**Read** Section 3.5 for next class. This is also **core** material.
We aren't covering 3.4.
Work through recommended homework questions.

**Quiz 5** will focus on 3.1, 3.2 and the first half of 3.3 (up to
and including Example 3.26).

**Midterm:** Saturday, October 25, 7-10pm.
It will cover the material up to and including the lecture on Monday, Oct 20.
Practice midterms are available on the exercises page.

**Office hour:** Today, 11:30-noon, MC103B.

**Help Centers:** Monday-Friday 2:30-6:30 in MC 106.

**Definition:** An **inverse** of an $n \times n$ matrix $A$ is an $n \times n$
matrix $A'$ such that
$$
A A' = I \qtext{and} A' A = I .
$$
If such an $A'$ exists, we say that $A$ is **invertible**.

**Theorem 3.6:** If $A$ is an invertible matrix, then its inverse is unique.

Because of this, we write $A^{-1}$ for **the** inverse of $A$,
when $A$ is invertible. We do *not* write $\frac{1}{A}$.

**Example:** If $A = \bmat{rr} 1 & 2 \\ 3 & 7 \emat$,
then $A^{-1} = \bmat{rr} 7 & -2 \\ -3 & 1 \emat$ is the inverse of $A$.

But the zero matrix and the matrix $B = \bmat{rr} -1 & 3 \\ 2 & -6 \emat$ are not invertible.

**Theorem 3.7:** If $A$ is an invertible matrix $n \times n$ matrix,
then the system $A \vx = \vb$ has the unique solution $\vx = A^{-1} \vb$
for any $\vb$ in $\R^n$.

**Remark:** This is **not** in general an efficient way to solve
a system.

**Theorem 3.8:** The matrix $A = \bmat{cc} a & b \\ c & d \emat$ is
invertible if and only if $ad - bc \neq 0$. When this is the case,
$$
A^{-1} = \frac{1}{ad-bc} \, \bmat{rr} \red{d} & \red{-}b \\ \red{-}c & \red{a} \emat .
$$

We call $ad-bc$ the **determinant** of $A$, and write it $\det A$.
It *determines* whether or not $A$ is invertible, and also shows
up in the formula for $A^{-1}$.

- $A^{-1}$ is invertible and $(A^{-1})^{-1} = \query{A}$
- If $c$ is a non-zero scalar, then $cA$ is invertible and $(cA)^{-1} = \query{\frac{1}{c} A^{-1}}$
- $AB$ is invertible and $(AB)^{-1} = \query{B^{-1} A^{-1}}$ (socks and shoes rule)
- $A^T$ is invertible and $(A^T)^{-1} = \query{(A^{-1})^T}$
- $A^n$ is invertible for all $n \geq 0$ and $(A^n)^{-1} = \query{(A^{-1})^n}$

To verify these, in every case you just check that the matrix shown is an inverse.

**Remark:** Property (c) is the most important, and generalizes to
more than two matrices, e.g. $(ABC)^{-1} = C^{-1} B^{-1} A^{-1}$.

**Remark:** For $n$ a positive integer, we define $A^{-n}$ to
be $(A^{-1})^n = (A^n)^{-1}$. Then $A^n A^{-n} = I = A^0$,
and more generally $A^r A^s = A^{r+s}$ for all integers $r$ and $s$.

**Remark:** There is no formula for $(A+B)^{-1}$.
In fact, $A+B$ might not be invertible, even if $A$ and $B$ are.

We can use these properties to solve a matrix equation for an unknown matrix.

There's no problem getting $A A' = I_2$. (Find an example.)

But it's not possible to have $A' A = I_3$ with the given sizes.

Suppose we did have $A' A = I_3$ with $A$ a $2 \times 3$ matrix.

Consider the homogenous system
$$A \colll x y z = \coll 0 0$$
Since $\rank\, A \leq 2$ and there are three variables, this system must
have infinitely many solutions.
But
$$
\kern-5ex
A \vx = \vec 0 \qimplies A' A \vx = A' \svec 0 \qimplies \vx = \vec 0 , $$
so there is only one solution. This is a contradiction.

More generally, unless $A$ is square, you can't find a matrix $A'$ that makes both $AA' = I$ and $A'A = I$ true.

**Theorem 3.12:**
Let $A$ be an $n \times n$ matrix. The following are equivalent:

a. $A$ is invertible.

b. $A \vx = \vb$ has a unique solution for every $\vb \in \R^n$.

c. $A \vx = \vec 0$ has only the trivial (zero) solution.

d. The reduced row echelon form of $A$ is $I_n$.

We'll use our past work on solving systems to show that (b) $\implies$ (c) $\implies$ (d) $\implies$ (b), which will prove that (b), (c) and (d) are equivalent.

We will only partially explain why (b) implies (a).

(b) $\implies$ (c): If $A \vx = \vb$ has a unique solution for every $\vb$, then it's true when $\vb$ happens to be the zero vector.

(c) $\implies$ (d):
Suppose that $A \vx = \vec 0$ has only the trivial solution.

That means that the rank of $A$ must be $n$.

So in reduced row echelon form, every row must have a leading $1$.

The only $n \times n$ matrix in reduced row echelon form with
$n$ leading $1$'s is the identity matrix.

(d) $\implies$ (b): If the reduced row echelon form of $A$ is $I_n$, then the augmented matrix $[A \mid \vb\,]$ row reduces to $[I_n \mid \vc\,]$, from which you can read off the unique solution $\vx = \vc$.

(b) $\implies$ (a) (partly):
Assume $A \vx = \vb$ has a solution for every $\vb$.

That means we can find $\vx_1, \ldots, \vx_n$ such that
$A \vx_i = \ve_i$ for each $i$.

If we let $B = [ \vx_1 \mid \cdots \mid \vx_n\,]$ be the matrix with the
$\vx_i$'s as columns, then
$$
\kern-8ex
AB = A \, [ \vx_1 \mid \cdots \mid \vx_n\,] = [ A \vx_1 \mid \cdots \mid A \vx_n\,]
= [ \ve_1 \mid \cdots \mid \ve_n \,] = I_n .
$$
So we have found a *right* inverse for $A$.

It turns out that $BA= I_n$ as well, but this is harder to see.
$\qquad\Box$

**Note:** We have omitted (e) from the theorem, since we aren't covering elementary matrices.
They are used in the text to prove the other half of (b) $\implies$ (a).

We will see many important applications of Theorem 3.12. For now, we illustrate one theoretical application and one computational application.

**Theorem 3.13:** Let $A$ be a square matrix. If $B$ is a square
matrix such that either $AB=I$ or $BA=I$, then $A$ is invertible
and $B = A^{-1}$.

**Proof:** If $BA = I$, then the system $A \vx = \vec 0$ has only
the trivial solution, as we saw in the challenge problem.
So (c) is true. Therefore (a) is true, i.e. $A$ is invertible. Then:
$$
\kern-6ex
B = BI = BAA^{-1} = IA^{-1} = A^{-1} .
$$
(The uniqueness argument again!)$\quad\Box$

This is very useful! It means you only need to check multiplication in one order to know you have an inverse.

**Theorem 3.14**: Let $A$ be a square matrix. If a sequence of row
operations reduces $A$ to $I$, then the **same** sequence of row
operations transforms $I$ into $A^{-1}$.

Why does this work? It's the combination of our arguments that (d) $\implies$ (b) and (b) $\implies$ (a). If we row reduce $[ A \mid \ve_i\,]$ to $[ I \mid \vc_i \,]$, then $A \vc_i = \ve_i$. So if $B$ is the matrix whose columns are the $\vc_i$'s, then $AB = I$. So, by Theorem 3.14, $B = A^{-1}$.

The trick is to notice that we can solve all of the
systems $A \vx = \ve_i$ **at once** by row reducing $[A \mid I\,]$.
The matrix on the right will be exactly $B$!

**Example on board:** Find the inverse of $A = \bmat{rr} 1 & 2 \\ 3 & 7 \emat$.

Illustrate proof of Theorem 3.14.

**Example on board:** Find the inverse of
$A = \bmat{rrr} 1 & 0 & 2 \\ 2 & 1 & 3 \\ 1 & -2 & 5 \emat$.
Illustrate proof of Theorem 3.14.

**Example on board:** Find the inverse of
$B = \bmat{rr} -1 & 3 \\ 2 & -6 \emat$.

So now we have a general purpose method for determining whether a matrix $A$ is invertible, and finding the inverse:

1. Form the $n \times 2n$ matrix $[A \mid I\,]$.

2. Use row operations to get it into reduced row echelon form.

3. If a zero row appears in the left-hand portion, then $A$ is not invertible.

4. Otherwise, $A$ will turn into $I$, and the right hand portion is $A^{-1}$.

The trend continues: when given a problem to solve in linear algebra, we usually find a way to solve it using row reduction!

Note that finding $A^{-1}$ is more work than solving a system $A \vx = \vb$.

We aren't covering inverse matrices over $\Z_m$.

**Question**:
Let $A$ be a $4 \times 4$ matrix with rank $3$.
Is $A$ invertible?
What if the rank is $4$?

**True/false**:
If $A$ is a square matrix, and the column vectors of $A$ are
linearly independent, then $A$ is invertible.

**True/false**:
If $A$ and $B$ are square matrices such that $AB$ is not invertible,
then at least one of $A$ and $B$ is not invertible.

**True/false**:
If $A$ and $B$ are matrices such that $AB = I$, then $BA = I$.

**Question:** Find invertible matrices $A$ and $B$ such that
$A+B$ is not invertible.