# 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...

• $2A^{-1}+I_5$ is upper triangular. The off diagonal terms of $A$ are irrelevant. Aug 10 at 13:13
• Eigenvalues of triangular matrices are on the diagonal Aug 10 at 13:14

As one comment notes, the fact that the $$A^{-1}$$ is upper triangular should be sufficient to figure things out.

Alternatively, here's an approach that doesn't require this knowledge or a familiarity with the idea of eigenvalues; it only requires that you know how to find the determinant of an upper triangular matrix. We have $$|2A^{-1} + I| = |(2I + A)A^{-1}| = |2I + A| \cdot |A^{-1}| = \frac{|2I + A|}{|A|}.$$

Hints:

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

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

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

4. $$\det A$$ 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}