TITLE: Show that if $a\neq b$, then we have for the $n\times n$-matrix $\textrm{det}=\frac{a^{n+1}-b^{n+1}}{a-b}$.
QUESTION [6 upvotes]: The Problem
Show that if $a\neq b$, then we have for the $n\times n$-matrix
$$\textrm{det}\begin{pmatrix} a+b & ab & 0 & \ldots & 0 & 0 \\
1 & a+b & ab & \ldots & 0 & 0 \\
0 & 1 & a+b & \ldots & 0 & 0 \\
\ldots & \ldots & \ldots & \ldots & \ldots & \ldots \\
0 & 0 & 0 & \ldots & a+b & ab \\
0 & 0 & 0 & \ldots & 1 & a+b\end{pmatrix} =\frac{a^{n+1}-b^{n+1}}{a-b}$$
What if $a=b$?

My Questions
I am not entirely sure how to begin this proof.  I suspect I am missing something that makes it quite simple.  I tried looking for similar problems, and I noticed a common theme being the use of row operations to rewrite the matrix, from which the determinant was found.  However, I still didn't understand many of the intermediate computations when it came to actually finding the determinant.  My questions are as follows.

Should I use row operations to rewrite this matrix?  If so, what would be an example of how that would look computationally?
In either case, how should I go about actually computing the determinant to show the statement is true?  Is there an algorithm that is helpful here?
I noticed for the follow up question that, if $a=b$, then $$\textrm{det}\begin{pmatrix} a+a & aa & 0 & \ldots & 0 & 0 \\
1 & a+a & aa & \ldots & 0 & 0 \\
0 & 1 & a+a & \ldots & 0 & 0 \\
\ldots & \ldots & \ldots & \ldots & \ldots & \ldots \\
0 & 0 & 0 & \ldots & a+a & aa \\
0 & 0 & 0 & \ldots & 1 & a+a\end{pmatrix}.$$  Am I right in my thinking there?


Other Details
The book used in the course is Abstract Linear Algebra by Curtis.  It has been of little help to me here...

REPLY [3 votes]: Let $L=\pmatrix{1\\ 1&\ddots\\ &\ddots&\ddots\\ &&1&1}$ and treat $a$ and $b$ as two indeterminates. Then the matrix in question is equal to $A=D^{-1}(aL+bL^T)D$, where $D=\operatorname{diag}(1,a,a^2,\ldots,a^{n-1})$. Since $\det L=1$, we get
\begin{align}
\det A&=\det(aL+bL^T)
=\det\left(aI+bL^TL^{-1}\right)\\
&=(-b)^n\det\left(-\frac abI-L^TL^{-1}\right)
=(-b)^nf\left(-\frac ab\right),\tag{1}
\end{align}
where $f(x)$ is the characteristic polynomial of $L^TL^{-1}$. Now, if you calculate $L^TL^{-1}$ directly, you will see that it is the transpose of a companion matrix. Hence you may read off the coefficients of $f$ directly from the entries of $L^TL^{-1}$. Then, using $(1)$, you should find that
$$
\det A=(-b)^nf(-a/b)=a^n+a^{n-1}b+\cdots+ab^{n-1}+b^n.\tag{2}
$$
Now the rest are straightforward when the indeterminates $a,b$ are specialised to two numbers.