square matrix

Square matrix

$A\in K^{n\times n}$ is a square matrix since it has the same number of rows and columns.

The determinant of a square matrix $A\in K^{n\times n}$ is non-zero $⟺$ its rank is $n$.

$$\det A\ne 0⟺\text{rank}(A)=n$$

Proof:
Row operations don’t change the determinant. The row echelon form of a square matrix is upper triangular, whose determinant is the product of diagonal entries. Full rank means all $n$ pivots are non-zero, so $\det A\ne 0$. If rank $<n$, at least one pivot is zero, so $\det A=0$.

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