Determinant of matrix that is diagonal, but for last row/column I would like to compute the determinant of a symmetric $(K+1)\times (K+1)$ matrix in which the upper left $K \times K$ matrix is diagonal but the $(K+1)$th row and column are complete. E.g.
$$
X = \begin{bmatrix} 
x_1 & 0 & \dots & 0 & y_1 \\
0 & x_2 & \dots & 0 & y_2 \\
\vdots & \vdots & \ddots & \vdots & \vdots \\
0 & 0 & \dots & x_K & y_K \\
y_1 & y_2 & \dots & y_K & y_{K+1}
\end{bmatrix}
$$ 
Is there a simple way to compute $\det(X)$?
Note, a similar question was asked for a matrix with the same form but with more constraints on the entries:
Determinant of an almost-diagonal matrix
I can't see that the answer to this question helps here though.
 A: For aesthetic considerations suggested by Michael Hoppe, I'll rename $K$ into $n$ and $y_{K+1}$ into $x_{n+1}$, so the matrix whose determinant you're searching is 
$$A=\begin{pmatrix} 
x_1 & 0 & \dots & 0 & y_1 \\
0 & x_2 & \dots & 0 & y_2 \\
\vdots & \vdots & \ddots & \vdots & \vdots \\
0 & 0 & \dots & x_n & y_n \\
y_1 & y_2 & \dots & y_n & x_{n+1}
\end{pmatrix}.
$$ 
First method. Use determinant expansion with respect to the last column to get  
$$\mathrm{det}(A) = x_1...x_n x_{n+1} + \sum_{i=1}^n(-1)^{n+1+i} \mathrm{det}(A_i)$$
where $A_i$ is the matrix $A$ deprived of its last column and $i$-th row. For example, 
$$A_1 = \begin{pmatrix}0& x_2 &0& ... &0 \\
0 & 0 & x_3 & ... & 0 \\
\vdots & & & \ddots& \vdots\\
0 & & ... & & x_n\\
y_1 & &... && y_n \\
 \end{pmatrix}. $$
 Using row expansion for $A_i$, it is easy to see that $\mathrm{det}(A_i)$ is equal to $(-1)^{n+i} y_i \prod_{j \neq i} x_j$, which yelds 
\begin{align*}\mathrm{det}(A) &= \prod_{i=1}^{n+1} x_i - \sum_{i=1}^n y_i^2 \prod_{j \neq i, j\leqslant n}x_j \\
&= \prod_{i=1}^{n+1}x_i \left( 1 - \sum_{i=1}^n \frac{y_i^2}{x_i}\right).
\end{align*}
Edit (second method, hint). This last expression suggests another method (maybe it's not working) : suppose that no $x_i$ is zero. Note $X = \mathrm{diag}(x_1, ..., x_{n+1})$ and $Y = A - X$, so that $A = X+Y = X(\mathrm{Id}+X^{-1}Y)$.  Then, $\mathrm{det}(A) = \mathrm{det}(X) \mathrm{det}(\mathrm{Id} - X^{-1}Y)$. Now, all you have to do is to compute $\mathrm{det}(\mathrm{Id} - X^{-1}Y)$. Maybe there's a simple ay of doing this (I don't know).
A: I'd do a Laplace expansion along the last row/column:
$$\det X=\sum_{j=1}^{K+1} (-1)^{j+1} y_j \det X_j,$$
where $X_j$ is obtained by deleting the $j$th row and last column from $X$. Then
$$\det X_{K+1}=x_1\cdot\ldots\cdot x_K$$
is easy. For the others, you have to do a second Laplace expansion:
$$X_1=\left( \begin{matrix} 0 & x_2 & 0 & \cdots & y_2 \\ 0 & 0 & x_3 & \cdots & y_3 \\ \vdots & & & & \vdots \\ y_1 & 0 & 0 & \cdots & y_{K+1}   \end{matrix} \right)$$
so
$$\det X_1=y_1 \det \left( \begin{matrix} x_2 & 0 & \cdots & y_2 \\ \vdots & & & \vdots \\ 0 & 0 & \cdots & y_{K+1}  \end{matrix} \right)=y_1x_2\ldots x_K y_{K+1}$$
since the remaining matrix is upper triangular. Works analogously for $X_2,\ldots,X_K$, but watch the signs.
A: Well yes, the answer provided in the other question should help you a lot. Assuming that the $x_i$ are not zero, you can use Gaussian elimination to turn your matrix into a triangular matrix and then read off the determinant.
If $x_i$ is zero for some $i$, a Laplace expansion using the $i$-th row or column should help you to reduce the problem.
A: $\def\m#1{\left[\begin{array}{r}#1\end{array}\right]}$Given a block matrix
$$X = \m{A&B\\C&D}$$
the determinant is
$$\det(X) = \det\left(D-CA^{-1}B\right)\cdot\det(A)$$
For the current problem, this results in
$$\det(X)
 = \left(y_{K+1}-\sum_{i=1}^K\frac{y^2_i}{x_i}\right)\cdot
   \left(\prod_{\ell=1}^K x_\ell\right)$$
