Suppose we have a partitioned matrix $A=(A_{ij}), i,j=1,\ldots,n$. If $A_{ij}$ are normal elements, we have the usual Leibniz formula to calculate the determinant of $A$:
$$\sum_{\sigma\in S_n}\bigl(\text{sgn}(\sigma)\prod_{i=1}^nA_{i,\sigma_i}\bigl)$$
where $S_n$ is the set of all permutations.
If $A_{ij}$ are general submatrix (they are square matrices of the same sizes), do we have the above Leibniz formula where $A_{i,\sigma_i}$ now denotes the determinant of this submatrix? For example, if
$$A=\left( \begin{array} &A_{11}&A_{12} \\A_{21}&A_{22} \end{array}\right),$$
do we have
$$|A|=|A_{11}||A_{22}|-|A_{12}||A_{21}|?$$
If it is the case, it would relieve my memory for math formulas. However, I searched google for such a formula but with no luck. So, is the above Leibniz formula true for general submatrices? Thanks.
Best Answer
No.
A counterexample is $\begin{pmatrix} 2&1&1&1 \\1&2&1&1 \\1&1&2&1 \\1&1&1&2 \end{pmatrix}$ with determinant $5$
while $\begin{pmatrix} 2&1 \\1&2 \end{pmatrix}$ has determinant $3$ and $\begin{pmatrix} 1&1 \\1&1 \end{pmatrix}$ has determinant $0$
since $5 \not = 3\times 3 - 0\times 0$.