Does this property of determinants generalize?

Question

Consider the following $2\times 2$ matrix $$ A = \begin{pmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{pmatrix}. $$

Let $\Delta = a_{11}a_{22}-a_{21}a_{12}$ be its determinant.

Then $A$ has full rank iff $\Delta \ne 0.$

I noticed that it holds that $$ A \begin{pmatrix} a_{22} \\ -a_{21} \end{pmatrix} = \begin{pmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{pmatrix} \begin{pmatrix} a_{22} \\ -a_{21} \end{pmatrix} = \begin{pmatrix} \Delta \\ 0 \end{pmatrix}. $$ That is: A linear combination of the columns of $A$ with coefficients that, up to their signs, consist of the elements of the second row of $A$ yields a vector that consists of a zero and the determinant of $A.$

This immediately yields that $A$ is singular if $\Delta=0:$ If all numbers in the second row of $A$ are zero, $A$ is singular. Otherwise we have found a nontrival vector in the kernel.

Question: Is there a similar intuition for the determinant of $n\times n$ matrices?

Answer 1

Yes, that generalizes...but not the way you think, probably.

1. You can use the first row instead (with alternating signs and shuffled order) and get a similar result. Alternatively, you could make a modified version of $A$ where we write $$ A = \pmatrix{a & b \\c & d}\\ A^+ = \pmatrix{d & -c \\-b & a} $$ where the elements are permuted a bit, and assigned +/- signs in a checkerboard pattern. When you do this, you find that $A^+ v = \pmatrix{\Delta\\0}$ when $v$ is the first row. The matrix $A^+$ is called the "classical adjoint" of $A$.

2. Generalization: For $n > 2$, the classical adjoint has, in position $i,j$, the determinant of the $(n-1) \times (n-1)$ matrix you get by deleting from $A$ the $i$th row and $j$th column, multiplied by $(-1)^{i+j}$ (assuming your indexing starts at "1"). Call this $A^+$. Then it's easy to see that the first entry of $A^+ v$, where $v$ is the first row of $A$, written as a column vector, is just the standard determinant expansion; it takes just a bit more work to see that you get $0$ if the entries are those of any other row. So that's the generalization of your observation.