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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.

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Have you tried induction? – bgins Apr 1 '12 at 15:30
Is there a name for such matrix?! – user2468 Apr 1 '12 at 18:37
@J.D. Yes. See my answer. – J. M. Apr 14 '12 at 10:57
up vote 3 down vote accepted

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

$$\begin{align*} &\begin{vmatrix}a&b&b&\cdots&b\\c&d&0&\cdots&0\\c&0&d&\ddots&\vdots\\\vdots &\vdots&\ddots&\ddots&0\\c&0&\cdots&0&d\end{vmatrix}\\ &=\det\left(\mathbf I+\color{magenta}{\begin{pmatrix}b&b&\cdots&b\\1&&&\end{pmatrix}}\color{red}{\begin{pmatrix}\frac1{a-b-c}&&&\\&\frac1d&&\\&&\ddots&\\&&&\frac1d\end{pmatrix}}\color{blue}{\begin{pmatrix}1&c\\&c\\&\vdots\\&c\end{pmatrix}}\right)\color{red}{(a-b-c)d^{n-1}}\\ &=\det\left(\mathbf I+\color{magenta}{\begin{pmatrix}b&b&\cdots&b\\1&&&\end{pmatrix}}\color{orange}{\begin{pmatrix}\frac1{a-b-c}&\frac{c}{a-b-c}\\&\frac{c}{d}\\&\vdots\\&\frac{c}{d}\end{pmatrix}}\right)\color{red}{(a-b-c)d^{n-1}}\\ &=\det\left(\mathbf I+\color{green}{\begin{pmatrix}\frac{b}{a-b-c}&bc\left(\frac1{a-b-c}+\frac{n-1}{d}\right)\\\frac1{a-b-c}&\frac{c}{a-b-c}\end{pmatrix}}\right)\color{red}{(a-b-c)d^{n-1}}\\ &=\frac{bc(1-n)+ad}{d(a-b-c)}(a-b-c)d^{n-1}\\ &=(bc(1-n)+ad)d^{n-2} \end{align*}$$

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What a fascinating solution. – Sunni Apr 19 '12 at 19:13

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 $d^{n-1}\left(a-(n-1)\frac{bc}d\right)=d^{n-2}(ad-(n-1)bc)$.

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Thanks, bgins's solution works also for general lambda matrices....I am looking for other approaches. – Sunni Apr 1 '12 at 16:46
You can see the matrix as a block matrix, and use the formula, if $a\neq 0$, $\det\pmatrix{a&b\mathbf 1^T\\\ c\mathbf 1&dI_{n-1}}=a\det(dI_{n-1}-\frac{bc}O)$, where $O$ is the matrix which has only ones. – Davide Giraudo Apr 1 '12 at 19:29

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.

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Thanks, but there is a small error in your calculation. ... – Sunni Apr 1 '12 at 16:45
Thanks @Sunni, I think it's fixed now. – bgins Apr 1 '12 at 18:24

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