linear algebra - Determinant of a special $ntimes n$ matrix

Compute the determinant of the nun matrix:
$$
\begin{pmatrix}
2 & 1 & \ldots & 1 \\
1 & 2 & \ldots & 1\\
\vdots & \vdots & \ddots & \vdots\\
1 & 1 &\ldots & 2
\end{pmatrix}
$$

For $n=2$, I have$$
\begin{pmatrix}
2 & 1 \\
1 & 2
\end{pmatrix}
$$

Then $det = 3$.

For $n=3$, we have

$$
\begin{pmatrix}
2 & 1 & 1\\
1 & 2 & 1\\
1 & 1 & 2 \\
\end{pmatrix}
$$

Then $det = 4$.

For $n=4$ again we have

$$
\begin{pmatrix}
2 & 1 & 1 & 1 \\
1 & 2 & 1 & 1\\
1 & 1 & 2 & 1\\
1 & 1 & 1 & 2
\end{pmatrix}
$$

Then $det = 5$

How can I prove that the determinant of nun matrix is $n+1$.

Answer

A standard result (http://en.wikipedia.org/wiki/Matrix_determinant_lemma) is $\det(I+AB) = \det(I+BA)$.

Since the matrix above can be written as $I+ e e^T$, where $e$ is a vector of ones, we have $\det(I+ e e^T) = \det(1+ e^T e) = 1+e^Te = n+1$.