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The outward flux of the Newtonian gravitational field through a closed surface is $-4\pi G$ times the enclosed mass. A sphere around a point mass establishes the coefficient; superposition extends the integral law to general mass distributions. The minus sign expresses attractive gravity. A symmetric Gaussian surface can turn the flux relation into a local field calculation, but the integral law alone does not imply that the field is uniform on an arbitrary surface.

For three mutually perpendicular axes through one point of a nonnegative mass distribution, $I_x+I_y-I_z=2\int z^{2}dm\ge 0$, with cyclic counterparts. Equality holds precisely when the mass distribution is supported in the plane $z=0$, giving the perpendicular axis theorem. The point need not be the centre of mass, and the axes need not be principal axes.

The principal moments of inertia are the eigenvalues of the symmetric inertia tensor. In units of $\mathrm{kg}{\mathrm{m}}^2$, its characteristic determinant isTherefore the requested algebraic answer isAn orthonormal basis of principal axes is $(1,1,0)/\sqrt{2}$, $(1,-1,0)/\sqrt{2}$, and $(0,0,1)$; the last two span the degenerate eigenspace and may be replaced by any orthonormal basis of it.

There is a physical inconsistency in calling the supplied matrix a rigid body's inertia tensor. The moment-of-inertia triangle inequalities require each principal moment to be no greater than the sum of the other two. Here $3>1+1$. In principal coordinates, $I_2+I_3-I_1=2\int x_1^{2}dm$ cannot be negative for a nonnegative mass distribution. Thus the eigenvalue calculation is well defined, but no ordinary rigid mass distribution realizes this tensor. The positive second-moment matrix test for an inertia tensor makes the same obstruction explicit.

Choose three mutually perpendicular axes through the same point, and label coordinates $x,y,z$. The moments of inertia of any rigid body areSubtracting givesFor a thin planar lamina lying in $z=0$, the right-hand side vanishes, proving the perpendicular axis theorem:The two in-plane axes and the perpendicular axis must intersect at one point; this need not be the centre of mass. For an extended three-dimensional mass distribution, the equality generally fails. The general moment-of-inertia triangle inequalities arewith equality in the first precisely when all mass lies in the plane perpendicular to its $z$ axis, and cyclically for the others. None of these identities requires principal axes.

A planar lamina is an ideal mass distribution confined to a plane, with surface mass density $\sigma$ and infinitesimal mass $dm=\sigma dA$. Its moment of inertia about an axis is the integral of squared perpendicular distance against this mass element. Three mutually perpendicular axes through one point, with two in the lamina's plane, obey the perpendicular axis theorem even when the density is nonuniform.

For a nonnegative mass distribution relative to an origin, the second-moment matrix is $\mathsf{S}=\int \mathbf{r}{\mathbf{r}}^{\mathsf{T}}dm$ and the inertia tensor is $\mathsf{I}=\mathrm{tr}(\mathsf{S})1-\mathsf{S}$. Taking the trace gives the displayed inverse formula, which must be positive semidefinite. Conversely, any such $\mathsf{S}$ can be realized by pairs of point masses along its orthogonal principal directions. Thus this is also a realizability test when total mass and dimensions are unrestricted. In principal coordinates it is equivalent to the moment-of-inertia triangle inequalities: $I_2+I_3-I_1=2S_{11}\ge 0$, with cyclic counterparts. A symmetric positive definite matrix alone need not be a physical inertia tensor.