To show that the matrix $A$ is nonsingular, it suffices to prove that $\det(A)\neq 0$.

One way is to compute the determinant of $A$ directly.

However, as the numbers in $A$ are quite large for hand computation, the direct calculation must be tedious.

So we consider an alternative method.
Note that we do not have to find the exact value of $\det(A)$, but we just need to know $\det(A)\neq 0$.

Thus, it suffices to show that $\det(A)$ is odd. ($0$ is an even number.)
This suggests that considering the matrix modulo $2$ is helpful.

Let $\bar{A}$ be the matrix whose $(i, j)$-entry is the $(i,j)$-entry of $A$ modulo $2$.
That is,
\begin{align*}
\bar{A}:=\begin{bmatrix}
1 & 0 & 0 & 0 \\
0 &1 & 0 & 0 \\
0 & 0 & 1 & 0 \\
0 & 0 & 0 & 1
\end{bmatrix}.
\end{align*}
(Remark that the diagonal entries of $A$ are odd, and off-diagonal entries are even.)

Since $\det(A)$ is a polynomial of entries of $A$, we have
\begin{align*}
\det(A) &\equiv \det(\bar{A}) \pmod{2}\\
&=1.
\end{align*}

It follows that $\det(A)$ is odd, and in particular $\det(A)\neq 0$.
Thus the matrix $A$ is nonsingular.

What’s $\det(A)$ anyway?

Just for the record, the determinant of $A$ is
\[\det(A)=-20330769121541702776233175.\]

Beautiful Formulas for $\pi$

This problem was nothing to do with the number $\pi$ (except we used the digits of $\pi$) and the matrix is far from beautiful.
(Although the method we used is beautiful.)

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