Show that $\begin{vmatrix} a^2 & 2ab & b^2\\ ac & ad+bc &bd \\ c^2& 2cd & d^2 \end{vmatrix}\geq 0$, where $a$, $b$, $c$ and $d$ are real numbers.
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I don't buy it. The determinant is equal to $(ad-bc)^3$, from which it follows that if $ad-bc<0$, the assertion is false.
Alex R.
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What if we put b^2 instead of d^2? I tell you this because I saw an exercise like this but with b^2 and I do not know which is good. They can not be both right. But I find it more logical to have d^2 in the right down corner. – Ghost Nov 14 '16 at 19:15
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Alex, yes, Goes back, at least, to Fricke and Klein (1897). There is a typo in Magnus (Noneuclidean Tesselations) on page 23, precisely where they give the incorrect exponent for $(ad-bc).$ – Will Jagy Nov 14 '16 at 19:25
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put in a short answer. Looked through F+K, they don't bother ever stating the determinant. – Will Jagy Nov 14 '16 at 19:38
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this is Important for fuchsian groups.
$$ H = \left( \begin{array}{rrr} 0 & 0 & -1 \\ 0 & 2 & 0 \\ -1 & 0 & 0 \end{array} \right) $$
Given $$ Q = \left( \begin{array}{ccc} \alpha^2 & 2 \alpha \beta & \beta^2 \\ \alpha \gamma & \alpha \delta + \beta \gamma & \beta \delta \\ \gamma^2 & 2 \gamma \delta & \delta^2 \end{array} \right), $$ there is a typographical error on page 23 of Magnus (Noneuclidean Tesselations and Their Groups), actually $\det Q = (\alpha \delta - \beta \gamma)^3.$ We have the identity $$ Q^T H Q = (\alpha \delta - \beta \gamma)^2 H. $$
Will Jagy
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