Since $\mu_I$ is proportional to the angular momentum, I'm guessing that $\mu_I$ is the magnetic moment of some particle. If so, $\mu_I$ is a constant vector; any derivative hitting on it will vanish.
Recall that magnetic moment is defined for a localized current distribution, which is the following integral,
\begin{equation}
{\bf m } = \frac{1}{2}\int {\bf x'} \times J({\bf x'}) dV'
\end{equation}
just like the total charge of a localized charge distribution, it has no spatial dependence.
Is there any physical significance to this matrix
The physical (geometric) relevance to the matrix
$$\left| \begin{matrix}
\vec{i} & \vec{j} & \vec{k} \\
a_i & a_j & a_k \\
b_i & b_j & b_k
\end{matrix}\right|$$
with regard to the cross product $\vec{a} \times \vec{b}$ is
1: that the three vectors $\vec{i}$, $\vec{j}$, and $\vec{k}$ constitute a vector basis that spans a space which is
either also spanned by $\vec{a}$, $\vec{b}$, and one additional vector which is perpendicular to $\vec{a}$ as well as $\vec{b}$;
or, in case that vectors $\vec{a}$ and $\vec{b}$ are parallel to each other, also spanned by $\vec{a}$ and two additional (non-parallel) vectors.
2: the three basis vectors $\vec{i}$, $\vec{j}$, and $\vec{k}$ are pairwise orthogonal (perpendicular) to each other.
Therefore vectors $\vec{a}$ and $\vec{b}$ as well as the cross product vector $\vec{a} \times \vec{b}$ can be completely and uniquely expressed in terms of the corresponding components:
$\vec{a} := a_i \vec{i} + a_j \vec{j} + a_k \vec{k}$,
$\vec{b} := b_i \vec{i} + b_j \vec{j} + b_k \vec{k}$, and
$\vec{a} \times \vec{b} := \{ab\}_i \vec{i} + \{ab\}_j \vec{j} + \{ab\}_k \vec{k}$.
Finally:
3: the three basis vectors $\vec{i}$, $\vec{j}$, and $\vec{k}$ have equal magnitudes:
$| \vec{i} | = | \vec{j} | = | \vec{k} |$.
As a consequence, the "mathematical trick" of expressing the cross product vector $\vec{a} \times \vec{b}$ as the above determinant "works":
The component of cross product vector $\vec{a} \times \vec{b}$ "along/parallel to" vector $\vec{a}$ vanishes explicitly:
$$\left( a_i (a_j b_k - a_k b_j) \frac{(| \vec{i} |)^2}{| \vec{a} \times \vec{b} |} \right) + \left( a_j (a_k b_i - a_i b_k) \frac{(| \vec{j} |)^2}{| \vec{a} \times \vec{b} |} \right) + \left( a_k (a_i b_j - a_j b_i) \frac{(| \vec{k} |)^2}{| \vec{a} \times \vec{b} |} \right) = $$
$$\frac{(| \vec{i} |)^2}{| \vec{a} \times \vec{b} |} \left( a_i (a_j b_k - a_k b_j) + a_j (a_k b_i - a_i b_k) + a_k (a_i b_j - a_j b_i) \right) = 0,$$
and likewise the component of cross product vector $\vec{a} \times \vec{b}$ "along/parallel to" vector $\vec{b}$ vanishes explicitly;
i.e. cross product vector $\vec{a} \times \vec{b}$ is expressed explicitly orthogonal to both vectors $\vec{a}$ and $\vec{b}$.
And, no less important the magnitude of cross product vector $\vec{a} \times \vec{b}$ "comes out correctly", i.e. such that
$$ \Big( | \vec{a} \times \vec{b} | \Big)^2 := $$
$$ \left( (a_j b_k - a_k b_j)^2 + (a_k b_i - a_i b_k)^2 + (a_i b_j - a_j b_i)^2 \right) ~ (| \vec{i} |)^4 = $$
$$ \left( (a_i^2 + a_j^2 + a_k^2) ~ (b_i^2 + b_j^2 + b_k^2) \right) ~ (| \vec{i} |)^4 - \left( (a_i b_i + a_j b_j + a_k b_k) ~ (| \vec{i} |)^2 \right)^2 := $$
$$ \Big( | \vec{a} | \Big)^2 \Big( | \vec{b} | \Big)^2 - | \vec{a} | ~ | \vec{b} | ~ a_b ~ b_a,$$
where $b_a$ denotes the component of vector $\vec{b}$ "along/parallel to" vector $\vec{a}$, and $a_b$ denotes the component of vector $\vec{a}$ "along/parallel to" vector $\vec{b}$.
Best Answer
As ACuriousMind has already noted, you can geometrically interpret the length of the cross product of two vectors as the area of the parallelogram (or as twice the area of the triangle) spanned by them, and (the absolute values of) its components as the areas of the projections of that parallelogram onto the coordinate planes.
As for the dot product of two vectors, based on the law of cosines, you can interpret it as half the difference between the sum of their squares and the square of their difference:
$$\|\vec a - \vec b\|^2 = \|\vec a\|^2 + \|\vec b\|^2 - 2(\vec a \cdot \vec b).$$
In other words, taking the vectors to be two sides of a triangle, the dot product measures (half) the amount by which Pythagoras' law fails for this triangle.
Another way to geometrically interpret (the absolute value of) the dot product is as half the area of the triangle formed by rotating one of the vectors by 90° in their common plane, and then taking the resulting vectors as two sides of a triangle:
This follows from the well known dot product formula $\vec a \cdot \vec b = \|\vec a\| \|\vec b\| \cos \gamma$, where $\gamma$ is the angle between $\vec a$ and $\vec b$, from the triangle area formula $T = \frac12 \|\vec a'\| \|\vec b\| \sin \gamma'$, where $T$ is the area of the triangle formed by the vectors $\vec a'$ and $\vec b$ and $\gamma'$ is the angle between them, and the fact that the angles $\gamma$ and $\gamma'$ are complementary, and so $|\cos \gamma| = |\sin \gamma'|$.
Note the similarity with the cross product here. In fact, we always have $\|\vec a \times \vec b\| = |\vec a' \cdot \vec b|$, where $\vec a'$ is $\vec a$ rotated by 90° in their common plane (or in any of the planes, if there are several)!
Ps. I did notice (after posting this answer) that you asked specifically about the units of the products and "not about geometric interpretations." Even so, these examples should at least show that both the dot and the cross product of two length vectors can, in fact, be meaningfully interpreted as areas, and it should therefore not be surprising that, if the original vectors have units of, say, meters, then their product will be measured in square meters.