An unusual cross product identity In this paper, the author gives a couple of unusual vector identities involving the cross product, namely (changing the notation to be a bit more standard), if $M$ is a $3 \times 3$ matrix and $a,b$ are three-dimensional,
$$ (Ma) \times (Mb) = (\operatorname{adj}{M})(a \times b) , \tag{1} $$
$$ b \times (Ma) - a \times (Mb) = M^T (a \times b) - (\operatorname{tr}{M})(a \times b) . \tag{2} $$
(1) is fairly well-known (and has been asked about before, albeit in slightly different form). I'm interested in proofs of (2). The paper says they plugged it into Mathematica, but I'd like to see something a bit more human-readable.
I've come up with the following summation convention proof, and would be interested in seeing proofs that use vector identities, rather than summation convention.

We start with
$$ T = \epsilon_{pij} \epsilon_{pmn}  \epsilon_{mkl} M_{kn} a_i b_j e_l , $$
and expand the triple $\epsilon$ in two ways:
$$ \epsilon_{pij} \epsilon_{pmn}  \epsilon_{mkl} = (\delta_{im}\delta_{jn}-\delta_{in}\delta_{jm}) \epsilon_{mkl} = \epsilon_{ikl}\delta_{jn} - \epsilon_{jkl}\delta_{in} \\
= \epsilon_{pij} (\delta_{nk}\delta_{pl}-\delta_{nl}\delta_{pk}) = \epsilon_{lij}\delta_{nk} - \epsilon_{kij}\delta_{nl} , $$
which gives
$$ T = (\epsilon_{ikl}\delta_{jn} - \epsilon_{jkl}\delta_{in}) M_{kn} a_i b_j e_l \\
= \epsilon_{ikl} a_i M_{kj} b_j e_l - \epsilon_{jkl} b_j M_{ki} a_i e_l \\
= a \times (Mb) - b \times (Ma) $$
and
$$ T = (\epsilon_{lij}\delta_{nk} - \epsilon_{kij}\delta_{nl}) M_{kn} a_i b_j e_l \\
= M_{kk} \epsilon_{ijl} a_i b_j e_l - M_{kl} e_l \epsilon_{ijk} a_i b_j \\
(\operatorname{tr}{M}) (a \times b) - M^T (a \times b) $$
But I got this proof by reverse-engineering the left-hand side, and I haven't got a satisfactory vectorial interpretation of $T$. Can someone provide either a more vectory proof, or a nice interpretation for $T$ itself?
 A: To prove an identity $x=y$ of vectors in $\Bbb R^3$, it suffices to prove that
$x\cdot c=y\cdot c$ for all $c\in\Bbb R^3$. So for $(2)$, I'll prove that
$$(b \times (Ma) - a \times (Mb))\cdot c = 
(M^T (a \times b) - (\operatorname{tr}{M})(a \times b))\cdot c . \tag{2'}$$
for all vectors $a$, $b$ and $c$.
Write
$$[a,b,c]=(a\times b)\cdot c$$
for the scalar triple product. This is a trilinear map, and is alternating:
$$[b,b,c]=[a,b,b]=0$$
for all $a$, $b$, $c$. There is only one alternating trilinear map up to scalar
multiplication; any alternating trilinear map is a constant multiple
of the scalar triple product.
Let $M$ be a matrix, and define
$$\phi(a,b,c)=[Ma,b,c]+[a,Mb,c]+[a,b,Mc].$$
It's easy to see that $\phi$ is alternating trilinear, so
$$\phi(a,b,c)=t[a,b,c]$$
where $t$ depends only on $M$. Taking $a$, $b$, $c$ to be the standard unit
vectors gives $t=\text{Tr}(M)$. Therefore
$$[Ma,b,c]+[a,Mb,c]+[a,b,Mc]=\text{Tr}(M)[a,b,c].\tag{3}$$
Observe that
$$[Ma,b,c]=-(b\times Ma)\cdot c,$$
$$[a,Mb,c]=(a\times Mb)\cdot c,$$
$$[a,b,Mc]=(a\times b)\cdot Nc=(M^T(a\times b))\cdot c$$
(using the identity $x\cdot My=M^Tx\cdot y$)
and
$$\text{Tr}(M)[a,b,c]=(\text{Tr}(M)(a\times b))\cdot c.$$
So $(3)$ becomes
$$(-b\times Ma+a\times Mb+M^T(a\times b))\cdot c=\text{Tr}(M)(a\times b)\cdot c$$
which is equivalent to $(2')$.
