Regularity of a basis over a local ring The following Theorem (Thm. 19.9 in Matsumura, CRT) will be necessary for the understanding of my question:

Theorem: Let $A$ be a Noetherian local ring and $I$ a proper ideal of $A$ with $\operatorname{projdim} I < \infty$. Then $I$ is generated by an $A$-sequence $\Leftrightarrow$ $I/I^2$ is a free module over $A/I$.

On to my question now. Let $A,I$ satisfy the hypothesis of the above theorem and let $x_1,\dots,x_n \in I$ be such that their images in $I/I^2$ form an $A/I$-basis. Then by the theorem $I=(y_1,\dots,y_n)$ where $y_1,\dots,y_n$ is an $A$-sequence. Now we can write $x_i = \sum a_{ij} y_j$ and the matrix $(a_{ij})$ is invertible. 

Question: Why does the invertibility of the matrix $(a_{ij})$ imply that $x_1,\dots,x_n$ is an $A$-regular sequence?

This claim is made in Matsumura, CRT, p. 161.
 A: We need only show $x_1, \dots x_n$ is quasi-regular.
Recall from page 125 of Matsumura that this is so iff the map $\phi: A/I[X_1, \dots X_n] \to gr_I(A)$ that takes a homogeneous form $F$ of degree $v$ to the image of $F(x_1, \dots, x_n)$ in $I^v/I^{v+1}$ is injective.
Let $\psi: A[X_1, \dots X_n] \to A[X_1, \dots X_n]$ be the map of free $A$-modules that sends a form $F(X_1, \dots X_N)$ to $F(\Sigma a_{1j}X_j, \dots, \Sigma a_{nj}X_j)$. Notice that $\psi$ is invertible; if $(b_{ij})$ is the inverse of $(a_{ij})$, the inverse of $\psi$ is given by sending a homogeneous form $F(X_1, \dots, X_n)$ to $F(\Sigma b_{1j}X_j, \dots, \Sigma b_{nj}X_j)$.
Notice further that, tensoring with $A/I$, we get an isomorphism of $A/I$ modules $\bar{\psi}:A/I[X_1, \dots X_n] \to A/I[X_1, \dots X_n]$. This may seem fishy at first, because $A/I$ is not flat over $A$; but we are okay, because functors take isomorphisms to isomorphisms. 
Now, let $\alpha: A/I[X_1, \dots X_n] \to gr_I(A)$ be the map that takes a homogeneous form $F$ of degree $v$ to the image of $F(y_1, \dots, y_n)$ in $I^v/I^{v+1}$. Since $\alpha$ is injective, $\alpha \circ \bar{\psi}$ is injective. Now, notice that $\phi = \alpha \circ \bar{\psi}$.
A: I'll give an elementary proof to the following:

Let $(R,\mathfrak m)$ be a local (not necessarily noetherian) ring, $x_1,\dots,x_n$ an $R$-sequence, $A\in M_n(R)$ an invertible matrix and $(y_1\dots y_n)^t=A(x_1\dots x_n)^t$. Then $y_1,\dots,y_n$ is an $R$-sequence. (Here ${\bf z}^t$ denotes the transpose of the vector ${\bf z}$.)

Proof. Since $A=(a_{ij})$ is invertible at least one element of $A$, say $a_{11}$, is not in $\mathfrak m$. (If all elements $a_{ij}\in\mathfrak m$, then $\det A\in\mathfrak m$, a contradiction.) Now we multiply the first row of $A$ by the inverse of $a_{11}$. (Note that on the LHS this performs a multiplication of $y_1$ by the inverse of $a_{11}$.) In this way we get a new invertible matrix $A$ whose element on the position $(1,1)$ is $1$. Now we use this element to make $0$ on the other entries of the first column. This means that we have to multiply the first row by some elements and subtract the first row from the others. But how this procedure affects the LHS? In the same way, of course, that is, instead of $y_i$ we will have $y_i+a_iy_1$, $a_i\in R$. After this first step we will have a matrix $A$ whose first column is $0$ except the position $(1,1)$ where we have $1$. Now what about the determinant of the matrix? This is given by the determinant of the submatrix $A'$ of $A$ obtained by removing the first row and the first column of $A$. Then in $A'$ we necessarily have an invertible element and now use the same procedure as before in order to make $0$ on an entire column of $A$ excepting one entry where we have $1$. In the end we will get instead of $A$ a permutation matrix, so $A(x_1\dots x_n)^t$ is nothing but a permutation vector of $(x_1\dots x_n)^t$. Since we are in a local ring a permutation of a regular sequence is also regular. 
How to prove now that $y_1,\dots,y_n$ is an $R$-sequence? Will do this backwards by noticing that we have performed on the vector $(y_1\dots y_n)^t$ only two types of elementary transforms: multiplication by an invertible element of $R$ and/or the subtraction of an entry by another one multiplied by an arbitrary element of $R$. Are  these transforms keeping the property of being an $R$-sequence? While the first one obviously does it, the second needs some explanation. Assume that we have $z_1,\dots,z_n$ an $R$-sequence and want to show that $z_1,\dots, z_j+az_i,\dots,z_n$ is also an $R$-sequence. Since we are in a local ring we can change the order of the elements of the sequence, so we can assume that we have only two terms $z_1,z_2$ and want to prove that $z_1,z_2+az_1$ (respectively $z_1+az_2,z_2$, which is basically the same thing) is an $R$-sequence: if $(z_2+az_1)b=z_1c$ then $z_2b=z_1(c-ab)\in(z_1)$, so $b\in(z_1)$. $\ \square$
A less elementary proof (which works for sequences contained in the Jacobson radical of a noetherian ring) goes as follows: 
Let $I=(x_1,\dots,x_n)$. Then the grade of $I$ is $n$. On the other side, $I=(y_1,\dots,y_n)$ and this shows that $y_1,\dots,y_n$ is also an $R$-sequence. (For the last claim see, for instance, Kaplansky, Commutative Rings, Theorem 129.)
