# Section 2.4 Matrix Inverses

## 2.4A

Definition: An $n \times n$ matrix $A$ is said to be invertible if there is an $n \times n$ matrix $C$ such that

$CA=I_{n}$ and $AC=I_{n}$

where $I_{n}$ is the $n \times n$ identity matrix.$C$ is the inverse of $A$

Theorem: If $B$ and $C$ are both inverses of $A$, then $B = C$

Fact: $C$ is uniquely determined by $A$. We can write the unique inverse of $A, A^{-1}$. Hence $AA^{-1} = I_{n}$ and $A^{-1}A = I_{n}$.

Example 1: Let $A=\begin{bmatrix} 1 & -2\\ 3 & -2 \end{bmatrix}$, show $A^{-1}=\frac{1}{4}\begin{bmatrix} -2 & 2\\ -3 & 1 \end{bmatrix}$.

Exercise 1: Let $A=\begin{bmatrix} 2 & 3\\ 1 & -1 \end{bmatrix}$, show $A^{-1}=\frac{-1}{5}\begin{bmatrix} -1 & -3\\ -1 & 2 \end{bmatrix}$.

Definition: Let $A=\begin{bmatrix} a & b\\ c & d \end{bmatrix}$ then $ad - bc$ is called the determinant of $A$. We write det$A = ad - bc$.

Theorem: Let $A=\begin{bmatrix} a & b\\ c & d \end{bmatrix}$. If det$A = ad - bc \neq 0$, then $A$ is invertible and

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

If $ad - bc = 0$ then $A$ is not invertible.

Fact: $A=\begin{bmatrix} a & b\\ c & d \end{bmatrix}$ is invertible if and only if $ad - bc \neq 0$

Theorem: If $A$ is an invertible $n \times n$ matrix then for each $\vec{b}$ in $\mathbb{R}^n$, the equation $A\vec{x} = \vec{b}$ has the unique solution $\vec{x} = A^{-1}\vec{b}$.

Proof:

Example 2: Find the unique solution of $A\vec{x} = \vec{b}$ without using the row operations, where $A=\begin{bmatrix} 4 & 3\\ -2 & -1 \end{bmatrix}$ and $\vec{b}=\begin{bmatrix}5\\-7\end{bmatrix}$.

Example 2: Find the unique solution of $A\vec{x} = \vec{b}$ without using the row operations, where $A=\begin{bmatrix} -2 & 1\\ 5 & 1 \end{bmatrix}$ and $\vec{b}=\begin{bmatrix}1\\2\end{bmatrix}$.

Theorem:

(a)If $A$ is an invertible matrix, then $A^{-1}$ is invertible and $(A^{-1})^{-1} = A$.

(b)If $A$ and $B$ are $n \times n$ invertible matrices, then so is $AB$ and the inverse of $AB$ is the product of the inverse of $A$ and $B$ in the reverse order, i.e. $(AB)^{-1} = B^{-1}A^{-1}$.

(c)If $A$ is an invertible matrix, then so is $A^T$ and the inverse of $A^T$ is the transpose of $A^{-1}$, i.e. $(A^T)^{-1} = (A^{-1})^T$.

Proof:

Example 3: Find the inverse of $AB$ by using the inverse of $A$ and the inverse of $B$ without finding $AB$ where $A=\begin{bmatrix} 1 & 3\\ 2 & 4 \end{bmatrix}$ and $B=\begin{bmatrix} 2 & 3\\ -1 & 2 \end{bmatrix}$.

Exercise 3: Find the inverse of $A^T$ by using the inverse of $A$ without finding $A^T$ where $A=\begin{bmatrix} 2 & 3\\ 1 & -1 \end{bmatrix}$.

Group Work 1: Mark each statement True or False. Justify each answer.

a. In order for a matrix $B$ to be the inverse of $A$, the equation $AB = I$ and $BA = I$ must both be true.

b. If $A$ and $B$ are $n \times n$ and invertible then the inverse of $AB$ is
$A^{-1}B^{-1}$.

c. If $A=\begin{bmatrix} a & b\\ c & d \end{bmatrix}$ and $ab - cd \neq 0$ then $A$ is invertible.

d. If $A$ is an invertible $n \times n$ matrix then the equation $A\vec{x} = \vec{b}$ is consistent for each $\vec{b}$ in $\mathbb{R}^n$.

e. If $A$ is invertible, then elementary row operations that reduce $A$ to the identity $I_{n}$ also reduce $A^{-1}$ to $I_{n}$.

f. If $A$ is invertible then the inverse of $A^{-1}$ is $A$.

Group. Work 2: Suppose $AB = AC$ where $B$ and $C$ are $n \times p$ matrices and $A$ is invertible $n \times n$ matrix. Show that $B = C$. Is this true in general? If this is not true in general, find an example.

Group Work 3: Let $A = \begin{bmatrix} 1 & 2\\ 1 & 3\\ 1 & 5 \end{bmatrix}$. Construct a $2 \times 3$ matrix $C$ using only 1, −1 and 0 as entries, such that $CA = I_{2}$. Then compute $AC$ and note that $AC \neq I_{3}$

Group Work 4: In each case either prove the assertion or give an example showing that it is false.

a. If $A \neq 0$ is a square matrix, then $A$ is invertible.

b. If $A$ and $B$ are both invertible, then $A + B$ is invertible.

c. If $A$ and $B$ are both invertible, then $(A^{-1}B)^T$ is invertible.

d. If $A^2$ is invertible, then $A$ is invertible.

## 2.4B

Question: How do we find the inverse of $A$ when $A$ is not a $2 \times 2$ matrix?

Theorem: An $n \times n$ matrix $A$ is invertible if and only if $A$ is row equivalent to $I_{n}$.

Example 1: Find $A^{-1}$ where $A=\begin{bmatrix} 1 & 2 & 3\\ 0 & 1 & 2\\ 1 & -1 & 0 \end{bmatrix}$.

Exercise 1: Find $A^{-1}$ where $A=\begin{bmatrix} 2 & 0 & 1\\ 1 & 3 & 2\\ 0 & -1 & 2 \end{bmatrix}$.

Theorem: Let $A$ be a $n \times n$ square matrix. Then the following statements are equivalent.

(a) $A$ is an invertible matrix.

(b) $A$ is row equivalent to the $n \times n$ identity matrix.

(c) $A$ has $n$ pivot positions or rank$A = n$.

(d) The equation $A\vec{x}=\vec{0}$ has only the trivial solution.

(e) The equation $A\vec{x}=\vec{b}$ has at least one solution for each $\vec{b}$ in $\mathbb{R}^n$.

(f) There is an $n \times n$ matrix $C$ such that $CA=I_{n}$.

(g) There is an $n \times n$ matrix $D$ such that $AD=I_{n}$.

(h) $A^T$ is an invertible matrix.

Remark: The theorem only applies to a square matrix. If $A$ and $B$ are square matrix. If $AB = I$ then both $A$ and $B$ are invertible with $A^{-1}=B$ and $B^{-1}=A$.

Proof: (1) We know (a) $\Leftarrow \Rightarrow$ (b), (b) $\Leftarrow \Rightarrow$ (c), (c) $\Leftarrow \Rightarrow$ (d), and (a) $\Rightarrow$ (f). Once we show (f) $\Rightarrow$ (d), then a, b, c, d, f are equivalent.

(2) We know (a) $\Leftarrow \Rightarrow$ (b), (a) $\Rightarrow$ (g), once we show (g) $\Rightarrow$ (e) and (e) $\Rightarrow$ (b), then a, b, e, g are equivalent.

(3) We know (a) $\Rightarrow$ (h), once we show (h) $\Rightarrow$ (a) then a, h are equivalent.

With (1) to (3), we have all are equivalent.

Example 2: Show (f) implies (d) in the above theorem, i.e when there is an $n \times n$ matrix $C$ such that $CA = I_{n}$ then the equation $A\vec{x} = \vec{0}$ only has trivial solution.

Exercise 2: Show (g) implies (e) in the above theorem, i.e when there is an $n \times n$ matrix $D$ such that $AD = I_{n}$ then for all $\vec{b}$ in $\mathbb{R}^n$ there is a solution for the equation $A\vec{x} = \vec{b}$.

Example 3: Show (h) implies (a), i.e. when $A^T$ is invertible then $A$ is invertible.

Exercise 3: Show (e) implies (b), i.e. when there is a solution of $A\vec{x}=\vec{b}$ for all \vec{b} in $\mathbb{R}^n$ then $A$ is row equivalent to $I_{n}$.

Example 4: Decide if $A$ is invertible $A=\begin{bmatrix} 1 & 0 & -1\\ 0 & 2 & 3\\ 2 & 1 & -2 \end{bmatrix}$.

Exercise 4: Decide if $A$ is invertible $A=\begin{bmatrix} 2 & -1 & 1\\ 1 & 0 & 3\\ 0 & 1 & -2 \end{bmatrix}$.

Definition: A transformation $T: \mathbb{R}^n \rightarrow \mathbb{R}^n$ is said to be invertible if there is a function $S: \mathbb{R}^n \rightarrow \mathbb{R}^n$ such that

$S(T(\vec{x})) = \vec{x}$ and $T(S(\vec{x})) = \vec{x}$

or all $\vec{x}$ in $\mathbb{R}^n$. $S$ is called the inverse of $T$ and it is uniquely determined by $T$, so we write $T^{-1}$.

Theorem: Let $T: \mathbb{R}^n \rightarrow \mathbb{R}^n$ be a transformation and let $A$ be the standard matrix of $T$ such that $T(\vec{x}) = A(\vec{x})$. Then $T$ is invertible if and only if $A$ is invertible. The inverse transform of $T$ is defined by $S(\vec{x})=A^{-1}\vec{x}$.

Proof:

Example 5: Let $T$ be a transformation from $\mathbb{R}^2$ to $\mathbb{R}^2$ defined by $T(x_{1}, x_{2}) = (2x_{1} - x_{2}, -x_{1} + 3x_{2})$. Show $T$ is invertible and find $T^{-1}$.

Exercise 5: Let $T$ be a transformation from $\mathbb{R}^2$ to $\mathbb{R}^2$ defined by $T(x_{1}, x_{2}) = (-x_{1} + 2x_{2}, x_{1} - 4x_{2})$. Show $T$ is invertible and find $T^{-1}$.

Theorem: All the following matrices are square matrices of the same size.

(a) $I_{n}$ is invertible and $I^{-1}=I$.

(b) If $A$ is invertible, so is $A^k$ for any $k > 0$, and $(A^k)^{-1} = (A^{-1})^k$.

(c) If $A$ is invertible and $a \neq 0$ is a number, then $aA$ is invertible and $(aA)^{-1} = \frac{1}{a}A^{-1}$.

Example 6: Show the statement (c) above.

Exercise 6: Show $(A^k)^{-1} = (A^{-1})^k$ if $A$ is invertible.

Group Work 1: Mark each statement True or False. Justify each answer. All
matrices are $n \times n$ matrices.

a. If $A\vec{x} = \vec{0}$ has only trivial solution, then $A$ is row equivalent to $I_{n}$.

b. If $A^T$ is not invertible then $A$ is not invertible.

c. If there is an $n \times n$ matrix $D$ such that $AD = I_{n}$ then $DA = I_{n}$.

d. If the transformation $\vec{x} \rightarrow A\vec{x}$ maps $\mathbb{R}^n$ into $\mathbb{R}^n$ then $A$ is row equivalent to $I_{n}$.

e. If $A\vec{x} = \vec{0}$ has nontrivial solution, then $A$ has fewer than $n$ pivot positions.

f. If there is a $\vec{b}$ in $\mathbb{R}^n$ such that the equation $A\vec{x} = \vec{b}$ is consistent then the solution is unique.

Group Work 2: Show that if there is a vector $\vec{b}$ such that $A\vec{x} = \vec{b}$ has more than one solution then the $n \times n$ matrix $A$ is not invertible.

Group Work 3: For square matrices $A$ and $B$. Show that $A = B$ if and only if $A^{-1}B = I$.

Group Work 4: In each case either prove the assertion or give an example
showing that it is false

a. $A$ has no inverse when $A$ has a row of zero.

b. If $AB = B$ for some $B \neq 0$ then $A$ is invertible.

c. If $A$ is invertible then $A^2 \neq 0$.

d. If $AB = 0$ then one of $A$ or $B$ must be zero.

e. If $AB = 0$ and $A$ is invertible then $B = 0$.