TITLE: Different approaches to evaluate this determinant
QUESTION [5 upvotes]: How to evaluate this determinant $$\det\begin{bmatrix}
a& b&b &\cdots&b\\ c  &d &0&\cdots&0\\c&0&d&\ddots&\vdots\\\vdots &\vdots&\ddots&\ddots& 0\\c&0&\cdots&0&d
\end{bmatrix}?$$
I am looking for the different approaches.

REPLY [9 votes]: Your (upper) arrowhead matrix can be decomposed as follows:
$$\begin{pmatrix}a&b&b&\cdots&b\\c&d&0&\cdots&0\\c&0&d&\ddots&\vdots\\\vdots &\vdots&\ddots&\ddots&0\\c&0&\cdots&0&d\end{pmatrix}=\color{red}{\begin{pmatrix}a-b-c&&&&\\&d&&&\\&&d&&\\&&&\ddots&\\&&&&d\end{pmatrix}}+\color{blue}{\begin{pmatrix}1&c\\&c\\&c\\&\vdots\\&c\end{pmatrix}}\cdot\color{magenta}{\begin{pmatrix}b&b&b&\cdots&b\\1&&&&\end{pmatrix}}$$
Now, one can then use the Sherman-Morrison-Woodbury formula for determinants:
$$\det(\color{red}{\mathbf A}+\mathbf{\color{blue}{U}\color{magenta}{V^\top}}) = \det(\mathbf I + \color{magenta}{\mathbf V^\top}\color{red}{\mathbf A}^{-1}\color{blue}{\mathbf U})\det(\color{red}{\mathbf A})$$
to yield
$$\begin{align*}
&\begin{vmatrix}a&b&b&\cdots&b\\c&d&0&\cdots&0\\c&0&d&\ddots&\vdots\\\vdots &\vdots&\ddots&\ddots&0\\c&0&\cdots&0&d\end{vmatrix}\\
&=\det\left(\mathbf I+\color{magenta}{\begin{pmatrix}b&b&\cdots&b\\1&&&\end{pmatrix}}\color{red}{\begin{pmatrix}\frac1{a-b-c}&&&\\&\frac1d&&\\&&\ddots&\\&&&\frac1d\end{pmatrix}}\color{blue}{\begin{pmatrix}1&c\\&c\\&\vdots\\&c\end{pmatrix}}\right)\color{red}{(a-b-c)d^{n-1}}\\
&=\det\left(\mathbf I+\color{magenta}{\begin{pmatrix}b&b&\cdots&b\\1&&&\end{pmatrix}}\color{orange}{\begin{pmatrix}\frac1{a-b-c}&\frac{c}{a-b-c}\\&\frac{c}{d}\\&\vdots\\&\frac{c}{d}\end{pmatrix}}\right)\color{red}{(a-b-c)d^{n-1}}\\
&=\det\left(\mathbf I+\color{green}{\begin{pmatrix}\frac{b}{a-b-c}&bc\left(\frac1{a-b-c}+\frac{n-1}{d}\right)\\\frac1{a-b-c}&\frac{c}{a-b-c}\end{pmatrix}}\right)\color{red}{(a-b-c)d^{n-1}}\\
&=\frac{bc(1-n)+ad}{d(a-b-c)}(a-b-c)d^{n-1}\\
&=(bc(1-n)+ad)d^{n-2}
\end{align*}$$