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Calculate the determinant of this matrix for $a, a_0,...,a_{n-1} \in K$ over any field $K$

$$ \begin{pmatrix} a & 0 & \cdots & 0 & -a_0 \\ -1 & \ddots & \ddots & \vdots & -a_1 \\ 0 & \ddots & \ddots & 0 & \vdots \\ \vdots & \ddots & \ddots & a & -a_{n-2} \\ 0 & \cdots & 0 & -1 & a-a_{n-1} \end{pmatrix} \in M(n \times n, K). $$

My tutor said it should be done with Laplace's formula, but I have no idea...

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  • $\begingroup$ What have you already tried? $\endgroup$ – Thomas Russell Dec 21 '13 at 19:16
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Multiply the second row by $a$, the the third by $a^2$, the fourth by $a^3$ and so on and add all this to the first row the determinant becomes: $$ \Delta=\left|\begin{array}\\ 0 & 0 & \cdots & 0 & P(a) \\ -1 & \ddots & \ddots & \vdots & -a_1 \\ 0 & \ddots & \ddots & 0 & \vdots \\ \vdots & \ddots & \ddots & a & -a_{n-2} \\ 0 & \cdots & 0 & -1 & a-a_{n-1} \end{array}\right| $$ where $$P(a)=a^n-a_{n-1}a^{n-1}-a_{n-2}a^{n-2}-\cdots-a_1a-a_0$$ now we develop relative to the first row we find: $$\Delta=(-1)^{n-1}\times (-1)^{n-1}P(a)=P(a)$$

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  • $\begingroup$ You forgot to update the rows after the first for the "new determinant" $\endgroup$ – Omnomnomnom Dec 21 '13 at 19:52
  • $\begingroup$ Have you ever seen this post? :) $\endgroup$ – mrs Dec 22 '13 at 3:16
  • $\begingroup$ Nice work, and +1 for all the formatting, too! $\endgroup$ – Namaste Dec 22 '13 at 15:57
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Looking at a few low-dimensional matrices we appear to get the following for $\mathbf{M}\in K^{n\times n}$ following your format:

$$\det(\mathbf{M})=a^{n}-\sum_{i=0}^{n}a^{i}a_{i}$$

We can prove this to be true using Laplace expansion and induction on $n$: We have our basis case $\mathbf{M}\in K^{1\times 1}$:

$$\det(\mathbf{M})=a$$

Therefore our proposition holds for $n=1$. We now assume it to be true for $n=k$ and show that it must therefore hold for $n=k+1$:

$$\det(\mathbf{M})=\begin{vmatrix} a & \cdots & -a_{0} \\ -1 & \cdots & -a_{1} \\ \vdots & \ddots & \vdots \\ 0 & \cdots & a-a_{n} \end{vmatrix}$$

Using Laplace expansion along the first row gives us:

$$\det(\mathbf{M})=a\begin{vmatrix}a & \cdots & -a_{1} \\ -1 & \cdots & -a_{2} \\ \vdots & \ddots & \vdots \\ 0 & \cdots & a-a_{n} \end{vmatrix}+(-1)^{n}a_{0}\begin{vmatrix}-1 & \cdots & 0 \\ \vdots & \ddots & \vdots \\ 0 & \cdots & -1 \end{vmatrix}=a^{n}-\sum_{i=1}^{n}a^{i}a_{i}-a_{0}$$

We get the $-a_{0}$ using the fact that the determinant of an upper triangular matrix $\mathbf{A}$ is $\prod_{i=1}^{n}A_{ii}$, which in this case is $(-1)^{n-1}$. Therefore our formula for $\det(\mathbf{M})$ holds $\forall n\in\mathbb{Z}$.

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