cz_3\\ \end{vmatrix}=(d-f)^2(d+2f)\end{align*}

Now:

  • A matrix has the same determinant as its transpose (so transpose the second of these two matrices)
  • The product of two determinants is the determinant of the product matrix.

So \begin{align*}(\text{equality to prove})&\Longleftrightarrow \begin{vmatrix} d&f&f\\f&d&f\\f&f&d \end{vmatrix}=(d-f)^2(d+2f)\end{align*}

Which is now an easy task.

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