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Let $a_{1}, \cdots, a_{n}$ and $b$ be real numbers. I like to know the determinant of the matrix $$\det\begin{pmatrix} a_{1}+b & b & \cdots & b \\ b & a_{2}+b & \cdots & b \\ \vdots & \vdots & \ddots & \vdots \\ b & b & \cdots & a_{n}+b \end{pmatrix}=?$$ I guess the answer is $$a_{1} \cdots a_{n}+ \sum_{i=1}^{n} a_{1} \cdots a_{i-1} b a_{i+1}\cdots a_{n} $$ after some direct calculations for $n=2,3$. The question is how to calculate it for general $n$. Thanks!

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    $\begingroup$ Maybe induction? $\endgroup$
    – Arpan
    Apr 24, 2015 at 7:15
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    $\begingroup$ Subtract the first row from each of the others, looks like you get a matrix that should be easier to deal with as it's mostly zeros. $\endgroup$ Apr 24, 2015 at 7:22

3 Answers 3

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Use the formula $\det(I+AB)=\det(I+BA)$.

First suppose $\Lambda = \operatorname{diag}(a_1,...,a_n)$ is invertible (that is, all the $a_k$ are non zero).

Then the matrix above can be written as $b e e^T + \Lambda = \Lambda (b \Lambda^{-1} e e^T + I)$, where $e = (1,...1)^T$. Then $\delta = c = \det \Lambda \det (\Lambda^{-1} e e^T + I) = \det \Lambda (1+b e^T \Lambda^{-1} e) $.

Expanding gives $\delta = (\prod_k a_k) (1 + b \sum_k {1 \over a_k})= \prod_k a_k + b \sum_k \prod_{i \neq k} a_i$.

Continuity shows that the formula holds for all $a_n$.

Addendum:

The functions $f_1(a) = \det(b e e^T + \operatorname{diag}(a_1,...,a_n))$, $f_2(a) = \prod_k a_k + b \sum_k \prod_{i \neq k} a_i$ are both continuous and defined for all $a$. We have shown above that $f_1(a) = f_2(a)$ for all $a$ such that no component of $a$ is zero.

Suppose $a$ is arbitrary, then there are $a_k \to a$ such that all components of $a_k$ are non zero. Continuity shows that $f_1(a) = f_2(a)$.

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  • $\begingroup$ "Continuity"? Maybe you meant "induction"? $\endgroup$
    – Timbuc
    Apr 24, 2015 at 7:40
  • $\begingroup$ No, I meant continuity. There is no explicit induction in my answer. $\endgroup$
    – copper.hat
    Apr 24, 2015 at 7:40
  • $\begingroup$ Ok...continuity of what in what topological space? Thanks. And there's still "hidden" induction. $\endgroup$
    – Timbuc
    Apr 24, 2015 at 7:41
  • $\begingroup$ Thanks. The condition of a_k non-zero is ok to me. $\endgroup$
    – user223515
    Apr 24, 2015 at 7:44
  • $\begingroup$ @Timbuc: $\mathbb{R}^n$ using the usual topology. If you accept the formula $\det(I+AB) = \det(I+BA)$ there is no induction. $\endgroup$
    – copper.hat
    Apr 24, 2015 at 7:46
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My solution is less elegant, but only involves basic properties of the determinant. Since the determinant is invariant under column operations, you have

$$\Delta := \det\begin{pmatrix} a_{1}+b & b & \cdots & b \\ b & a_{2}+b & \cdots & b \\ \vdots & \vdots & \ddots & \vdots \\ b & b & \cdots & a_{n}+b \end{pmatrix}= \det\begin{pmatrix} a_{1} & 0 & 0 & \cdots & b \\ -a_2 & a_{2} & 0 & \cdots & b \\ 0 & -a_{3} & a_3 & \cdots & b \\ \vdots & \vdots &\vdots & \ddots & \vdots \\ 0 & 0 & 0 &\cdots & a_{n}+b \end{pmatrix},$$

by subtracting the second column from the first, the third column from the second, ...

Using Laplace expansion of the last column you get

$$ \Delta = \sum_{k=1}^{n-1} \left[(-1)^{n+k}b\right] \left[\prod\limits_{i=1}^{k-1} a_i \right] \left[\prod\limits_{i=k+1}^n (-1)a_i \right] + (a_n+b) \left[\prod\limits_{i=1}^{n-1} a_i \right]. $$

Simplifying the first summand yields

$$ \Delta = \sum_{k=1}^{n-1} \left[(-1)^{n+k}(-1)^{n-k}b\right] \prod\limits_{\substack{i=1\\i\neq k}}^{n} a_i + (a_n+b) \prod\limits_{i=1}^{n-1} a_i, $$

Which finally gives the desired formula

$$ \Delta = b \sum_{k=1}^{n} \prod\limits_{\substack{i=1\\i\neq k}}^{n} a_i + \prod\limits_{i=1}^{n} a_i. $$

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Applying multilinearity and an inductive argument:

$$\begin{vmatrix} a_{1}+b & b & \cdots & b \\ b & a_{2}+b & \cdots & b \\ \vdots & \vdots & \ddots & \vdots \\ b & b & \cdots & a_{n}+b \end{vmatrix}=\begin{vmatrix} a_{1} & b & \cdots & b \\ 0 & a_{2}+b & \cdots & b \\ \vdots & \vdots & \ddots & \vdots \\ 0 & b & \cdots & a_{n}+b \end{vmatrix}+\begin{vmatrix} b & b & \cdots & b \\ b & a_{2}+b & \cdots & b \\ \vdots & \vdots & \ddots & \vdots \\ b & b & \cdots & a_{n}+b \end{vmatrix}$$

$$=\begin{vmatrix} a_{1} & 0 & \cdots & b \\ 0 & a_{2} & \cdots & b \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \cdots & a_{n}+b \end{vmatrix}+ \begin{vmatrix} a_{1} & b & \cdots & b \\ b & b & \cdots & b \\ \vdots & \vdots & \ddots & \vdots \\ b & b & \cdots & a_{n}+b \end{vmatrix}+ \begin{vmatrix} b & 0 & \cdots & b \\ b & a_{2} & \cdots & b \\ \vdots & \vdots & \ddots & \vdots \\ b & 0 & \cdots & a_{n}+b \end{vmatrix}+ \begin{vmatrix} b & b & \cdots & b \\ b & b & \cdots & b \\ \vdots & \vdots & \ddots & \vdots \\ b & b & \cdots & a_{n}+b \end{vmatrix}=\ldots$$

Note the very last determinant is zero.

For example:

$$\begin{vmatrix}a_1+b&b&b\\ b&a_2+b&b\\ b&b&a_3+b\end{vmatrix}= \begin{vmatrix}a_1&b&b\\ 0&a_2+b&b\\ 0&b&a_3+b\end{vmatrix}+ \begin{vmatrix}b&b&b\\ b&a_2+b&b\\ b&b&a_3+b\end{vmatrix}=$$

$$=\begin{vmatrix}a_1&0&b\\ 0&a_2&b\\ 0&0&a_3+b\end{vmatrix}+ \begin{vmatrix}a_1&b&b\\ 0&b&b\\ 0&b&a_3+b\end{vmatrix}+ \begin{vmatrix}b&0&b\\ b&a_2&b\\ b&0&a_3+b\end{vmatrix}+ \begin{vmatrix}b&b&b\\ b&b&b\\ b&b&a_3+b\end{vmatrix}=\ldots$$

$$=a_1a_2a_3+b(a_1a_2+a_1a_3+a_2a_3)$$

and you were right.

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