I would like to show that
$$ \det \pmatrix{ a^2+1 & ab & ac \\ ab & b^2+1 & bc \\ ac & bc & c^2+1 } = a^2+b^2+c^2+1 $$
Is there a trick or do I need to calculate the determinant the ugly way?
The $3\times3$ matrix $M$ with $M_{ij}=a_ia_j+\mathbf 1_{i=j}$ has determinant $a_1^2+a_2^2+a_3^2+1$
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determinant
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5[trick](https://en.wikipedia.org/wiki/Sylvester's_determinant_identity). – 2017-02-05
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1What 'tricks' have you tried so far? – 2017-02-05
3 Answers
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It's just $$\prod\limits_{cyc}(a^2+1)+2a^2b^2c^2-\sum_{cyc}a^2b^2(c^2+1)=a^2+b^2+c^2+1$$
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If $A$ is that matrix, and we let $v=\begin{pmatrix}a\\b\\c\end{pmatrix}$, then we notice that $Ax = (v\cdot x)v+x$ for all vectors $x$. In particular, $Av=(|v|^2+1)v$ whereas $Aw=w$ for $w\perp v$. Thus, we can express $A$ with respect to a suitable basis $v,w_1,w_2$ as $$\begin{pmatrix}|v|^2+1&0&0\\0&1&0\\0&0&1\end{pmatrix} $$ which obviously has determinant $|v|^2+1=a^2+b^2+c^2+1$.
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Actually the "ugly way" is not too tricky, because there are some easy intermediate cancellations. Expanding using the first row gives $$(a^2+1)(b^2+c^2+1)-ab(ab)+ac(-ac)=a^2+b^2+c^2+1$$
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0You are totally right. In mathematics, not everything is as compilcated as it looks like. – 2017-02-05
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1@Marcel Of course the formula generalises to many dimensions using the "tricks" people have pointed out - there is a limit to computing such things by hand! – 2017-02-05