Invertibility test: $\det A\neq0$ for square matrices

For square $A$, $\det A\neq 0$ is equivalent to invertibility on finite-dimensional spaces: trivial nullspace, independent columns, and a genuine inverse $A^{-1}$ . Bijective linear maps preserve dimension, so volume cannot collapse.

When $\det A=2$ in $\mathbb{R}^2$, every planar region scales in area by factor $2$ while orientation stays positive. Inverse scaling follows $\det(A^{-1})=1/\det(A)$.

Do not over-import this test to infinite-dimensional operators: determinants as volume forms need finite matrices. In data science you still meet determinant thinking through covariance volumes and change-of-variables Jacobians, but the formal test is square and finite. When $\det A=2$ in $\mathbb{R}^2$, a region of area $5$ maps to area $10$; when $\det A=0$, even a large region can collapse to a line or point in the image. Invertibility, full rank, and trivial nullspace are equivalent for square matrices . Bijective linear maps on finite-dimensional spaces preserve dimension, so volume cannot collapse to zero. Inverse volume scaling follows $\det(A^{-1})=1/\det(A)$. Areas in $\mathbb{R}^2$ scale by $\lvert\det A\rvert$.