Different approaches to evaluate this determinant 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.
 A: 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
$$\begin{align*}
&\begin{vmatrix}a&b&b&\cdots&b\\c&d&0&\cdots&0\\c&0&d&\ddots&\vdots\\\vdots &\vdots&\ddots&\ddots&0\\c&0&\cdots&0&d\end{vmatrix}\\
&=\det\left(\mathbf I+\color{magenta}{\begin{pmatrix}b&b&\cdots&b\\1&&&\end{pmatrix}}\color{red}{\begin{pmatrix}\frac1{a-b-c}&&&\\&\frac1d&&\\&&\ddots&\\&&&\frac1d\end{pmatrix}}\color{blue}{\begin{pmatrix}1&c\\&c\\&\vdots\\&c\end{pmatrix}}\right)\color{red}{(a-b-c)d^{n-1}}\\
&=\det\left(\mathbf I+\color{magenta}{\begin{pmatrix}b&b&\cdots&b\\1&&&\end{pmatrix}}\color{orange}{\begin{pmatrix}\frac1{a-b-c}&\frac{c}{a-b-c}\\&\frac{c}{d}\\&\vdots\\&\frac{c}{d}\end{pmatrix}}\right)\color{red}{(a-b-c)d^{n-1}}\\
&=\det\left(\mathbf I+\color{green}{\begin{pmatrix}\frac{b}{a-b-c}&bc\left(\frac1{a-b-c}+\frac{n-1}{d}\right)\\\frac1{a-b-c}&\frac{c}{a-b-c}\end{pmatrix}}\right)\color{red}{(a-b-c)d^{n-1}}\\
&=\frac{bc(1-n)+ad}{d(a-b-c)}(a-b-c)d^{n-1}\\
&=(bc(1-n)+ad)d^{n-2}
\end{align*}$$
A: If the dimension of the matrix is $2$, it's only $ad-bc$. If it's $\geq 3$, and $d=0$ the determinant is $0$. If $d\neq 0$, do $C_1\leftarrow C_1-\frac cdC_j$, $2\leq j\leq n$. The first column becomes $\pmatrix{a-(n-1)\frac{bc}d\\\ 0\\\ \vdots\\\ 0}$, and the determinant is $d^{n-1}\left(a-(n-1)\frac{bc}d\right)=d^{n-2}(ad-(n-1)bc)$. 
A: Let $A_n=(a_{ij})$ be the $n\times n$ matrix with
$$
a_{ij}=\left\{\matrix{a&i=j=1\\b&i=1\ne j\\c&i\ne1=j\\d&i=j\ne1\\0&\text{otherwise}}\right.
$$
and $\Delta_n=\det A_n$. Then ($\Delta_1=a$ according to my definition above, which may differ from your implicit definition), $\Delta_2=ad-bc$ and for $n\ge2$, 
$$
\Delta_{n+1}=d\,\Delta_n-bc\,d^{n-1}=\left(\Delta_n-bcd^{n-2}\right)d
$$
expanding on the bottom row (or equivalently the rightmost column, but with the matrix below transposed and with $c$ in stead of $b$).
This is because the $n\times n$ below has determinant
$$\det\begin{bmatrix}
b&b &\cdots&b&b\\
d &0&\cdots&0&0\\
0&d&\ddots&\vdots&0\\
\vdots&\ddots&\ddots&0&0\\
0&\cdots&0&d&0
\end{bmatrix}
= (-1)^{n-1}bd^{n-1}\,,
$$
and is multiplied by $(-1)^nc$ from $a_{n+1,1}=c$,
giving the second term above. Thus
$$
\eqalign{
\frac{\Delta_{n+1}}{d^{n-1}}&=
\frac{\Delta_{n}  }{d^{n-2}}-bc
\quad
\text{and}
\quad
\frac{\Delta_1}{d^{-1}}=ad
\quad
\implies
\\\\
\frac{\Delta_{n}}{d^{n-2}}&=
ad-(n-1)bc
\qquad
\implies
\\\\
\Delta_{n}&=\big[ad-(n-1)bc\big]d^{n-2}
}
$$
for $n\ge1$ by induction.
