- #1
Charles49
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Let $$A$$ be the $$n\times n$$ matrix:
\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.
\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.