I am trying to follow the proof of the transformation behaviour of theta functions associated to positive definite integer-valued quadratic forms in Iwaniec's Topics in Classical Automorphic Forms. One detail that I am a bit stuck on is how to arrive at expression (10.46) in the proof of Theorem 10.9.
To summarise the situation: We are given a positive definite symmetric integer matrix $A\in M_r(\mathbb{Z})$ of even rank $r\equiv 0\pmod{2}$ with even diagonal entries $\operatorname{Diag}A \equiv 0\pmod{2}$ and odd determinant $|A|\equiv 1\pmod{2}$. I am trying to show the equivalence $$|A|\equiv (-1)^{r/2}\pmod{4}.$$ The claim is clear for $r=2$ since for a matrix $A=\begin{pmatrix} a & b \\ b & c \end{pmatrix}$ satisfying the above conditions $$|A|=ac-b^2\equiv -b^2\equiv -1 \pmod{4}.$$ However I am not sure how to proceed so any help or hints would be greatly appreciated.
The author seems to suggest that there is some neat trick (probably some parity argument), but I'll prove the statement by mathematical induction on $r$. In the inductive step, as $A$ has an even diagonal but an odd determinant, some of its off-diagonal entries must be odd. By permuting the rows and columns of $A$ if necessary, we may assume that $$ A =\left(\begin{array}{c|c}X&Y^T\\ \hline Y&Z\end{array}\right) =\left(\begin{array}{cc|c}a&b&u^T\\ b&c&v^T\\ \hline u&v&Z\end{array}\right) $$ where $b$ is odd and $Z$ an $(r-2)\times(r-2)$ symmetric integer matrix. Thus $$ x:=\det(X)=ac-b^2\equiv-b^2\equiv-1\pmod 4,\tag{1} $$ meaning that $x=\det(X)$ is odd and $X$ is nonsingular. (This also proves the base case where $r=2$.) Therefore, using Schur complement, we obtain \begin{align} x^{r-3}\det(A) &=x^{r-3}\det(X)\det(Z-YX^{-1}Y^T)\\ &=x^{r-2}\det(Z-YX^{-1}Y^T)\\ &=\det\left(xZ-Y\operatorname{adj}(X)Y^T\right)\\ &=\det\left(xZ-\pmatrix{u&v}\pmatrix{c&-b\\ -b&a}\pmatrix{u^T\\ v^T}\right)\\ &=\det\underbrace{\left(xZ-cuu^T-avv^T+b(uv^T+vu^T)\right)}_S.\tag{2} \end{align} Since $a,c$ and the diagonal entries of $Z$ are all even, $S$ has an even diagonal. Also, $\det(S)=x^{r-3}\det(A)$ is odd because both $x$ and $\det(A)$ are odd. Therefore, by induction assumption, $\det(S)\equiv(-1)^{(r-2)/2}\pmod4$. Consequently, $$ (-1)^{r-3}\det(A)\equiv x^{r-3}\det(A)=\det(S)\equiv (-1)^{(r-2)/2}\pmod4, $$ i.e. $\det(A)\equiv (-1)^{r-3}(-1)^{(r-2)/2}=(-1)^{r/2}\pmod4$.