# Prove $\det(A+I)=1$

Need help with my homework.

$A \in M_{nxn}(\mathbb{R})$ is upper-triangular and $A^{n}=0$

Please hint how to prove, that $\det(A+I)=1$

I dont know how it do, know laplace equation

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Do you know how to calculate the determinant of an upper triangular matrix? What is the $i$-th diagonal entry of $T^p$ for a triangular matrix $T$? What can you conclude about the diagonal entries of $A$? –  Olivier Bégassat Feb 1 '13 at 0:32

1) Show that for an upper triangular matrix $A$, if one entry on the main diagonal is non-zero then $A^n$ is never the zero matrix (hint: split the matrix as $D+T$ where $D$ is diagonal and $T$ is strictly upper triangular).
2) Conclude that for the given $A$ the main diagonal consists of zeros only.
3) What form does $A+I$ have then? What is the determinant?
I dont understand what mean that the $A^n=0$ –  aiki93 Feb 1 '13 at 0:51
You wrote $A^n=0$ in your own question. Do you not know what that means? –  Ittay Weiss Feb 1 '13 at 0:58
yea, I dont know what mean the $A^n = 0$ in this exercise –  aiki93 Feb 1 '13 at 1:02