TITLE: Determinant of a matrix of the form $M = I + xx^t$
QUESTION [5 upvotes]: $$M=\begin{pmatrix} 1+x_1^2 & x_1x_2 &...&x_1x_n \\ x_2x_1 & 1+x_2^2 &...&x_2x_n \\...&...& &...& \\x_nx_1 & x_nx_2& ...&1+x_n^2&\end{pmatrix}.$$
So I noticed that $M$ is a symmetric matrix and $ M=I+(x_1, x_2,...,x_n)^T(x_1, x_2,...,x_n)$. That's all I can get.

REPLY [2 votes]: Denote by $M_n$ the matrix we want to compute the determinant. Letting $\mathbf{x}=(x_1,\dots,x_{n-1})$, we express $M_n$ as a block matrix:
$$
M_n=\pmatrix{M_{n-1}& x_n\mathbf x^t\\
x_n\mathbf{x}&1+x_n^2}.
$$
Since $\pmatrix{  x_n\mathbf x^t\\1+x_n^2}=\pmatrix{  0\\1}+\pmatrix{  \mathbf x_nx^t\\x_n^2}$, it follows by multi-linearity of the determinant that 
$$
\det\left(M_n\right)=\det\pmatrix{M_{n-1}& 0\\
x_n\mathbf{x}&1}+\det\pmatrix{M_{n-1}& x_n\mathbf x^t\\
x_n\mathbf{x}&x_n^2}.
$$
The first term of the right hand side is $\det\left(M_{n-1}\right)$; for the second one, using two times the multi-lineartiy of the determinant (with respect to lines and then columns), we derive that 
$$
\det\left(M_{n}\right)=\det\left(M_{n-1}\right)+x_n^2\det\pmatrix{M_{n-1}&  \mathbf x^t\\
 \mathbf{x}&1}.
$$
The last determinant is one: one can make the linear combinations $C_i\leftarrow x_iC_n$ to see this.