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 $\det(I+ e e^T) = \det(1+ e^T e) = 1+e^Te = n+1$.