<s>[INST] Upper triangular determinants Given that
$$A=
\begin{bmatrix}
1&211&311&327&337\\
0&2&427&438&491\\
0&0&3&547&551\\
0&0&0&2&672\\
0&0&0&0&3
\end{bmatrix}
$$
Find $|2A^{-1}+I_5|$. The solution is $\frac{100}{3}$ but I can't figure out how to do it. Any hints? My best guess is to use properties of the determinants of upper triangular matrices, but then I got lost in the algebra...
 [/INST] Hints:

*

*Eigenvalues of an upper triangular matrix are precisely the entries of the main diagonal.


*If $A$ is invertible then $\lambda$ is an eigenvalue of $A$ iff $\frac{1}{\lambda}$ is an eigenvalue of $A^{-1}$


*Eigenvalues are polynomial invariant i.e $\lambda$ is an eigenvalue of $A$ implies $p(\lambda) $ is an eigenvalue of $p(A) $ where $p(x) \in K[x]$


*$\det A$ is the product of all eigenvalues taking care of multiplicity.


Eigenvalues of $A$ are $1, 2,3,2,3$.


Hence eigenvalues of $A^{-1}$ are $1, \frac{1}{2}, \frac{1}{3}, \frac{1}{2},  \frac{1}{3}$


 Now eigenvalues of $2A^{-1}+I_5$ are $2\cdot 1+1, 2\cdot\frac{1}{2}+1, 2\cdot \frac{1}{3}+1, 2\cdot\frac{1}{2}+1, 2\cdot\frac{1}{3}+1$


 $\begin{align}&|2A^{-1}+I_5|\\&=3\cdot 2\cdot \frac{5}{3}\cdot 2\cdot \frac{5}{3}\\&=\frac{100}{3}\end{align}$

</s>