Matrix Inverses

Recall that the Identity matrix $I_n \in M_n(\mathbb{R})$ is the matrix $$ I_n = \left[ \begin{array}{cccc} 1 & 0 & \cdots & 0 \\ 0 & 1 & \cdots & 0 \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \cdots & 1 \\ \end{array} \right] $$ It's called the identity because $A I_n = I_n A = A$ for every matrix $A\in M_n(\mathbb{R});$ in other words, it does for matrices what the number $1$ does for multiplication of numbers.

A matrix $A \in M_n(\mathbb{R})$ is Invertible if there exists another matrix $B \in M_n(\mathbb{R})$ such that $$A B = B A = I_n.$$

If $A$ is invertible, we call $B$ the Inverse of $A$ and denote it $B = A^{-1}.$ Another term for invertible is Nonsingular. If a matrix does not have an inverse, it is called Singular.

Here are some properties of the inverse:

Let $A$ and $B \in M_n(\mathbb{R})$ be invertible. Then

The inverse of $A,$ $A^{-1},$ is unique.
$\left( A^{-1} \right)^{-1} = A.$
$\left( A B \right)^{-1} = B^{-1} A^{-1}.$
$A {\bf x} = {\bf b}$ has a unique solution ${\bf x} = A^{-1} {\bf b}$ for each ${\bf b} \in \mathbb{R}^n.$

There is an easy formula that is important to remember for computing the inverse of a nonsingular $2 \times 2$ matrix.

Let $$A = \left[ \begin{array}{cc} a & b \\ c & d \\ \end{array} \right] \in M_2(\mathbb{R}).$$

The Determinant of $A$ is $$\textrm{det}(A) = a d - b c.$$
$A$ is nonsingular if and only if det$(A) \neq 0.$
If $A$ is nonsingular, then $$A^{-1} = \frac{1}{\textrm{det}(A)} \left[ \begin{array}{cc} d & -b \\ -c & a \\ \end{array} \right].$$

Let's use the inverse to solve a linear system.

Problem: Solve $$ \begin{array}{rcc}2x + y & = \ 2 \, , \\ 6x + 4y & = \ 4 \, . \\ \end{array}$$
Solution: Instead of converting the system into an augmented matrix and row reducing, we consider the matrix equation $$A {\bf x} = \left[ \begin{array}{cc} 2 & 1 \\ 6 & 4 \\ \end{array} \right] {\bf x} = \left[ \begin{array}{c} 2 \\ 4 \\ \end{array} \right].$$

Here $$\textrm{det}(A) = 2 \cdot 4 - 1 \cdot 6 = 2 \neq 0$$ so $A$ is invertible with inverse $$A^{-1} = \frac{1}{2} \left[ \begin{array}{cc} 4 & -1 \\ -6 & 2 \\ \end{array} \right] = \left[ \begin{array}{cc} 2 & -\frac{1}{2} \\ -3 & 1 \\ \end{array} \right].$$ Thus the solution to the system is $${\bf x} = A^{-1} \left[ \begin{array}{c} 2 \\ 4 \\ \end{array} \right] = \left[ \begin{array}{cc} 2 & -\frac{1}{2} \\ -3 & 1 \\ \end{array} \right] \left[ \begin{array}{c} 2 \\ 4 \\ \end{array} \right] = \left[ \begin{array}{c} 2 \\ -2 \\ \end{array} \right].$$

An $n \times n$ matrix $A$ is invertible if and only if $\text{RREF}(A) = I_n$. There are many such criteria equivalent to the invertibility of a matrix. Here are a few.

Theorem: For $A \in M_n(\mathbb{R}),$ the following are equivalent:

$A$ is invertible, i.e., there exists a matrix $A^{-1} \in M_n(\mathbb{R})$ such that $A A^{-1} = A^{-1} A = I_n.$
The matrix equation $A {\bf x} \, = \, {\bf b}$ has a unique solution for every ${\bf b} \in {\mathbb R}^n.$
The matrix equation $A {\bf x} \, = \, {\bf 0}$ has only the trivial solution.
The reduced row echelon form of $A$ is $I_n,$ i.e., $\textrm{RREF}(A) = I_n.$

The last criterion gives us an algorithm for computing inverses of matrices that are bigger than $2 \times 2.$

To compute the inverse of a nonsingular matrix $A \in M_n(\mathbb{R}),$ create the augmented matrix with $A$ on the left of the vertical bar and the identity matrix $I_n$ on the right of the vertical bar and row reduce. $$\left[ \begin{array}{c|c} A & I_n \\ \end{array} \right] \rightarrow \textrm{RREF} \left( \left[ \begin{array}{c|c} A & I_n \\ \end{array} \right] \right) = \left[ \begin{array}{c|c} I_n & A^{-1} \\ \end{array} \right].$$ The resulting matrix to the right of the vertical bar will be $A^{-1}.$

Problem: Find $A^{-1}$ when $$A \ = \ \left[\begin{array}{ccc} 0 & 1 & 2 \\ 1 & 0 & 3 \\ 4 & -3 & 8 \\ \end{array} \right].$$
Solution: $$\left[ \begin{array}{c|c} A & I_3 \\ \end{array} \right] = \left[ \begin{array}{ccc|ccc} 0 & 1 & 2 & 1 & 0 & 0 \\ 1 & 0 & 3 & 0 & 1 & 0 \\ 4 & -3 & 8 & 0 & 0 & 1 \\ \end{array} \right]$$

row reduces to $$\left[ \begin{array}{c|c} A & I_3 \\ \end{array} \right] = \left[ \begin{array}{ccc|ccc} 1 & 0 & 0 & -9/2 & 7 & -3/2 \\ 0 & 1 & 0 & -2 & 4 & -1 \\ 0 & 0 & 1 & 3/2 & -2 & 1/2 \\ \end{array} \right]$$ thus $$ A^{-1} \ = \ \left[\begin{array}{ccc} -9/2 & 7 & - 3/2 \\ -2 & 4 & -1 \\ 3/2 & -2 & 1/2 \\ \end{array}\right].$$