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Is the following matrix invertible?

$ \begin{bmatrix} x & a & a & \dots & a \\ a & x & a & \dots & a \\ a & a & x & \dots & a \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ a & a & a & \dots & x \end{bmatrix} $

I cannot use "determinants" or other stuff but the rank of the matrix. I really tried hard this problem making a lot of algebra and cases but I didn't find the solution; can someone help me for solving it?

Note: The cases when $x=a$ and $a=0$ were covered yet.

Thanks.

rowcol
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    when invertible, the inverse has the same shape, all $y$ on the main diagonal and all else $b.$ Given the dimension $n,$ what are $y$ and $b,$ and when is it impossible to solve for them? – Will Jagy Jan 14 '19 at 03:26
  • This post is relevant, but I don't think that any of the link posts avoid the use of determinants or eigenvalues. The answer to your question will ultimately be that the matrix is invertible unless we have $x = a$ or $x = -(n-1)a$, where $n$ is the size of the matrix. – Ben Grossmann Jan 14 '19 at 03:26
  • If you’re supposed to use the rank of the matrix, then have you examined when this matrix doesn’t have full rank? Have you tried decomposing it into the sum of a diagonal matrix and one that has all elements equal? That might give you some insights. – amd Jan 14 '19 at 20:26

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