Characteristic Equation of Gauss Transform From Richard Bellman's Introduction to Matrix Analysis

Consider the Gauss transform $B=(b_{ij})$ of the matrix $A=(a_{ij})$,
$$b_{i1} = \delta_{i1} a_{i1},\;\;\; b_{ik} = a_{11}^{-1}(a_{11}a_{ik} - a_{i1}a_{ik}),\;\;\; k>1$$
Let $A_{11}=(a_{ij})$ for $2\leq i,j\leq N$. Show that
$$|\lambda I - B| = a_{11}^{-1}\left[\lambda|\lambda I - A_{11}| - |\lambda I - A|\right].$$

where $|A|$ represents the determinant of the matrix $A$ and $\delta_{ij}$ is the Kronecker delta.
Work/Question: The first column of $B$ is $a_{11}$ followed by all zeros, so by cofactor expansion along the first column, we get the determinant of $\lambda I - B$ is $(\lambda - a_{11})$ times the determinant of $B$ with the first column and row removed which I'll denote by $B_{11}$. From here, the goal would be to find $|\lambda I - B_{11}|$. By distributing the $a_{11}^{-1}$, we get
$$b_{ik} = a_{ik} - a_{11}^{-1}a_{i1}a_{ik}$$
or equivalently that $B_{11} = A_{11} - a_{11}^{-1}\text{diag}(a_{21},\dots,a_{N1})A_{11}$. But this leaves me with
$$(\lambda - a_{11})\left|\lambda I - \left(A_{11} - a_{11}^{-1}\text{diag}(a_{21},\dots,a_{N1})A_{11}\right)\right|$$
which doesn't seem to be on the right track. Hints would be appreciated - thanks.
 A: The author is wrong. Note that the RHS of the equality is a scalar multiple of the difference of two monic degree-$n$ polynomials in $\lambda$. The result is thus a polynomial of degree at most $n-1$. However, the LHS is a monic polynomial of degree $n$. Hence the two sides are unequal.
Let $A=\pmatrix{a_{11}&y^T\\ x&A_{11}}$. According to the given definition, $B$ is actually a block matrix in the form of $\pmatrix{a_{11}&0\\ 0&B_{11}}$, where $B_{11}=\frac{a_{11}A_{11}-xy^T}{a_{11}}$. The problem statement can be corrected by replacing the LHS by $\det(\lambda I_{n-1}-B_{11})$, because
\begin{aligned}
\det(\lambda I_n-A)
&=\det\pmatrix{\lambda-a_{11}&-y^T\\ -x&\lambda I_{n-1}-A_{11}}\\
&=\det\pmatrix{\lambda&-y^T\\ 0&\lambda I_{n-1}-A_{11}}
-\det\pmatrix{a_{11}&-y^T\\ x&\lambda I_{n-1}-A_{11}}\\
&=\lambda\det(\lambda I_{n-1}-A_{11})
-\det\left[\pmatrix{a_{11}&-y^T\\ x&\lambda I_{n-1}-A_{11}}\pmatrix{1&\frac{1}{a_{11}}y^T\\ 0&I_{n-1}}\right]\\
&=\lambda\det(\lambda I_{n-1}-A_{11})
-\det\pmatrix{a_{11}&0\\ x&\lambda I_{n-1}-A_{11}+\frac{1}{a_{11}}xy^T}\\
&=\lambda\det(\lambda I_{n-1}-A_{11})
-a_{11}\det\left(\lambda I_{n-1}-\frac{a_{11}A_{11}-xy^T}{a_{11}}\right)\\
&=\lambda\det(\lambda I_{n-1}-A_{11})
-a_{11}\det(\lambda I_{n-1}-B_{11}).\\
\end{aligned}
