= Solution
A Cartesian <tensor> of order $n$ is a multilinear geometric object whose components under an orthogonal change of axes obey the <tensor component transformation law>,
$$
T'_{i_1\cdots i_n}=Q_{i_1j_1}\cdots Q_{i_nj_n}T_{j_1\cdots j_n},\qquad Q^TQ=I.
$$
An <isotropic tensor> has the same components under every proper rotation. For the <Kronecker delta>,
$$
\delta'_{ij}=Q_{ip}Q_{jq}\delta_{pq}=Q_{ip}Q_{jp}=\delta_{ij}.
$$
Thus \b[$\delta_{ij}$ is isotropic], even under improper orthogonal transformations.
To prove <multiplicativity of the determinant> from the given formula, first consider $C_{pqr}=\epsilon_{ijk}A_{ip}A_{jq}A_{kr}$. Exchanging any two of $p,q,r$ changes its sign, so it is proportional to the <Levi-Civita symbol>. Its $123$ component is the determinant, as follows by expanding the given double-epsilon expression. Therefore
$$
\epsilon_{ijk}A_{ip}A_{jq}A_{kr}=(\det A)\epsilon_{pqr}.
$$
Substitute $(AB)_{il}=A_{ip}B_{pl}$ into that same determinant formula:
$$
\begin{aligned}
\det(AB)&=\frac16\epsilon_{ijk}\epsilon_{lmn}A_{ip}B_{pl}A_{jq}B_{qm}A_{kr}B_{rn}\\
&=\frac{\det A}{6}\epsilon_{pqr}\epsilon_{lmn}B_{pl}B_{qm}B_{rn}=(\det A)(\det B).
\end{aligned}
$$
Hence
$$
\boxed{\det(AB)=\det A\det B.}
$$
No invertibility assumption was used, so this identity includes singular matrices.
Now differentiate the double-epsilon formula for a differentiable $A(t)$. Its three differentiated terms are equal after renaming the paired dummy indices: the simultaneous swaps in both epsilon symbols cancel their signs. Thus
$$
\frac d{dt}\det A=\frac12\epsilon_{ijk}\epsilon_{lmn}\dot A_{il}A_{jm}A_{kn}=C_{il}\dot A_{il},
$$
where $C_{il}$ is the cofactor of entry $A_{il}$. For an invertible matrix, the <adjugate matrix> identity gives $C_{il}=(\det A)(A^{-1})_{li}$. Consequently the <Jacobi determinant derivative formula> is
$$
\boxed{\frac d{dt}\det A=(\det A)\sum_{i,l}(A^{-1})_{li}\dot A_{il}=(\det A)\operatorname{tr}(A^{-1}\dot A).}
$$
The order of the factors inside the trace follows from the reversed cofactor indices; no commutativity of $A$ and $\dot A$ is assumed. The cofactor version is also <Jacobi's formula> and remains valid when $A$ is singular, although the inverse version then cannot be used.
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