21. Inverses: examples

Method for finding the inverse

Theorem:

0.30 Given an $n$ -by- $n$ matrix $A$ , form the augmented matrix $(A|I)$ (i.e. you have the $n$ -by- $n$ identity matrix on the right hand side of the bar). Apply row operations to this augmented matrix until you obtain an augmented matrix $(B|C)$ with $B$ in reduced echelon form.

If $B=I$ then $A$ is invertible with inverse $C$ .
If $B\neq I$ then $A$ is not invertible.

3.17 So if $A$ is invertible, you start with $(A|I)$ and you end up with $(I|A^{-1})$ . We're going to postpone the proof for now, and work out some examples before developing the theory necessary for the proof.

Examples

3.42 Let $A=\begin{pmatrix}1&-1\\ 1&1\end{pmatrix}$ . The augmented matrix is

$\begin{pmatrix}1&-1&|&1&0\\ 1&1&|&0&1\end{pmatrix}.$ To put the left-hand block into reduced echelon form, we do

$R_{2}\mapsto R_{2}-R_{1}$ :

$\begin{pmatrix}1&-1&|&1&0\\ 0&2&|&-1&1\end{pmatrix},$ then we do

$R_{2}\mapsto\frac{1}{2}R_{2}$ to get:

$\begin{pmatrix}1&-1&|&1&0\\ 0&1&|&-1/2&1/2\end{pmatrix}.$ Next we do

$R_{1}\mapsto R_{1}+R_{2}$ , which gives:

$\begin{pmatrix}1&0&|&1/2&1/2\\ 0&1&|&-1/2&1/2\end{pmatrix}.$ Now the left-hand block is in reduced echelon form; it is the identity matrix, so

$A$ is invertible and its inverse is the right-hand block:

$\begin{pmatrix}1/2&1/2\\ -1/2&1/2\end{pmatrix}.$ You can check this agrees with the 2-by-2 inverse we talked about last time.

7.42 Here's a messy example. Let's invert $A=\begin{pmatrix}1&2&3\\ 0&1&4\\ 1&5&0\end{pmatrix}$ . We have

$(A|I)=\begin{pmatrix}1&2&3&|&1&0&0\\ 0&1&4&|&0&1&0\\ 1&5&0&|&0&0&1\end{pmatrix}.$ Perform

$R_{3}\mapsto R_{3}-R_{1}$ to clear the first column:

$\begin{pmatrix}1&2&3&|&1&0&0\\ 0&1&4&|&0&1&0\\ 0&3&-3&|&-1&0&1\end{pmatrix}.$ Next,

$R_{3}\mapsto R_{3}-3R_{2}$ gets us into echelon form:

$\begin{pmatrix}1&2&3&|&1&0&0\\ 0&1&4&|&0&1&0\\ 0&0&-15&|&-1&-3&1\end{pmatrix}.$ 10.34 Now

$R_{3}\mapsto-R_{3}/15$ gives:

$\begin{pmatrix}1&2&3&|&1&0&0\\ 0&1&4&|&0&1&0\\ 0&0&1&|&1/15&1/5&-1/15\end{pmatrix}.$ Clear the last column with

$R_{2}\mapsto R_{2}-4R_{3}$ and

$R_{1}\mapsto R_{1}-3R_{3}$ to get:

$\begin{pmatrix}1&2&0&|&4/5&-3/5&1/5\\ 0&1&0&|&-4/15&1/5&4/15\\ 0&0&1&|&1/15&1/5&-1/15\end{pmatrix}.$ 13.52 Finally,

$R_{1}\mapsto R_{1}-2R_{2}$ yields:

$\begin{pmatrix}1&0&0&|&4/3&-1&-1/3\\ 0&1&0&|&-4/15&1/5&4/15\\ 0&0&1&|&1/15&1/5&-1/15\end{pmatrix}.$ 15.53 The left-hand block is the identity matrix, so

$A$ is invertible and the right-hand block is

$A^{-1}$ .

16.20 This means $A^{-1}=\begin{pmatrix}4/3&-1&-1/3\\ -4/15&1/5&4/15\\ 1/15&1/5&-1/15\end{pmatrix}$ . We can check this by computing $A^{-1}A$ : we get $I$ .