What is the first interesting matric Toda bracket in the stable homotopy of the sphere? Feel free to gloss ‘interesting’ as you see fit.  One way:

1. What is the lowest degree matric Toda bracket in $\pi_\ast(S)$ that doesn't contain zero?

By ‘degree’ I mean total homotopical degree, i.e. the $\ast$ in $\pi_\ast$.  By ‘matric’ I mean to exclude ordinary Toda brackets, that is matric Toda brackets all of whose entries are one-by-one matrices, and also to exclude brackets that are trivially determined by ordinary Toda brackets.
I'm also interested in the title question with ‘first’ replaced by ‘simplest’.  For instance:

2. What is the lowest order matric Toda bracket in $\pi_\ast(S)$, all of whose matrix entries are sums of products of Hopf elements, that doesn't contain zero?

By ‘order’ I mean the number of entries in the bracket, i.e. whether it is 3-fold or 4-fold or 5-fold or...
Now I'd like to know the same, but with the proviso that the bracket be detectable in the classical Adams spectral sequence:

3. What is the first or simplest, interesting matric Toda bracket in $\pi_\ast(S)$ that is detectable in the $\mathrm{E}_2$ term of the classical Adams spectral sequence?

Some remarks:


*

*I believe the matric Massey product $\langle h_2^2, h_0, \left(\begin{array}{cc} h_1 h_3 & h_2^2 \end{array} \right), \left(\begin{array}{c} h_1 \\ h_2 \end{array}\right)\rangle $ in the $\mathrm{E}_2$ term of the Adams spectral sequence is the class $e_0$, but that is not a permanent cycle.

*Kochman lists the matric Toda bracket $\langle \sigma, \left(\begin{array}{cc} A[31] & \nu\end{array}\right),\left(\begin{array}{cc} \eta & 0 \\ 0 & \eta \end{array}\right),\left(\begin{array}{c} \nu \\ \eta A[30] \end{array}\right)\rangle$ in degree 44 as containing an element of order 8, where $A[30]$ and $A[31]$ refer to certain elements of order 2 in those degrees.
 A: I know this post is quite old, but in case you are still interested, or anyone else is, I thought about sharing my recent thoughts about the topic. After all, this is the second result on "matric toda bracket" on google, and it frustrated me several times that this is still unanswered.
The bracket André suggested sadly uses relations which are only valid in $tmf$: in $\pi(S)$, we have $\nu^3 + \eta \epsilon = \eta^2 \sigma \neq 0$.
We want to construct a nontrivial matric Toda bracket, so we have to use a relation which cannot be written as a single product. The first one of those is given by $4\nu + \eta^3 = 0$.
Multiplying this by two relates this in some sense to the easier relation $8\nu = 0$, so we could wonder about this enabling us to exhibit the Toda bracket $\langle \nu, 8, \nu\rangle$ as twice the matric Toda bracket 
$$\left\langle \left(\begin{array}{cc} \nu & \eta\end{array}\right), \left(\begin{array}{cc} 4 & \eta^2\\ \eta^2 & 0\end{array}\right), \left(\begin{array}{c} \nu \\ \eta\end{array}\right)\right\rangle$$
This yields a degree $7$ class of indeterminacy $\nu \cdot \pi_4 + \pi_4 \cdot \nu + \eta \cdot\pi_6 + \pi_6 \cdot \eta = 0$.
We want to multiply the middle term by $2$. In order to make sense of that, consider the following setting: In a good model for spectra, we can think of bimodules over a ring spectrum $E$, i.e. there are maps $E \wedge X\rightarrow X$ and $X\wedge E \rightarrow X$ plus respective commutative diagrams. Then we can talk about Toda brackets $\langle a, x, b\rangle$, where $a,b\in \pi_* E$ and $x\in \pi_* X$. Now given a bimodule map $f:X \rightarrow Y$, a defining system for the Toda bracket $\langle a, x, b\rangle$ can be pushed forward using $f$ to obtain a defining system for the Toda bracket $\langle a, f_* x, b\rangle$, so we obtain
$$
f_* \langle a, x, b \rangle \subseteq \langle a, f_* x, b\rangle
$$
This statement immediately generalizes to matric Toda brackets.
Notice that we can use this here: $S\xrightarrow{2} S$ has a representative which commutes with the right and the left action of $S$ on itself, since it can be factored naturally through the pinching map $S\rightarrow S\vee S$. 
This can be applied as above to see
$$
2\cdot\left\langle \left(\begin{array}{cc} \nu & \eta\end{array}\right), \left(\begin{array}{cc} 4 & \eta^2\\ \eta^2 & 0\end{array}\right), \left(\begin{array}{c} \nu \\ \eta\end{array}\right)\right\rangle \subseteq \left\langle \left(\begin{array}{cc} \nu & \eta\end{array}\right), \left(\begin{array}{cc} 8 & 0\\ 0 & 0\end{array}\right), \left(\begin{array}{c} \nu \\ \eta\end{array}\right)\right\rangle
$$
This actually is an equality, since the indeterminacy is zero on both sides.
In fact, the right hand side equals the usual Toda bracket $\langle \nu, 8, \nu\rangle$.
This element is usually known as $8\cdot \sigma$, which is nonzero. We obtain
$$
\left\langle \left(\begin{array}{cc} \nu & \eta\end{array}\right), \left(\begin{array}{cc} 4 & \eta^2\\ \eta^2 & 0\end{array}\right), \left(\begin{array}{c} \nu \\ \eta\end{array}\right)\right\rangle = 4\sigma
$$
What this example tells us is that matric Toda brackets really should be expected all over the place: Whenever a simple relation (consisting of a single product) lifts to somewhere as a more complicated relation (with more than one summand), we should be able to lift Toda brackets built from the simple relation to matric Toda brackets by adding additional rows which account for the new summands, in our case the $\eta^3$. Note also that the exact way in which we did that did not matter that much, we could also have chosen something like
$$\left\langle \left(\begin{array}{cc} \nu & \eta^2\end{array}\right), \left(\begin{array}{cc} 4 & \eta\\ \eta & 0\end{array}\right), \left(\begin{array}{c} \nu \\ \eta^2\end{array}\right)\right\rangle
$$
or similar. The only way this generally might go wrong is that you are unable to do this in a way which actually reduces back to the bracket you wanted (we used $2\cdot \eta^2$ to end up with a diagonal matrix).
If you look closely into the reference André posted, there a similar thing happened, only that the lifting did not happen along an actual map, but from the $E^{\infty}$-page of a spectral sequence to the actual thing, because of nontrivial extensions.
