= Solution
The <principal moments of inertia> are the <eigenvalues> of the symmetric <inertia tensor>. In units of $\mathrm{kg\,m^2}$, its characteristic <determinant> is
$$
\det(\mathsf I-\lambda\mathsf 1)
=(1-\lambda)[(2-\lambda)^2-1]
=(1-\lambda)^2(3-\lambda).
$$
Therefore the requested algebraic answer is
$$
\boxed{I_1=3\ \mathrm{kg\,m^2},\qquad I_2=I_3=1\ \mathrm{kg\,m^2}.}
$$
An <orthonormal basis> of <principal axes> is $(1,1,0)/\sqrt2$, $(1,-1,0)/\sqrt2$, 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.
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