= Solution
In Cartesian coordinates a rank-$r$ <tensor> is an array whose components transform under a change of orthonormal axes by one <rotation matrix> factor per index:
$$
T'_{i_1\cdots i_r}=Q_{i_1j_1}\cdots Q_{i_rj_r}T_{j_1\cdots j_r}.
$$
An <isotropic tensor> has unchanged components under every proper <rotation>. The <Levi-Civita symbol> obeys
$$
Q_{ia}Q_{jb}Q_{kc}\epsilon_{abc}=(\det Q)\epsilon_{ijk}=\epsilon_{ijk}
$$
for $Q\in SO(3)$, proving <rotation invariance of the Levi-Civita symbol>. Under an orthogonal transformation of determinant $-1$ its sign reverses; including such transformations requires the <pseudotensor> distinction. Here isotropy is with respect to proper rotations.
Contract the determinant identity with $\epsilon_{ijk}$. Since $\epsilon_{ijk}\epsilon_{ijk}=3!=6$, one obtains
$$
\boxed{\det A=\frac16\epsilon_{ijk}\epsilon_{lmn}A_{il}A_{jm}A_{kn}.}
$$
Every index on the right is contracted, so the <tensor component transformation law> leaves no free rotation factor: the <determinant> is a <scalar>. Equivalently, $A'=QAQ^T$ gives $\det A'=(\det Q)^2\det A=\det A$.
For the <matrix inverse>, define
$$
C_{ab}=\frac12\epsilon_{ajk}\epsilon_{bmn}A_{jm}A_{kn}.
$$
This is the <cofactor matrix>. To verify the inverse formula rather than just recognize it, contract with $A_{ib}$ and use the determinant identity again:
$$
A_{ib}C_{ab}=\frac12\epsilon_{ajk}(\det A)\epsilon_{ijk}=(\det A)\delta_{ia}.
$$
Here the <contraction of two Levi-Civita symbols> gives $\epsilon_{ajk}\epsilon_{ijk}=2\delta_{ai}$. Therefore $AC^T=(\det A)I$ and, when $\det A\ne0$,
$$
\boxed{(A^{-1})_{ba}=\frac{C_{ab}}{\det A}=\frac1{2\det A}\epsilon_{ajk}\epsilon_{bmn}A_{jm}A_{kn}.}
$$
These are the <Levi-Civita formulas for a three-dimensional determinant and inverse>.
Now differentiate the cubic determinant formula with respect to the individual entry $A_{ab}$. The three differentiated factors give equal contributions: exchanging two paired indices changes the sign of both epsilon symbols, leaving their product unchanged. Thus
$$
\frac{\partial\det A}{\partial A_{ab}}=\frac36\epsilon_{ajk}\epsilon_{bmn}A_{jm}A_{kn}=C_{ab}.
$$
Substituting the inverse expression proves the entrywise <derivative of the determinant>:
$$
\boxed{\frac{\partial\det A}{\partial A_{ab}}=(\det A)(A^{-1})_{ba}=(\det A)(A^{-1})^T_{ab}.}
$$
The cofactor version remains valid for singular matrices; only the inverse version requires invertibility.
For the time-dependent <Jacobian matrix>, the stated smooth expansion implies
$$
A_{ij}(t,x)=\delta_{ij}+t\partial_i u_j(x)+O(t^2),\qquad A(0,x)=I.
$$
Applying the <chain rule> and the entrywise determinant derivative gives <Jacobi's formula> near $t=0$:
$$
\frac d{dt}\det A=\sum_{ab}C_{ab}\dot A_{ab}=(\det A)\operatorname{tr}(A^{-1}\dot A).
$$
At zero, $C=I$, so $\left.\partial_t\det A\right|_0=\sum_a\partial_a u_a=\nabla\cdot u$. A <Taylor expansion> therefore yields
$$
\boxed{\det A(t,x)=1+t\nabla\cdot u(x)+O(t^2)=1+O(t^2)\quad\text{if }\nabla\cdot u=0.}
$$
The <Jacobian determinant> measures local volume change, so this establishes volume preservation to first order. An arbitrary family specified only through its first-order expansion need not preserve volume exactly at higher orders.
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