There is a proof of Auslander-Buchsbaum formula in Matsumura's Commutative Ring Theory page 155. I am trying to understand the case $\operatorname{pd} M = 1$. He says take a short exact sequence $$ 0 \to A^{\oplus m} \stackrel{\varphi}{\to} A^{\oplus n} \to M \to 0.$$ This I am fine with. Then he says consider the induced map on Ext:
$$\operatorname{Ext}^i_A(k,A)^{\oplus m} \stackrel{\varphi_\ast}{\to} \operatorname{Ext}^i_A(k,A)^{\oplus n}. $$
Then he says that because $\varphi$ is given by a matrix in coefficients in $\mathfrak{m}$, $\varphi_\ast$ is given by the same matrix and thus is the zero map.
Firstly, why should $\varphi_\ast$ be given by the same matrix?
Second, why are the coefficients of $\varphi$ in $\mathfrak{m}$?
Third, why should $\varphi_\ast$ be the zero map?
Thank you.
For the first point: You have to check that for any two modules $M,N$ over a commutative ring $R$ and any $x\in R$, the map on $\text{Ext}_R^j(N,M)$ induced, by functoriality, from the multiplication map $\cdot x: M\to M$, is the multiplication map for the $R$-module $\text{Ext}_R^j(N,M)$.
For the second point: You can choose $\varphi$ to be a minimal projective resolution of $M$, and then its image lies in ${\mathfrak m}^n$.
For the third point: use the dual of the first point to see that the map under consideration is also the one induced from the multiplication map $\cdot x: N\to N$. If $N=k$ and $x\in{\mathfrak m}$, then the latter map vanishes, and hence so does the map on $\text{Ext}_R^j(N,M)$.