# How can we prove the equality of the following determinants? [closed]

Assume that $A$ is an $n\times n$ real matrix whose entries are all $1$. Then how can we show the following determinant equality for any $x$?

$\det(A-xI)$=\begin{vmatrix} 1 -x & 1 & 1 & \cdots & 1 \\ 1 & 1 -x & 1 & \cdots & 1 \\ 1 & 1 & 1 -x & \cdots & 1\\ \vdots & \vdots & \vdots & \ddots & \vdots\\ 1 & 1 & 1 & \cdots & 1 -x \end{vmatrix}=

\begin{vmatrix} n-x & n-x & n-x & \cdots & n -x \\ 1 & 1 -x & 1 & \cdots & 1 \\ 1 & 1 & 1 -x & \cdots & 1\\ \vdots & \vdots & \vdots & \ddots & \vdots\\ 1 & 1 & 1 & \cdots & 1 -x \end{vmatrix}

Thanks.

## closed as off-topic by Did, Davide Giraudo, Dragonemperor42, hardmath, HenrikMay 20 '17 at 13:44

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Recall that if an $n\times n$ matrix $B$ is obtained from an $n\times n$ matrix $A$ by adding a scalar multiple of one row of $A$ to another row, then $\det(B)=\det(A)$.
In your problem the second matrix is obtained from the first by adding rows $2$ through $n$ to row $1$. So the determinants are the same.
Just add row 2, row 3,... up to row $n$ to row 1.
Note that $$\begin{bmatrix} n-x & n-x & n-x & \cdots & n -x \\ 1 & 1 -x & 1 & \cdots & 1 \\ 1 & 1 & 1 -x & \cdots & 1\\ \vdots & \vdots & \vdots & \ddots & \vdots\\ 1 & 1 & 1 & \cdots & 1 -x \end{bmatrix} = \\ \begin{bmatrix} 1&1&1&\cdots&1\\ &1&0&\cdots&0\\ &&\ddots&\ddots&\vdots\\ &&&&0\\ &&&&1 \end{bmatrix} \begin{bmatrix} 1 -x & 1 & 1 & \cdots & 1 \\ 1 & 1 -x & 1 & \cdots & 1 \\ 1 & 1 & 1 -x & \cdots & 1\\ \vdots & \vdots & \vdots & \ddots & \vdots\\ 1 & 1 & 1 & \cdots & 1 -x \end{bmatrix}$$