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For which $a,b \in\Bbb R$ does there exist the inverse of $A$: $$A=\begin{pmatrix} a&b& b &\ldots &b\\ b& a &b &\ldots&b\\ b& b & a &\ldots &b\\ \vdots&\vdots&\vdots& \ddots&\vdots\\ b&b& b&\ldots & a \end{pmatrix}\in M_n$$

So we want to see for which $a,b$ will the $\det(A)\neq 0$ .

$$\begin{vmatrix} a&b &b &\ldots &b\\ b& a &b &\ldots&b\\ b& b & a &\ldots &b\\ \vdots&\vdots&\vdots& \ddots&\vdots\\ b&b& b&\ldots & a \end{vmatrix}=\begin{vmatrix}a &b &b &\ldots &b\\ b-a& a-b &0 &\ldots&0\\ b-a& 0 & a-b &\ldots &0\\ \vdots&\vdots&\vdots& \ddots&\vdots\\ b-a&0& 0&\ldots & a-b \end{vmatrix}$$

This is almost a lower triangular matrix, only there should be zeros instead of $b$s in the 1st row. So if I put $b=0$, then I get a lower triangular matrix. The main diagonal will only have $a$s on it, so then $a\neq 0$? Is that it?

lmc
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    You could calculate the determinant by developing it w.r.t. the last column. Notice that by doing so the factor $(b-a)$ appears. So surely we need that $b\neq a$ which is obvious by just looking at the matrix. – Mathematician 42 Jan 22 '17 at 09:57
  • @Mathematician42 I don't understand what you mean by developing it w.r.t the last column, can you explain? – lmc Jan 22 '17 at 12:27
  • There is this recursive way of defining determinant, I meant using that formula w.r.t. the last column of your second matrix. – Mathematician 42 Jan 22 '17 at 13:16

2 Answers2

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$$\mathrm A = (a - b) \mathrm I_n + b 1_n 1_n^{\top}$$

If $a = b$, then $\mathrm A$ is rank-$1$ and, thus, non-invertible. If $a \neq b$, using the matrix determinant lemma,

$$\det (\mathrm A) = \left(1 + \frac{n b}{a-b}\right) (a-b)^n = (a + (n-1) b) \, (a-b)^{n-1}$$

Thus, if $a \neq b$ and $a + (n-1) b \neq 0$, then $\mathrm A$ is invertible.

Rodrigo de Azevedo
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1.if a=b, this matrix is not invertible.

  1. Add all rows to first. In first rows now $(n-1)b + a$ at any columns=> if $(n-1)b + a$ - matrix is not invertible

else, we can create upper-triangle matrix without diagonal zeroes

kotomord
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