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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...

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

2 Answers 2

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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 $$ |2A^{-1} + I| = |(2I + A)A^{-1}| = |2I + A| \cdot |A^{-1}| = \frac{|2I + A|}{|A|}. $$

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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 $2\cdot 1+1, 2\cdot\frac{1}{2}+1, 2\cdot \frac{1}{3}+1, 2\cdot\frac{1}{2}+1, 2\cdot\frac{1}{3}+1$

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

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