Krdytkyu

For an integer $n$ ($2$ in your case) the application:

$$begin{array}{l|rcl}
varphi : & mathbb Z & longrightarrow & mathbb Z_n\
& x & longmapsto & overline{x}
end{array}$$

is a ring homomorphism.

Hence for a multivariate polynomial $P(x_1, dots, x_m)$ you have $varphi(P) = overline{P}$ where $overline{P}$ is the polynomial obtained by taking the modulus of all coefficients.

Now $$det A = sum_{sigma in mathcal S_n} (-1)^{epsilon(sigma)} a_{1 sigma(1)} dots a_{n sigma(n)}$$ is a multivariate polynomial of the coefficients of $A$. Therefore the conclusion. See Wikipedia - determinant for above formula on the determinant.

If $x equiv y pmod N$ and if ${a_i}_{i=0}^n subseteq mathbb Z$, then

$$sum_{i=0}^n a_i x^i equiv sum_{i=0}^n a_i y^i pmod N$$

An example should help.

Let $A = left[begin{array}{c} 11 & 13 \ 17 & 19end{array}right]$.

Then $det A = 11 times 19 - 13 times 17 = -12 equiv 2 pmod 7$

Also $bar A = A pmod 7 =
left[begin{array}{c}
4 & 6 \
3 & 5
end{array}right]$

and $det bar A =4 times 5 - 6 times 3
equiv 11 times 19 - 13 times 17
equiv det A pmod 7$

That is because you have a matrix in $mathcal M_n(bf Z)$ and the canonical map
begin{align}
pi:mathbf Z&longrightarrow mathbf Z/2mathbf Z \
n&longmapsto nbmod 2
end{align}
is a ring homomorphism, i.e. it is compatible with addition and multiplication.

Some details:

Let's use the general formula for an $ntimes n$ determinant,denoting $overline x$ the reduction of $xbmod2$ (or any modulus):

$$overline{det(a_{ij})}=sum_{sigmainmathfrak S_n}varepsilon(sigma)overline{a_{sigma(1),1} a_{sigma(2),2}dotsm a_{sigma(n),n}}=sum_{sigmainmathfrak S_n}varepsilon(sigma)overline{a_{sigma(1),1}}:overline{a_{sigma(2),2}}dotsm overline{a_{sigma(n),n}}=det(overline{a_{i,j}}).$$

If $,f(x,y),$ is a polynomial with integer coefficients then by the polynomial congruence rule

$$bmod 2!:, f(m,n),equiv, f(overbrace{mbmod 2}^{large overline{m}},, overbrace{nbmod 2}^{largeoverline{n}})qquad $$

So $,f(m,n) = 0,Rightarrow, f(bar m,bar n) equiv 0pmod{!2} $ so contra+, $ f(bar m,bar n) notequiv 0pmod{!2},Rightarrow, f(m,n)neq 0$

i.e. any root $,(m,n),$ of $f$ persists as a root $!bmod 2.,$ OP is the special case when $f$ is the determinant function - which is indeed a polynomial function of its entries (with integer coefficients). Concretely

$$d = left|begin{array}{}color{#c00}{11} & color{#0a0}{22}\ color{#0a0}{33} & color{#c00}{55}end{array}right| = left{begin{align}&f(11,22,33,55) = color{#c00}{11(55)}!-!color{#0a0}{22(33)}\
equiv &f( 1, 0, 1, 1) equiv color{#c00}{ 1( 1)}-color{#0a0}{0( 1)}equiv 1!!!pmod{!2}end{align}right.qquad$$

therefore $,dequiv 1pmod{!2},,$ i.e. $,d,$ is odd, so $,dneq 0.$

Remark $ $ The same works for any modulus $,m,,$ e.g. we could use $,m = 9,$ ("casting out nines") exactly as above (or more generally as a check of any such polynomial computation), or we could compare unit digits $bmod m!=!10!: ,dequiv color{#c00}{1(5)}-color{#0a0}{2(3)}equiv -1 $ so $,dneq 0.$

For motivation you might find instructive this post on the parity root test for polynomials. Such parity arguments are a prototypical way of solving ring (algebraic) problems via information gleaned from (simpler) modular images (an algebraic way of "dividing and conquering").

Here is perhaps the simplest answer to the specific question, avoiding the use of ring homomorphisms and residue classes. (Note: You should learn about ring homomorphisms and residue classes if you want to get anywhere in abstract algebra; you won't always be able to work as concretely as below.)

Fix a nonnegative integer $n$. We let $left[nright]$ denote the set $left{1,2,ldots,nright}$.

Lemma 1. Let $A = left(a_{i,j}right)_{i, j in left[nright]}$ and $B = left(b_{i,j}right)_{i, j in left[nright]}$ be two $ntimes n$-matrices of integers. Let $N$ be an integer. Assume that
begin{equation}
a_{i,j} equiv b_{i,j} mod N
label{darij.eq.l1.1}
tag{1}
end{equation}
for all $i, j in left[nright]$. Then, $det A equiv det B mod N$.

Proof of Lemma 1. The Leibniz formula for determinants yields $det A = sumlimits_{sigma in S_n} left(-1right)^{sigma} prodlimits_{i=1}^n a_{i, sigmaleft(iright)}$ and $det B = sumlimits_{sigma in S_n} left(-1right)^{sigma} prodlimits_{i=1}^n b_{i, sigmaleft(iright)}$ (where $S_n$ denotes the set of all permutations of $left[nright]$, and where $left(-1right)^{sigma}$ denotes the sign of a permutation $sigma$). Thus,
begin{align}
det A
& = sumlimits_{sigma in S_n} left(-1right)^{sigma} prodlimits_{i=1}^n underbrace{a_{i, sigmaleft(iright)}}_{substack{equiv b_{i, sigmaleft(iright)} mod N \ left(text{by eqref{darij.eq.l1.1}}right)}} \
& equiv sumlimits_{sigma in S_n} left(-1right)^{sigma} prodlimits_{i=1}^n b_{i, sigmaleft(iright)}
= det B mod N
end{align}
(here, we tacitly used the fact that everything in our expressions is an integer, and that congruences modulo $N$ can be multiplied).
This proves Lemma 1. $blacksquare$

Corollary 2. Let $A$ be an $n times n$-matrix of integers such that all diagonal entries of $A$ are odd and all non-diagonal entries of $A$ are even. Then, $det A$ is odd.

Proof of Corollary 2. Write the matrix $A$ as $A = left(a_{i,j}right)_{i, j in left[nright]}$. Also, write the identity $ntimes n$-matrix $I_n$ as $I_n = left(b_{i,j}right)_{i, j in left[nright]}$. Then, it is easy to prove the congruence $a_{i,j} equiv b_{i,j} mod 2$ for all $i, j in left[nright]$. (Indeed, if $i = j$, then $a_{i,j}$ is a diagonal entry of $A$, whereas $b_{i,j}$ is a diagonal entry of $I_n$ and thus equals $1$; hence, this congruence says that all diagonal entries of $A$ are odd. This is true by assumption. On the other hand, if $i neq j$, then $a_{i,j}$ is a non-diagonal entry of $A$, whereas $b_{i,j}$ is a non-diagonal entry of $I_n$ and thus equals $0$; hence, this congruence says that all non-diagonal entries of $A$ are even. This is again true by assumption. Thus, the congruence holds in both cases $i = j$ and $i neq j$.)

Thus, Lemma 1 (applied to $N=2$ and $B=I_n$) yields $det A equiv detleft(I_nright) = 1 mod 2$. In other words, $det A$ is odd. $blacksquare$

Corollary 2 can be directly applied to your matrix $A$.