# 3.5 Invertible matrices

###### Definition 3.5.1.

An $n\times n$ matrix $A$ is called invertible if and only if there exists an $n\times n$ matrix $B$ such that $AB=BA=I_{n}$.

If there is such a matrix $B$, we can prove that there is only one such matrix $B$:

###### Proposition 3.5.1.

If $AB=BA=I_{n}$ and $AC=CA=I_{n}$ then $B=C$.

###### Proof.
 $\displaystyle B$ $\displaystyle=BI_{n}$ Theorem 3.4.2 $\displaystyle=B(AC)$ $\displaystyle=(BA)C$ associativity $\displaystyle=I_{n}C$ $\displaystyle=C$ Theorem 3.4.2

This means that when a matrix is invertible we can talk about the inverse of $A$. We write $A^{-1}$ for the inverse of $A$ when it exists.

## 3.5.1 Matrices with rows or columns of zeroes are not invertible

###### Theorem 3.5.2.

If an $n\times n$ matrix $A$ has a row of zeroes, or a column of zeroes, then it is not invertible.

###### Proof.

Suppose $A$ has a column of zeroes and that $B$ is any other $n\times n$ matrix. By Theorem 3.2.3, the columns of $BA$ are $B$ times the columns of $A$. In particular, one of these columns is $B$ times the zero vector, which is the zero vector. Since one of the columns of $BA$ is all zeroes, $BA$ is not the identity.

If $A$ has a row of zeroes, we can make a similar argument using Theorem 3.2.6. ∎

## 3.5.2 Inverse of a product of matrices

If you multiply any number of invertible matrices together, the result is invertible. Recall the shoes-and-socks result about the inverse of a composition of two functions: exactly the same thing is true.

###### Theorem 3.5.3.

If $A_{1},\ldots,A_{k}$ are invertible $n\times n$ matrices then $A_{1}\cdots A_{k}$ is invertible with inverse $A_{k}^{-1}\cdots A_{1}^{-1}$.

The proof is the same as for functions: you can simply check that $A_{k}^{-1}\cdots A_{1}^{-1}$ is a two sided inverse to $A_{1}\cdots A_{k}$ using the associativity property for matrix multiplication.

This theorem has a useful corollary about when matrix products are invertible.

###### Corollary 3.5.4.

Let $A$ and $E$ be $n\times n$ matrices with $E$ invertible. Then $EA$ is invertible if and only if $A$ is invertible, and $AE$ is invertible if and only if $A$ is invertible.

###### Proof.

If $A$ is invertible then the theorem tells us that so are $EA$ and $AE$.

Suppose $EA$ is invertible. Certainly $E^{-1}$ is invertible (its inverse is $E$), so by the theorem $E^{-1}EA$ is invertible, that is, $A$ is invertible. The argument for $AE$ is similar. ∎