Proof of Proposition 2.4 in Atiyah-MacDonald I'm struggling here with the proof. To be honnest i need a really concrete explanation because i have been on it for a long time and i can not find it nowhere else.

Please can anyone help me with this? 
And thank you in advance.
 A: I will assume that you are fine up to equation
$(1): \,  \sum_{j=1}^n (\delta_{ij} \phi -a_{ij} ) x_j=0, \, \, \, i=1,\dots,n$.
These equations can be represented in matrix form as
$(2): \, \boldsymbol{C} \boldsymbol{x}=0$, where $\boldsymbol{x}=(x_1,\dots,x_n)^\top$ and $\boldsymbol{C}$ is an $n \times n$ matrix with $\boldsymbol{C}_{ij} = \delta_{ij} \phi -a_{ij}$.
If you wonder why (2) makes sense, note that the ring of endomorphisms of $M$ acts on $M$ in a way such that $M$ is an $\operatorname{End}_A(M)$-module. If you multiply from the left equation $(2)$ with the adjoint of $\boldsymbol{C}$, you will get
$(3): \,  \boldsymbol{C}^{\dagger} \boldsymbol{C} \boldsymbol{x}=0$.
But matrix algebra extends to matrices over commutative rings (see Serge Lang's "Algebra" for this) and the well-known relation
$(4): \,  \boldsymbol{C}^{\dagger} \boldsymbol{C} = \operatorname{det}(\boldsymbol{C}) I$
still holds (again see Lang). Substitute this into $(3)$ and get that 
$\operatorname{det}(\boldsymbol{C}) \boldsymbol{x} = 0$, which implies that $\operatorname{det}(\boldsymbol{C}) M=0$.
A: Consider the commutative subring $R=A[\phi]\subset\mathrm{End}_A(M)$ generated by $\phi$; then $R$ acts on $M$, and thus $M_n(R)$ acts on $M\oplus\cdots\oplus M$, the direct sum of $n$ copies of $M$. The equations
$$\phi(x_j)=\sum_{i=1}^na_{ij}x_i\,,$$
for $j=1,\dots, n$, can be reinterpreted with the action of $M_n(R)$ on $M^{\oplus n}$: write
$$B=\begin{pmatrix}
a_{11}-\phi & a_{12} & a_{13} & \cdots\\
a_{21} & a_{22}-\phi & a_{23} & \cdots\\
a_{31} & a_{32} & a_{33}-\phi & \cdots\\
\vdots&\vdots&\vdots&\ddots
\end{pmatrix}\in M_n(R)\qquad\text{and}\qquad X=\begin{pmatrix}x_1\\x_2\\\vdots\\x_n\end{pmatrix}\in M^{\oplus n}$$
(To be very precise, we shoud actually write $a_{ij}\mathrm{id}_M$ everywhere instead of $a_{ij}$).
Then the $n$ equations we wrote at the start are equivalent to the one equation
$$B X=0$$
Since $R$ is commutative, we have $\mathrm{Adj}(B)\times B=\det(B)I_n=B\times\mathrm{Adj}(B)$, which is an equation which holds in $M_n(R)$. If we multiply the previous equation on the left by $\mathrm{Adj}(B)$, we get that
$$0=\mathrm{Adj}(B) B X=\begin{pmatrix}\det(B)\\&\det(B)\\&&\ddots\\&&&\det(B)\end{pmatrix}\begin{pmatrix}x_1\\x_2\\\vdots\\x_n\end{pmatrix}=\begin{pmatrix}\det(B)x_1\\\det(B)x_2\\\vdots\\\det(B)x_n\end{pmatrix}$$
Since the $x_i$ generate $M$, this is equivalent to saying that $\det(B)$, which is an element of $R$, hence an endomorphism of $M$, is the zero endomorphism of $M$. The determinant $\det(B)\in R\subset\mathrm{End}_A(M)$ can be calculated by the standard formula
$$\det(B)=\sum_{\sigma\in\mathfrak{S}_n}(-1)^\sigma\prod_{j=1}^nB_{\sigma(j),j}$$
which, upon inspection, is polynomial in $\phi$ of degree $n$ with coefficients in the ideal $\mathfrak a$. The coefficient in front of $\phi^n$ is $(-1)^n$, and since $\det(B)=0$, you get a relation
$$\phi^n+a_1\phi^{n-1}+\cdots+a_{n-1}\phi+a_n\mathrm{id}_M=0$$
