Let $$A$$ be the $$n\times n$$ matrix:(adsbygoogle = window.adsbygoogle || []).push({});

\begin{equation}

A= \begin{bmatrix} % or pmatrix or bmatrix or Bmatrix or ...

2 & 1&\dots & 1 & 1 \\

1 & 2&\dots & 1 & 1 \\

\vdots&\ddots & \ddots & 2 & 1 \\

1 & & \dots & 1&2 \\

\end{bmatrix}

\end{equation}

(2s along the diagonal and 1s everywhere else.) Compute the determinant of $$A$$.

Here is what I did.

As usual, let $$I$$ be the identity matrix. Observe that $$A=I+B$$ where $$B$$ is a square matrix such that every element is 1. Next we can use the Taylor series expansion of the determinant

\begin{align} \det(I + B) = \sum_{k=0}^{\infty} \frac{1}{k!} \left( - \sum_{j=1}^{\infty} \frac{(-1)^j}{j}\mathrm{tr}(B^j) \right) ^k.\end{align}

Note that $$\mathrm{tr}\left(B^j\right)$$ is always $$n^j$$, and so it follows that

\begin{align} \det(A)& = \sum_{k=0}^{\infty} \frac{1}{k!} \left( - \sum_{j=1}^{\infty} \frac{(-1)^j}{j}n^j \right) ^k\, , \end{align}

The well known series expansions

\begin{equation}

-\sum _{j=1}^{\infty }{\frac { \left( -1 \right) ^{j}}{j}}{n}^{j}=\log(n+1)

\end{equation}

and

\begin{equation}

\sum_{k=0}^\infty\frac{f(n)^k}{k!}=\exp{f(n)}

\end{equation}

allow us to conclude that the determinant is $$n+1.$$

I was wondering if there was a simpler way to do this problem.

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# How to compute the determinant of this matrix?

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