# Does this property of determinants generalize?

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?

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.