Prove $|C+\overrightarrow{Y}\overrightarrow{Y}'|=(1+\overrightarrow{Y}'C^{-1}\overrightarrow{Y})|C|$ where $Y$ is a $p \times 1$ vector and $C$ is a non-singular $p \times p$ matrix. See Exercise $4.11$ in $JW$.
$\mathbf{Proof}$
Define the block matrix
$$\mathbf{A}=\begin{pmatrix} C & -\overrightarrow{Y} \\ \overrightarrow{Y}^{'} & 1 \end{pmatrix}$$
Problem 4.11 in JW asserts that for any square block matrix $\mathbf{B}$:
$$|\mathbf{B}|=|\mathbf{B}_{11}||\mathbf{B}_{22}-\mathbf{B}_{21}\mathbf{B}_{11}^{-1}\mathbf{B}_{12}|$$
Using the result of $4.11$ we have that:
$$|\mathbf{A}|=|C||1-\overrightarrow{Y}^{'}C^{-1}\overrightarrow{Y}|=|C|(1-\overrightarrow{Y}^{'}C^{-1}\overrightarrow{Y})$$
I understand that for any block matrix $\mathbf{B}=\begin{pmatrix} A & B \\ C & D \end{pmatrix}$ it is not always the case that $|\mathbf{B}|$=AD-BC. Knowing this, I am confused on how I can evaluate the determinant of $\mathbf{A}$ without making use of the result of 4.11 so that $|\mathbf{A}|=|C+\overrightarrow{Y}\overrightarrow{Y}'|$ as desired.
You can factorise the underlying matrix as $$ \pmatrix{C & 0\\0 &1} \pmatrix{I & 0\\y^{T}&1} \pmatrix{I & -C^{-1}y\\0^{T} & y^{T}C^{-1}y+1} $$ and then all is clear.