Determinant of enlarged matrix

Question

Let $A$ be the $n\times n$ matrix with entries $A_{ij}$ and consider the enlarged $2n\times 2n$ matrix $$\tilde A=\begin{pmatrix}A_{11} & 0 & A_{12} & 0 & \dots & A_{1n} & 0 \\ 0 & A_{11} & 0 & A_{12} & \dots & 0 & A_{1n} \\ A_{21} & 0 & A_{22} & 0 & \dots & A_{2n} & 0 \\ 0 & A_{21} & 0 & A_{22} & \dots & 0 & A_{2n} \\ \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots \\ A_{n1} & 0 & A_{n2} & 0 & \dots & A_{nn} & 0 \\ 0 & A_{n1} & 0 & A_{n2} & \dots & 0 & A_{nn}\end{pmatrix} $$ that is, the matrix where every row is duplicated and padded with zeros. Is there a simple formula for the determinant of $\tilde A$ given the determinant of $A$?

Answer 1

Note that $$\widetilde{A} = A \otimes I,$$ for $\otimes$ the Kronecker product, see https://en.wikipedia.org/wiki/Kronecker_product. The determinant of Kronecker products admits a closed form, which is also presented in the same wikipedia article.

Answer 2

Consider small case for $n=2$. You can generalize the process for any $n$.

Say we have $A=\begin{pmatrix} a & b \\ c & d\end{pmatrix}$.

Then $\bar{A}=\begin{pmatrix} a & 0 & b & 0 \\ 0 & a & 0 & b \\ c & 0 & d & 0 \\ 0 & c & 0 & d \end{pmatrix}$.

Now interchange $C_{2}$ with $C_{3}$ to get $\begin{pmatrix} a & b & 0 & 0 \\ 0 & 0 & a & b \\ c & d & 0 & 0 \\ 0 & 0 & c & d \end{pmatrix}$.

Now interchange $R_{2}$ with $R_{3}$ to get $\begin{pmatrix} a & b & 0 & 0 \\ c & d & 0 & 0 \\ 0 & 0 & a & b \\ 0 & 0 & c & d \end{pmatrix}=\begin{pmatrix} A & 0 \\ 0 & A\end{pmatrix}$.

For any $n\times n$ matrix $A$, we can perform such row and column operations to get $\bar{A}=\begin{pmatrix} A & 0 \\ 0 & A\end{pmatrix}$.

Also note that total row and column operations required to simply $\bar{A}$ are even in numbers for any $n$. So, we have $\det(\bar{A})=(\det(A))^2$