Inverse Matrices and Invertibility

Definition Let $A$ be an $n \times n$ matrix. An $n \times n$ matrix $C$ is an inverse of $A$ if

When an inverse exists, $A$ is called invertible, or nonsingular, and the inverse is unique. Indeed, if $C$ and $D$ are both inverses of $A$, then

The inverse is written $A^{-1}$. The equation $CA = I_n$ says that applying $A$ and then $C$ returns every input. The equation $AC = I_n$ says that for every output $\mathbf{b}$ the input $C\mathbf{b}$ is sent to $\mathbf{b}$. Here inverses are defined only for square matrices.

Definition A square matrix that has no inverse is singular. An $n \times n$ matrix $A$ is singular exactly when some nonzero vector $\mathbf{v}$ satisfies $A\mathbf{v} = \mathbf{0}$. One direction is short. If $A\mathbf{v} = \mathbf{0}$ with $\mathbf{v} \ne \mathbf{0}$ and $C$ were an inverse of $A$, then

which is a contradiction. The other direction follows from the pivot test below. A singular matrix has fewer than $n$ pivots, so the system $A\mathbf{x} = \mathbf{0}$ has a free variable, and setting it equal to $1$ gives a nonzero solution. The words singular and nonsingular are used only for square matrices.

Definition Let $A$ be an $n \times n$ matrix. Form the $n \times 2n$ matrix $[A \mid I_n]$ and apply elementary row operations to whole rows. If the left block can be reduced to $I_n$, the right block is then $A^{-1}$. If an echelon form of the left block has fewer than $n$ pivots, then $A$ is singular. The method works because the row operations together amount to multiplying on the left by one invertible matrix $E$, the product of their elementary matrices, so they turn $[A \mid I_n]$ into $[EA \mid E]$. If $EA = I_n$, then $A = E^{-1}$, so $AE = E^{-1}E = I_n$ as well, and $E$ is a two-sided inverse. Column by column, the $j$th column of $A^{-1}$ is the solution of $A\mathbf{x} = \mathbf{e}_j$, and the method solves these $n$ systems at once.

Definition For an $n \times n$ matrix $A$ the following statements are equivalent.

1. The matrix $A$ has a pivot in every column, that is, $n$ pivots.
2. For every $\mathbf{b}$ in $\mathbb{R}^n$ the system $A\mathbf{x} = \mathbf{b}$ has exactly one solution.
3. The matrix $A$ is invertible.

The reasons form a cycle. With $n$ pivots, $A$ row reduces to $I_n$, and row reduction of $[A \mid I_n]$ produces $A^{-1}$. If $A$ is invertible, every system has exactly one solution, namely $A^{-1}\mathbf{b}$. If every system has exactly one solution, then $A\mathbf{x} = \mathbf{0}$ has only $\mathbf{x} = \mathbf{0}$, so no variable is free and every column has a pivot. The test needs $A$ to be square.

Definition If $A$ is an invertible $n \times n$ matrix, then for every $\mathbf{b}$ in $\mathbb{R}^n$ the system $A\mathbf{x} = \mathbf{b}$ has exactly one solution,

It is a solution because $A(A^{-1}\mathbf{b}) = (AA^{-1})\mathbf{b} = \mathbf{b}$. It is the only one because any solution satisfies $\mathbf{x} = A^{-1}(A\mathbf{x}) = A^{-1}\mathbf{b}$. To solve a single system, row reducing $[A \mid \mathbf{b}]$ takes less work than computing $A^{-1}$ first, and both methods give the same vector.