The $3\times3$ matrix $M$ with $M_{ij}=a_ia_j+\mathbf 1_{i=j}$ has determinant $a_1^2+a_2^2+a_3^2+1$

Question

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?

Answer 1

Actually the "ugly way" is not too tricky, because there are some easy intermediate cancellations. Expanding using the first row gives $$(a^2+1)(b^2+c^2+1)-ab(ab)+ac(-ac)=a^2+b^2+c^2+1$$

Answer 2

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 $|v|^2+1=a^2+b^2+c^2+1$.

Answer 3

It's just $$\prod\limits_{cyc}(a^2+1)+2a^2b^2c^2-\sum_{cyc}a^2b^2(c^2+1)=a^2+b^2+c^2+1$$