3 Matrices 3.12 Invertibility and RREF Further reading

3.13 Finding inverses

Let $A$ be a square matrix. We now have a method of determining whether or not $A$ is invertible: do row operations to $A$ until you reach a matrix in RREF. Then by Theorem 3.12.1 $A$ is invertible if and only if the RREF matrix is invertible.

What if we actually want to know what the inverse matrix is? You probably already know that a $2\times 2$ matrix $A=\begin{pmatrix}a&b\\ c&d\end{pmatrix}$ is invertible if and only if $ad-bd\neq 0$ , and in this case

A^{-1}=\frac{1}{ad-bc}\begin{pmatrix}d&-b\\ -c&a\end{pmatrix}

This formula does generalise to larger matrices, but not in a way which is easy to use: for example, the general formula for the inverse of a $3\times 3$ invertible matrix $A=(a_{ij})$ is

A^{-1}=\frac{1}{\Delta}\begin{pmatrix}\begin{vmatrix}a_{22}&a_{23}\\ a_{32}&a_{33}\end{vmatrix}&-\begin{vmatrix}a_{12}&a_{13}\\ a_{32}&a_{33}\end{vmatrix}&\begin{vmatrix}a_{12}&a_{13}\\ a_{22}&a_{23}\end{vmatrix}\\ -\begin{vmatrix}a_{21}&a_{23}\\ a_{31}&a_{33}\end{vmatrix}&\begin{vmatrix}a_{11}&a_{13}\\ a_{31}&a_{33}\end{vmatrix}&-\begin{vmatrix}a_{11}&a_{13}\\ a_{21}&a_{23}\end{vmatrix}\\ \begin{vmatrix}a_{21}&a_{22}\\ a_{31}&a_{32}\end{vmatrix}&-\begin{vmatrix}a_{11}&a_{12}\\ a_{31}&a_{32}\end{vmatrix}&\begin{vmatrix}a_{11}&a_{12}\\ a_{21}&a_{22}\end{vmatrix}\end{pmatrix}

where $\begin{vmatrix}a&b\\ c&d\end{vmatrix}$ means $ad-bc$ and

\Delta=a_{11}a_{22}a_{33}+a_{12}a_{23}a_{31}+a_{13}a_{21}a_{32}-a_{11}a_{23}a_% {32}-a_{12}a_{21}a_{33}-a_{13}a_{22}a_{31}.

This isn’t a formula that you want to use. Luckily we can use RREF techniques to determine invertibility and find inverses.

3.13.1 How to determine invertibility and find inverses

Let $A$ be an $n\times n$ matrix, and suppose we want to find out whether $A$ is invertible and if so what its inverse is. Let $I_{n}$ be the $n\times n$ identity matrix. Here is a method:

1.

Form the super-augmented matrix $(A\mid I_{n})$ .
2.

Do row operations to put this into RREF.
3.

If you get $(I_{n}\mid B)$ then $A$ is invertible with inverse $B$ .
4.

If the first part of the matrix isn’t $I_{n}$ then $A$ isn’t invertible.

It works because the first part of the matrix is a RREF matrix resulting from doing row operations to $A$ , so if it is $I_{n}$ then by Theorem 3.12.1 $A$ is invertible, and if it is not $I_{n}$ then $A$ is not invertible. It just remains to explain why, in the case $A$ is invertible, you end up with $(I_{n}\mid A^{-1})$ .

Think about the columns $\mathbf{c}_{1},\ldots,\mathbf{c}_{n}$ of the inverse of $A$ . We have $A(\mathbf{c}_{1}\;\cdots\;\mathbf{c}_{n})=I_{n}$ , so $A\mathbf{c}_{1}=\mathbf{e}_{1}$ , $A\mathbf{c}_{2}=\mathbf{e}_{2}$ , etc, where $\mathbf{e}_{i}$ is the ith column of $I_{n}$ . So $\mathbf{c}_{1}$ is the unique solution of the matrix equation $A\mathbf{x}=\mathbf{e}_{1}$ . You find that by putting $(A\mid\mathbf{e}_{1})$ into RREF, and you must get $(I_{n}\mid\mathbf{c}_{1})$ since $\mathbf{c}_{1}$ is the only solution.

Repeating that argument for every column, when we put $(A\mid\mathbf{e}_{1}\;\cdots\;\mathbf{e}_{n})$ into RREF we get $(I_{n}\mid\mathbf{c}_{1}\;\cdots\;\mathbf{c}_{n})$ , that is, $(I_{n}\mid A^{-1})$ .

Example 3.13.1.

Let $A=\begin{pmatrix}1&2\\ 3&4\end{pmatrix}$ . To find whether $A$ is invertible, and if so what its inverse is, we put $(A\mid I_{2})$ into RRE form:

	$\displaystyle\begin{pmatrix}1&2&1&0\\ 3&4&0&1\end{pmatrix}$	$\displaystyle\xmapsto{\textbf{r}_{2}\mapsto\textbf{r}_{2}-3\textbf{r}_{1}}% \begin{pmatrix}1&2&1&0\\ 0&-2&-3&1\end{pmatrix}$
		$\displaystyle\xmapsto{\textbf{r}_{2}\mapsto(-1/2)\textbf{r}_{2}}\begin{pmatrix% }1&2&1&0\\ 0&1&3/2&-1/2\end{pmatrix}$
		$\displaystyle\xmapsto{\textbf{r}_{1}\mapsto\textbf{r}_{1}-2\textbf{r}_{2}}% \begin{pmatrix}1&0&-2&1\\ 0&1&3/2&-1/2\end{pmatrix}$

This is in RRE form, so the inverse of $A$ is

\begin{pmatrix}-2&1\\ 3/2&-1/2\end{pmatrix}

as you can check by multiplying them together.