Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
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
add comment

3 Answers

up vote 2 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

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

share|improve this answer
What a fascinating solution. –  Sunni Apr 19 '12 at 19:13
add comment

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

share|improve this answer
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
add comment

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.

share|improve this answer
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
add comment

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.