Help with Cayley-Hamilton determinant trick theorem from Matsumura's Commutative Algebra. Theorem 2.1. Suppose that $M$ is an $A$-module generated by $n$ elements, and that $\varphi \in \text{Hom}_A(M,M)$; let $I$ be an ideal of $A$ such that $\varphi(M) \subset IM$.  Then there is a relation of the form:
$$
 \varphi^n + a_1 \varphi^{n-1} + \dots + a_{n-1} \varphi + a_n = 0, 
 $$
with $a_i \in I$ for $1 \leq i \leq n$ (where both sides are considered as endomorphisms of $M$).
Proof. Let $M = A w_1 + \dots Aw_n$.  By the assumption $\varphi(M) \subset IM$ there exist $a_{ij} \in I$ such that $\varphi(w_i) = \sum_{j=1}^n a_{ij} w_j$.  This can be rewritten $\sum_{j=1}^n (\varphi \delta_{ij} - a_{ij}) w_j = 0, \ i=1..n$.  The coefficients of this systme of linear equations can be viewed as a square matrix $(\varphi \delta_{ij} - a_{ij})$ of elements of $A'[\varphi]$, the commutative subring of the endomorphisms ring $E = \text{Hom}_A(M,M)$ generated by the image $A'$ of $A$ under $a \mapsto (x \mapsto ax)$, together with $\varphi$.  Let $b_{ij}$ denote its $(i,j)$th cofactor, and $d$ its determinant.  Multiply the equation through by $b_{ik}$ and sum over $i$:
$$
\sum_{i=1}^n b_{ik}(\sum_{j=1}^n  (\varphi \delta_{ij} - a_{ij}) w_j ) = 0
$$
For example, if $B = (\varphi \delta_{ij} - a_{ij})$ is a $3 \times 3$ matrix we have:
$$
B = -\begin{pmatrix} 
a_{11} - \varphi & a_{12} & a_{13} \\ 
a_{21} & a_{22} -\varphi & a_{23} \\
a_{31} & a_{32} & a_{33} - \varphi
\end{pmatrix}
$$
Then for example at $i = 1$ we have the following cofactors:
$$
b_{11} = -\det\begin{pmatrix} a_{22} - \varphi & a_{23} \\ a_{32} & a_{33} - \varphi \end{pmatrix}, b_{12} = \det\begin{pmatrix} a_{21} & a_{23} \\ a_{31} & a_{33} - \varphi \end{pmatrix}, b_{13} = -\det\begin{pmatrix} a_{21}  & a_{22} - \varphi \\ a_{31} & a_{32} \end{pmatrix}
$$
Then the sum:
$$
\sum_{i=1}^n b_{ik} \sum_{j=1}^n B_{ij} w_j = 0 \iff \\
\sum_{j=1}^n \sum_{i=1}^n b_{ik} B_{ij} w_j = 0 \iff \\
\sum_{j=1}^n (\det B) w_j = 0 \iff  \\
(\det B) \sum_{j=1}^n w_j = 0 \ \ \textbf{( wrong here )}
$$
I believe it should be $(\det B) w_j = 0, \ \forall j=1..n$ instead, as in this lecture video.
I followed the recipe to multiply by $b_{ik}$ and sum over $i$ and I get the wrong answer.  Please help me find where I've made a mistake.

I shall defer slightly from your notation. $b_{ij}=(i,j)$th entry of $\operatorname{Adj}B$. 
We know that $\displaystyle{}\sum_{j=1}^n B_{ij}w_j=0 \ \forall \ i=1,2,\dots, n$. Then as you did , we get $\displaystyle{}\sum_{i=1}^nb_{ki}\sum_{j=1}^n B_{ij}w_j=0 \ \forall \ k=1,2,\dots, n \\ 
\implies \displaystyle{}\sum_{j=1}^n\left( \sum_{i=1}^n b_{ki}B_{ij}\right )w_j=0  \ \forall \ k=1,2,\dots, n    \\ 
\implies\displaystyle{ \sum_{j=1}^n(\det B)\delta_{kj}w_j=0 \ \forall \ k=1,2,\dots, n }\\ 
\implies \displaystyle{(\det B)w_k=0 \ \forall \ k=1,2,\dots, n }$
which is what you wanted.
Edit: Your mistake was writing $\displaystyle {\sum_{i}b_{ik}B_{ij}=\det B}$ instead of $(\det B)\delta_{jk}$. After all, you should expect the sum to depend on $k$ in some way since you are not summing over $k$.
