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 $A{A}^{-1}=I_n$ and $A^{-1}A=I_n$.

 

 

Example 1: Let $A=\left[\begin{array}{cc}1& -2\\ 3& -2\end{array}\right]$, show $A^{-1}=\frac{1}{4}\left[\begin{array}{cc}-2& 2\\ -3& 1\end{array}\right]$.

 

 

Exercise 1: Let $A=\left[\begin{array}{cc}2& 3\\ 1& -1\end{array}\right]$, show $A^{-1}=\frac{-1}{5}\left[\begin{array}{cc}-1& -3\\ -1& 2\end{array}\right]$.

 

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

 

Theorem: Let $A=\left[\begin{array}{cc}a& b\\ c& d\end{array}\right]$. If det$A=ad-bc\ne 0$, then $A$ is invertible and

 

$A^{-1}=\frac{1}{ad-bc}\left[\begin{array}{cc}d& -b\\ -c& a\end{array}\right]$.

 

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

Fact: $A=\left[\begin{array}{cc}a& b\\ c& d\end{array}\right]$ is invertible if and only if $ad-bc\ne 0$

 

 

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

Proof:

 

Example 2: Find the unique solution of $A\overrightarrow{x}=\overrightarrow{b}$ without using the row operations, where $A=\left[\begin{array}{cc}4& 3\\ -2& -1\end{array}\right]$ and $\overrightarrow{b}=\left[\begin{array}{c}5\\ -7\end{array}\right]$.

 

 

Exercise 2: Find the unique solution of $A\overrightarrow{x}=\overrightarrow{b}$ without using the row operations, where $A=\left[\begin{array}{cc}-2& 1\\ 5& 1\end{array}\right]$ and $\overrightarrow{b}=\left[\begin{array}{c}1\\ 2\end{array}\right]$.

 

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 $A=\left[\begin{array}{cc}1& 3\\ 2& 4\end{array}\right]$ and $B=\left[\begin{array}{cc}2& 3\\ -1& 2\end{array}\right]$.

 

 

Exercise 3: Find the inverse of $A^T$ by using the inverse of $A$ without finding $A^T$ where $A=\left[\begin{array}{cc}2& 3\\ 1& -1\end{array}\right]$.

 

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 $A=\left[\begin{array}{cc}a& b\\ c& d\end{array}\right]$ and $ab-cd\ne 0$ then $A$ is invertible.

 

d. If $A$ is an invertible $n\times n$ matrix then the equation $A\overrightarrow{x}=\overrightarrow{b}$ is consistent for each $\overrightarrow{b}$ 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 $A=\left[\begin{array}{cc}1& 2\\ 1& 3\\ 1& 5\end{array}\right]$. 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 $AC\ne I_3$

 

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

a. If $A\ne 0$ 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 $A=\left[\begin{array}{ccc}1& 2& 3\\ 0& 1& 2\\ 1& -1& 0\end{array}\right]$.

 

 

Exercise 1: Find $A^{-1}$ where $A=\left[\begin{array}{ccc}2& 0& 1\\ 1& 3& 2\\ 0& -1& 2\end{array}\right]$.

 

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 $A\overrightarrow{x}=\overrightarrow{0}$ has only the trivial solution.

(e) The equation $A\overrightarrow{x}=\overrightarrow{b}$ has at least one solution for each $\overrightarrow{b}$ 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 $A\overrightarrow{x}=\overrightarrow{0}$ 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 $\overrightarrow{b}$ in ${\mathbb{R}}^n$ there is a solution for the equation $A\overrightarrow{x}=\overrightarrow{b}$.

 

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 $A\overrightarrow{x}=\overrightarrow{b}$ for all \vec{b} in ${\mathbb{R}}^n$ then $A$ is row equivalent to $I_n$.

 

Example 4: Decide if $A$ is invertible $A=\left[\begin{array}{ccc}1& 0& -1\\ 0& 2& 3\\ 2& 1& -2\end{array}\right]$.

 

 

Exercise 4: Decide if $A$ is invertible $A=\left[\begin{array}{ccc}2& -1& 1\\ 1& 0& 3\\ 0& 1& -2\end{array}\right]$.

 

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

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

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

 

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 $a\ne 0$ 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 $A\overrightarrow{x}=\overrightarrow{0}$ 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 $\overrightarrow{x}\to A\overrightarrow{x}$ maps ${\mathbb{R}}^n$ into ${\mathbb{R}}^n$ then $A$ is row equivalent to $I_n$.

 

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

 

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

 

Group Work 2: Show that if there is a vector $\overrightarrow{b}$ such that $A\overrightarrow{x}=\overrightarrow{b}$ 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 $B\ne 0$ then $A$ is invertible.

 

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

 

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

 

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