Inverse and Singular Matrices¶

Definition¶

For a square matrix $\boldsymbol{A}$, an inverse matrix is a matrix $\boldsymbol{A}^{-1}$ such that

If such a matrix exists, then $\boldsymbol{A}$ is invertible, nonsingular, or nondegenerate. If no inverse exists, then $\boldsymbol{A}$ is singular, non-invertible, or degenerate.

Interpretation¶

An inverse matrix undoes the action of a linear map: if $\boldsymbol{A}$ sends $\boldsymbol{v}$ to $\boldsymbol{A}\boldsymbol{v}$, then $\boldsymbol{A}^{-1}$ sends $\boldsymbol{A}\boldsymbol{v}$ back to $\boldsymbol{v}$. Singular matrices are precisely those whose action cannot be undone: they send some nonzero vector to $\boldsymbol{0}$, so different inputs share an output.

$\boldsymbol{A}$ maps the unit square onto the parallelogram spanned by its columns and $\boldsymbol{v}$ onto $\boldsymbol{A}\boldsymbol{v}$; the arrow labelled $\boldsymbol{A}^{-1}$ leads back, $\boldsymbol{A}^{-1}(\boldsymbol{A}\boldsymbol{v})=\boldsymbol{v}$ for every $\boldsymbol{v}$. This is possible because the parallelogram is not flattened: $\boldsymbol{A}$ is invertible if and only if $\det \boldsymbol{A}\neq0$$0$.

The defining equalities use matrix products, which represent composition of linear maps, and the identity matrix, which leaves every vector unchanged. So $\boldsymbol{A}^{-1}\boldsymbol{A}=\boldsymbol{I}$ says that first $\boldsymbol{A}$ and then $\boldsymbol{A}^{-1}$ changes nothing, and $\boldsymbol{A}\boldsymbol{A}^{-1}=\boldsymbol{I}$ says the same in the other order. For square matrices either equality already implies the other, and an inverse, when it exists, is unique.

An inverse makes every system $\boldsymbol{A}\boldsymbol{x}=\boldsymbol{b}$ uniquely solvable, with $\boldsymbol{x}=\boldsymbol{A}^{-1}\boldsymbol{b}$${\mathbf{A}}^{-1}\mathbf{b}$. To compute it, row-reduce $(\boldsymbol{A}\mid\boldsymbol{I})$ by Gauss-Jordan elimination to $(\boldsymbol{I}\mid\boldsymbol{A}^{-1})$${\mathbf{A}}^{-1})$, or use the cofactor formula of the adjugate matrix.

Two slips are common. The inverse of a product reverses the order, $(\boldsymbol{A}\boldsymbol{B})^{-1}=\boldsymbol{B}^{-1}\boldsymbol{A}^{-1}$${\mathbf{B}}^{-1}{\mathbf{A}}^{-1}$: to undo "first $\boldsymbol{B}$, then $\boldsymbol{A}$", undo $\boldsymbol{A}$ first. And $\boldsymbol{A}^{-1}$ is not the matrix of reciprocals $1/a_{ij}$: $\begin{pmatrix}1&1\\0&1\end{pmatrix}$ has a zero entry, yet its inverse is $\begin{pmatrix}1&-1\\0&1\end{pmatrix}$.

Key Results¶

• If $\boldsymbol{A}$ and $\boldsymbol{B}$ are invertible, then $(\boldsymbol{A}\boldsymbol{B})^{-1}=\boldsymbol{B}^{-1}\boldsymbol{A}^{-1}$${\mathbf{B}}^{-1}{\mathbf{A}}^{-1}$.
• A square matrix is invertible if and only if its determinant is nonzero, if and only if it has full rank.
• If $\boldsymbol{A}$ is invertible, then $(\boldsymbol{A}^{\top})^{-1}=(\boldsymbol{A}^{-1})^{\top}$$({\mathbf{A}}^{-1}{)}^{\top }$.

Examples¶

• $\begin{pmatrix}1&0\\0&1\end{pmatrix}$ is its own inverse.
• $\begin{pmatrix}1&2\\2&4\end{pmatrix}$ is singular because its rows are dependent.

Prerequisites¶

• matrix
• Identity Matrix
• Matrix Product

Related Concepts¶

• Determinant
• Adjugate Matrix
• Gauss-Jordan Elimination
• Linear Map
• Rank

Sources¶

• Math II: Linear Algebra · Week 4
• Math V: Advanced Linear Algebra · Week 1 (revisited)

Aliases: matrix inverse Keywords: invertible matrix, nonsingular matrix, non-singular matrix, singular matrix, non-invertible matrix, degenerate matrix, nondegenerate matrix, inverse matrix

Last updated: 20 October 2026 03:02