Given $a_1, \cdots, a_n \in \mathbb R$, how can we evaluate this? $$\det \left(\delta_{ij}+a_ia_j\right)_{ij}$$
I think the answer should be $$1+a_1^2+\cdots a_n^2$$ and this question is motivated by computing the volume form of a Riemannian submanifold whose parametrization can be represented by a graph. (See Riemannian volume form on surface of a smooth function. In the last answer of the link, $\partial f/\partial x_i$ is just replaced to $a_i$ in my question.)
Aside: I am also curious whether there is an another way to find the volume form not by evaluating the above determinant.
The determinant of a matrix is the product of its eigenvalues, and adding $\delta^i_j$ shifts all eigenvalues by $1$. Therefore, if $\lambda_1,\dots, \lambda_n$ are the eigenvalues of the matrix $(a^ia_j)$, we obtain $\det(\delta^i_j + a^ia_j) = \Pi_{k=1}^n (\lambda_k+1)$. Let $(v_k)_{k=1}^n$ be an orthogonal base, and $v_1 = a$. Then $(a^ia_j) v_k = a^ia_jv^j_k e_i= a^i<a,e_k>e_i = \delta^1_k |a|^2 a$, where $(e_k)_{k=1}^n$ denotes the standard base. Thus the eigenvalues are $|a|^2,0,\dots,0$, which yields $\det(\delta_{ij} + a_ia_j) = \det(\delta^i_j + a^ia_j) = 1 + |a|^2$.
The above computation can be interpreted in the context of the volume element of a graph as follows: Let $\nabla f(x), v_2, \dots, v_n$ be an orthogonal base of $T_x\mathbb{R}^n$. Now, $v_2,\dots,v_n$ are orthogonal to $\nabla f$, hence their pushforwards on the graph are just parallel transports of themselves, with the same length. The first vector, $\nabla f(x)$, transforms to $(\nabla f(x), |\nabla f(x)|^2)^T$, thus its stretched by a factor of $1+|\nabla f|^2$. Therefore, the volume element is stretched by a total factor of $1+|\nabla f|^2$.