Let $A$ be a Noetherian ring, $M$ a finite $A$-module and $I=(y_1,\cdots,y_n)$ an ideal of $A$ such that $M \neq IM$. Denote by $H_i(y_1,\cdots,y_n;M)$ the homology at dimension $i$ of the augmented Koszul complex $K_{\cdot}(y_1,\cdots,y_n) \otimes M$.

Theorem: If $y_1,\cdots,y_n$ is a regular sequence, then $H_i(y_1,\cdots,y_n;M)=0, \, \forall i>0$. (see for example Matsumura, CRT, Theorem 16.5 (i)).

Question: Is the converse true? I.e. if $H_i(y_1,\cdots,y_n;M)=0, \, \forall i>0$, is it true that the generators of $I$, $y_1,\cdots,y_n$ are a regular sequence under some permutation?

Remark 1: The converse is true if $A$ is local or if $A$ and $M$ are $\mathbb{N}$-graded with all generators of $I$ having positive degree.

Remark 2: If $H_i(y_1,\cdots,y_n;M)=0, \, \forall i>0$, then by the characterization of the $I$-depth of $M$ by means of the Koszul complex (e.g. Matsumura, CRT, Theorem 16.8), we have that $I$ contains an $M$-regular sequence of length $n$.

Remark 3: In the proof of the Corollary to Theorem 16.8, Matsumura claims that $H_i(y_1,\cdots,y_n;M)=0, \forall i>0 \Leftrightarrow y_1,\cdots,y_n$ is an $M$-sequence. I can't see why this is true, and this is the motivation for my question.

Edit:

My question might become clearer to those who do not have Matsumura's book, if i state the Corollary to Theorem 16.8, to which i refer in my Remark 3:

Let $A$ be a Noetherian ring, $I=(y_1,\cdots,y_n)$ an ideal of $A$, $M$ a finite $A$-module such that $IM \neq M$. Then $y_1,\cdots,y_n$ is an $M$-sequence if and only if $\operatorname{depth}(I,M)=n$.

Matsumura's proof goes as follows: "$\operatorname{depth}(I,M)=n \Leftrightarrow H_i(y_1,\cdots,y_n;M)=0, \forall i>0 \Leftrightarrow y_1,\cdots,y_n$ is an $M$-sequence."

Remark 4: Certainly $\operatorname{depth}(I,M)=n \Leftrightarrow H_i(y_1,\cdots,y_n;M)=0, \forall i>0$ and certainly $y_1,\cdots,y_n$ is $M$-sequence $\Rightarrow H_i(y_1,\cdots,y_n;M)=0, \forall i>0$. But why $H_i(y_1,\cdots,y_n;M)=0, \forall i>0 \Rightarrow y_1,\cdots,y_n$ is an $M$-sequence?