How to compute the determinant of this matrix?

[Charles49]

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.

 

[DonAntonio]

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.

1) It's easy to see, inductively, that
$$\begin{equation}\left|\begin{pmatrix} 1&1&1&...&1\\1&2&1&...&1\\...&...&...&...&...\\1&1&1&...&2\end{pmatrix}\right|\end{equation}=1$$Say, substract first row from second, develop by minors of the new 2nd row, etc.

So substracting the third row from the 2nd one in the original matrix, we get:

$$\begin{equation}\left|\begin{matrix} 2&1&1&...&1\\1&2&1&...&1\\0&\!\!\!-1&1&...&0\\...&...&...&...&...\\1&1&1&...&2 \end{matrix}\right|\end{equation}$$
Developing wrt the third row, using the above fact and induction we get what we want.

DonAntonio

 

[Charles49]

Thanks DonAntonio

 

[AlephZero]

Another way is to factorize A = LU where L is unit lower triangular and U is upper triangular. Use Crout's verison of Gaussian elimination.

The diagonals of U are 2, 3/2, 4/3, 5/4, ... (n+1)/n. The determinant is the product of the diagonals = n+1

 

[Charles49]

Thanks AlephZero, I am going to learn about Crouts version.