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Calculate the determinant

$$\begin{align*}D[n]=\begin{array}{cccccc} b & b & b & \dots & b & a \\ b & b & b & \dots & a & b \\ \vdots & \vdots & \vdots & & \vdots & \vdots \\ b & b & a & b & b & b \\ b & a & b & b & b & b \\ a & b & b & b & b & b \\\end{array}\end{align*}$$

$$\begin{align*}\left(\begin{array}{ccccc} b & b & b & b & a \\ b & b & b & a & b \\ b & b & a & b & b \\ b & a & b & b & b \\ a & b & b & b & b \\\end{array}\right)\end{align*}$$

I calculated the value of $n=2$, and $n=3$, $(a-b)^2(a+2 b)$

I guess the result is $(a-b)^{n-1}(a+(n-1)b)(-1)^{n-1}$

So, I need a prove by induction? how to do that?

I'm not sure whether this is easier to do than think out one new method.

Community
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SEuser2013
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  • When you write $\backslash$ begin{array}{cccccc} b & b & b & $\backslash$text{...} & b & a $\backslash$ end{array} it looks like this: $\displaystyle\begin{array}{cccccc} b & b & b & \text{...} & b & a \end{array}$. But when you write $\backslash$ begin{array}{cccccc} b & b & b & \dots & b & a $\backslash$ end{array}, it looks like this: $\displaystyle\begin{array}{cccccc} b & b & b & \dots & b & a \end{array}$. You'd get better results with \text{ . . . } (with spaces between the dots.) (I changed it to \dots in some places and \vdots in others.) – Michael Hardy Aug 17 '13 at 17:51

1 Answers1

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You may swap the rows to bring the $a$s to the main diagonal (what effect does that have on the sign?). Then replace $a$ with $a-X$ to obtain the characteristic polynomial $\chi_M(X)$ of your matrix while we are looking for $\chi_M(0)$. Its roots are the eigenvalues with corresponding multiplicity. Note that $(1\;1\;\ldots\;1)$ is an obvious eigenvector of eigenvalue $a+(n-1)b$ and the $(n-1)$-dimensional space orthogonal to it consists of eigenvectors of eigenvalue $a-b$. Hence your conjecture follows.

Hagen von Eitzen
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  • Hi, how to decide the sign of swapping the rows of a n-determinant? I know the sign sequence is $-1,-1,1,1,-1,-1,...$, how to express as a function of n something like $(-1)^n$? – SEuser2013 Sep 02 '13 at 13:57