10.6 Inverting a Matrix

We have seen that we can solve the system of equations described by $$\begin{pmatrix} 3&-4 \ 2&-1\end{pmatrix}\begin{pmatrix} x \ y\end{pmatrix} = \begin{pmatrix} 5 \ -10\end{pmatrix}$$ with Cramer’s rule. Rather than grinding through(打破砂锅问到底) Cramer’s rule, it sure would be nice to be able to divide by the matrix on the left. Unfortunately, division by a matrix is not defined. However, we can take another look solving equations with division by real numbers to get an idea how we might accomplish our goal. Consider the equation $$3x = 9.$$ To solve this equation we can “divide both sides by 3.” What we’re really doing when we “divide both sides by 3” is “multiplying both sides by $\frac{1}{3}$.” More precisely, we are multiplying by the multiplicative inverse of 3, which eliminates the $3$ on the left side. Multiplying both sides of $3x=9$ by $\frac{1}{3}$ gives $$\frac{1}{3} \cdot 3 x =\frac{1}{3}\cdot 9.$$ The $\frac{1}{3}$ and 3 cancel, and we’re left with $x=\frac{1}{3}\cdot 9=3$.

Let’s see if we can accomplish something similar with matrices.

Problems

Problem 10.33

[!question]
Suppose that $\mathbf{\text{A}}\mathbf{\text{B}}\mathbf{\text{v}}=\mathbf{\text{v}}$ for all vectors $\mathbf{\text{v}}$. What matrix must $\mathbf{\text{A}}\mathbf{\text{B}}$ equal?

Problem 10.34

[!question]
If $\mathbf{\text{A}}\mathbf{\text{B}}$ equals the answer you found in ## Problem 10.33, then we say that $\mathbf{\text{A}}$ is the inverse of $\mathbf{\text{B}}$ and that $\mathbf{\text{B}}$ is invertible. We often denote the inverse of a matrix $\mathbf{\text{B}}$ as ${\mathbf{\text{B}}}^{-1}$. Let $\mathbf{\text{M}}=\left(\begin{array}{cc}2& 3\\ 6& 7\end{array}\right)$. Find ${\mathbf{\text{M}}}^{-1}$.

Problem 10.35

[!question]
Does the matrix $\mathbf{\text{M}}=\left(\begin{array}{cc}3& 5\\ -6& -10\end{array}\right)$ have an inverse? If it does, then find it. If it doesn’t, explain why not.

Problem 10.36

[!question]
If $\mathbf{\text{A}}$ is invertible, then how are $det(\mathbf{\text{A}})$ and $det({\mathbf{\text{A}}}^{-1})$ related?

Problem 10.37

[!question]
Let $\mathbf{\text{A}}=\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)$. In this problem, we find the conditions under which $\mathbf{\text{A}}$ is invertible, and we find the inverse in terms of $a$, $b$, $c$, and $d$.

(a) Suppose $\mathbf{\text{A}}$ is invertible, and let ${\mathbf{\text{A}}}^{-1}=\left(\begin{array}{cc}p& q\\ r& s\end{array}\right)$. What matrix must ${\mathbf{\text{A}}}^{-1}\mathbf{\text{A}}$ equal?

(b) Use your answer to (a) to find a system of equations in terms of $p$, $q$, $a$, $b$, $c$, and $d$. Solve the system for $p$ and $q$ in terms of the components of $\mathbf{\text{A}}$.

(c) Find $r$ and $s$ in terms of the components of $\mathbf{\text{A}}$.

(d) Under what conditions is it possible to find an inverse of $\mathbf{\text{A}}$, and what is that inverse in terms of $a$, $b$, $c$, and $d$?

(e) Confirm your result from part (d) by multiplying $\mathbf{\text{A}}$ by the matrix you found in part (d) .

Problem 10.38

[!question]
(a) Find the inverse of $\left(\begin{array}{cc}4& -3\\ 5& 7\end{array}\right)$.

(b) Use your answer to part (a) to solve the system of equations $4x-3y=-2$, $5x+7y=-3$.

(c) Find the inverse of $\left(\begin{array}{cc}-0.1& 5\\ -0.4& 30\end{array}\right)$.

(d) Use your answer to part (c) to solve the system of equations $-0.1t+5u=-3$, $-0.4t+30u=7$.

Problem 10.39

[!question]
If $\mathbf{\text{A}}=\left(\begin{array}{cc}4& -3\\ 2& 1\end{array}\right)$ and $\mathbf{\text{B}}=\left(\begin{array}{cc}-5& 3\\ -8& 2\end{array}\right)$, then find all matrices $\mathbf{\text{M}}$ such that $\mathbf{\text{A}}\mathbf{\text{M}}=\mathbf{\text{B}}$.

Problem 10.40

[!question]
Earlier we saw that $\mathbf{\text{A}}\mathbf{\text{B}}$ is not necessarily equal to $\mathbf{\text{B}}\mathbf{\text{A}}$. Must $\mathbf{\text{A}}{\mathbf{\text{A}}}^{-1}$ equal ${\mathbf{\text{A}}}^{-1}\mathbf{\text{A}}$?

Problem 10.414

[!question]
Suppose $\mathbf{\text{A}}$ and $\mathbf{\text{B}}$ are invertible. Prove that $(\mathbf{\text{A}}\mathbf{\text{B}}{)}^{-1}={\mathbf{\text{B}}}^{-1}{\mathbf{\text{A}}}^{-1}$.

PROBLEMS

Problem 10.33

[!question]
Suppose the product $\mathbf{\text{A}}\mathbf{\text{B}}\mathbf{\text{v}}=\mathbf{\text{v}}$ for all vectors $\mathbf{\text{v}}$. What matrix must $\mathbf{\text{A}}\mathbf{\text{B}}$ equal?

Solution for Problem 10.33:

In ## Problem 10.13, we discovered the identity matrix $\mathbf{\text{I}}$, which is the unique matrix such that $\mathbf{\text{I}}\mathbf{\text{v}}=\mathbf{\text{v}}$ for all vectors $\mathbf{\text{v}}$. If $\mathbf{\text{A}}\mathbf{\text{B}}\mathbf{\text{v}}=\mathbf{\text{v}}$ for all vectors $\mathbf{\text{v}}$, then $\mathbf{\text{A}}\mathbf{\text{B}}$ must be this matrix. That is, we must have $\mathbf{\text{A}}\mathbf{\text{B}}=\left(\begin{array}{cc}1& 0\\ 0& 1\end{array}\right)$. □

Problem 10.34

[!question]
If $\mathbf{\text{A}}\mathbf{\text{B}}=\mathbf{\text{I}}$, then we say that $\mathbf{\text{A}}$ is the inverse of $\mathbf{\text{B}}$ and that $\mathbf{\text{B}}$ is invertible. We often denote the inverse of a matrix $\mathbf{\text{B}}$ as ${\mathbf{\text{B}}}^{-1}$. Let $\mathbf{\text{M}}=\left(\begin{array}{cc}2& 3\\ 6& 7\end{array}\right)$. Find ${\mathbf{\text{M}}}^{-1}$.

Solution for Problem 10.34:

We seek the matrix such that$$\begin{pmatrix} a&b \ c&d\end{pmatrix}\begin{pmatrix} 2&3 \ 6&7\end{pmatrix} = \begin{pmatrix} 1&0 \ 0&1\end{pmatrix}.$$Expanding the product on the left, we have$$\begin{pmatrix} 2a+6b&3a +7b \ 2c+ 6d&3c+7d\end{pmatrix} = \begin{pmatrix} 1&0 \ 0&1\end{pmatrix}.$$From the entries in the first rows, we have $2a+6b=1$ and $3a+7b=0$. Solving this system gives $a=-\frac{7}{4}$ and $b=\frac{3}{4}$. From the entries in the second rows, we have $2c+6d=0$ and $3c+7d=1$. Solving this system gives $c=\frac{3}{2}$ and $d=-\frac{1}{2}$. Checking, we find that$$\begin{pmatrix} -7/4&3/4 \ 3/2&-1/2\end{pmatrix}\begin{pmatrix} 2&3 \ 6&7\end{pmatrix}= \begin{pmatrix} 1&0 \ 0&1\end{pmatrix},$$so ${\mathbf{\text{M}}}^{-1}=\left(\begin{array}{cc}-7/4& 3/4\\ 3/2& -1/2\end{array}\right)$. □

Problem 10.35

[!question]
Does the matrix $\mathbf{\text{M}}=\left(\begin{array}{cc}3& 5\\ -6& -10\end{array}\right)$ have an inverse? If it does, then find it. If it doesn’t, explain why not.

Solution for Problem 10.35:

Suppose $\mathbf{\text{A}}=\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)$ is the inverse of $\mathbf{\text{M}}$. Then we must have $\mathbf{\text{A}}\mathbf{\text{M}}=\mathbf{\text{I}}$, or$$\begin{pmatrix} a&b \ c&d\end{pmatrix} \begin{pmatrix} 3&5 \ -6&-10\end{pmatrix} = \begin{pmatrix} 1&0 \ 0&1\end{pmatrix}.$$Expanding the product on the left gives$$\begin{pmatrix} 3a-6b&5a-10b \ 3c-6d&5c-10d\end{pmatrix} = \begin{pmatrix} 1&0 \ 0&1\end{pmatrix}.$$From the entries in the first rows, we have $3a-6b=1$ and $5a-10b=0$. The second equation gives $a=2b$, but substituting this into the first equation gives $6b-6b=1$. Uh-oh; this equation has no solution! Therefore, we cannot find a matrix $\mathbf{\text{A}}$ such that $\mathbf{\text{A}}\mathbf{\text{M}}=\mathbf{\text{I}}$, which means that $\mathbf{\text{M}}$ does not have an inverse. □

Why doesn’t $\mathbf{\text{M}}$ have an inverse? Here’s a clue:

Problem 10.36

[!question]
If $\mathbf{\text{A}}$ is invertible, then how are $det(\mathbf{\text{A}})$ and $det({\mathbf{\text{A}}}^{-1})$ related?

Solution for Problem 10.36:

We have ${\mathbf{\text{A}}}^{-1}\mathbf{\text{A}}=\mathbf{\text{I}}$, so$$ \det(\textbf{A}^{-1}\textbf{A}) = \det(\textbf{I}) = (1)(1)-(0)(0) = 1. $$We also have $det({\mathbf{\text{A}}}^{-1}\mathbf{\text{A}})=det({\mathbf{\text{A}}}^{-1})\cdot det(\mathbf{\text{A}})$, so we must have $det({\mathbf{\text{A}}}^{-1})\cdot det(\mathbf{\text{A}})=1$. □

Now we see why the matrix $\mathbf{\text{M}}=\left(\begin{array}{cc}3& 5\\ -6& -10\end{array}\right)$ in ## Problem 10.35 has no inverse. Since$$\det(\textbf{M}) = (3)(-10)-(-6)(5) = 0,$$we know that there can be no matrix ${\mathbf{\text{M}}}^{-1}$ such that $det({\mathbf{\text{M}}}^{-1})\cdot det(\mathbf{\text{M}})=1$.

Problem 10.37

[!question]
Let $\mathbf{\text{A}}=\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)$. Under what conditions does $\mathbf{\text{A}}$ have an inverse? If $\mathbf{\text{A}}$ has an inverse, what is that inverse in terms of $a$, $b$, $c$, and $d$?

Solution for Problem 10.37:

Suppose $\mathbf{\text{A}}$ is invertible. We let ${\mathbf{\text{A}}}^{-1}=\left(\begin{array}{cc}p& q\\ r& s\end{array}\right)$, so we must have$$\begin{pmatrix} p&q \ r&s\end{pmatrix}\begin{pmatrix} a&b \ c&d\end{pmatrix} = \begin{pmatrix} 1&0 \ 0&1\end{pmatrix}.$$Expanding the product on the left gives$$\begin{pmatrix} pa+qc&pb + qd \ ra+sc&rb+sd\end{pmatrix} =\begin{pmatrix} 1&0 \ 0&1\end{pmatrix}.$$Equating the corresponding entries on the first rows gives the system $pa+qc=1$, $pb+qd=0$, and equating the corresponding entries on the second rows gives $ra+sc=0$, $rb+sd=1$. If $\left|\begin{array}{cc}a& c\\ b& d\end{array}\right|\ne 0$, then applying Cramer’s rule to solve the system $pa+qc=1$, $pb+qd=0$ for $p$ and $q$ gives$$p = \frac{\begin{vmatrix} 1&c \ 0&d\end{vmatrix}}{\begin{vmatrix} a&c \ b&d\end{vmatrix}} = \frac{d}{\begin{vmatrix} a&c \ b&d\end{vmatrix}}\qquad \text{ and }\qquad q = \frac{\begin{vmatrix} a&1 \ b&0\end{vmatrix}}{\begin{vmatrix} a&c \ b&d\end{vmatrix}} =\frac{-b}{\begin{vmatrix} a&c \ b&d\end{vmatrix}}.$$Similarly, if $\left|\begin{array}{cc}a& c\\ b& d\end{array}\right|\ne 0$, then we can apply Cramer’s rule to $ra+sc=0$, $rb+sd=1$ to give$$r = \frac{\begin{vmatrix} 0&c \ 1&d\end{vmatrix}}{\begin{vmatrix} a&c \ b&d\end{vmatrix}} = \frac{-c}{\begin{vmatrix} a&c \ b&d\end{vmatrix}} \qquad\text{ and } \qquad s = \frac{\begin{vmatrix} a&0 \ b&1\end{vmatrix}}{\begin{vmatrix} a&c \ b&d\end{vmatrix}} = \frac{a}{\begin{vmatrix} a&c \ b&d\end{vmatrix}}.$$Since $\left|\begin{array}{cc}a& c\\ b& d\end{array}\right|=ad-bc=det(\mathbf{\text{A}})$, the denominator in each of the results for $p$, $q$, $r$, and $s$ equals $det(\mathbf{\text{A}})$, and we have$$\textbf{A}^{-1} = \frac{1}{\det(\textbf{A})}\begin{pmatrix} d&-b \ -c&a\end{pmatrix}.$$

Checking, we find that $$\begin{align*}\textbf{A}^{-1}\textbf{A} &=\frac{1}{\det(\textbf{A})}\begin{pmatrix} d&-b \ -c&a\end{pmatrix}\begin{pmatrix} a&b \ c&d\end{pmatrix} \&=\frac{1}{\det(\textbf{A})}\begin{pmatrix} da-bc&db-bd \ -ca + ac&-cb + ad\end{pmatrix} \&=\frac{1}{\det(\textbf{A})}\begin{pmatrix} ad-bc&0 \ 0&ad-bc\end{pmatrix} \&= \begin{pmatrix} 1&0 \ 0&1\end{pmatrix},\end{align*}$$as expected.

Our work above shows that we can find an inverse of any $2\times 2$ matrix with a nonzero determinant. We say that such a matrix is invertible. Our solution also tells us that if $\mathbf{\text{A}}$ is invertible, then its inverse is unique. That is, if $\mathbf{\text{A}}$ is invertible, there is only one matrix ${\mathbf{\text{A}}}^{-1}$ such that ${\mathbf{\text{A}}}^{-1}\mathbf{\text{A}}=\mathbf{\text{I}}$. □

[! Danger] Important:
Let $\mathbf{\text{A}}=\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)$. If $det(\mathbf{\text{A}})=0$, then $\mathbf{\text{A}}$ does not have an inverse. Otherwise, we have$$\textbf{A}^{-1} = \frac{1}{\det(\textbf{A})} \begin{pmatrix} d&-b \ -c&a\end{pmatrix}.$$

Problem 10.38

[!question]
(a) Find the inverse of $\left(\begin{array}{cc}4& -3\\ 5& 7\end{array}\right)$.

(b) Use your answer to part (a) to solve the system of equations $4x-3y=-2$, $5x+7y=-3$.

(c) Find the inverse of $\left(\begin{array}{cc}-0.1& 5\\ -0.4& 30\end{array}\right)$.

(d) Use your answer to part (c) to solve the system of equations $-0.1t+5u=-3$, $-0.4t+30u=7$.

Solution for Problem 10.38:

(a) The determinant of the matrix is $(4)(7)-(-3)(5)=43$, so the inverse of the matrix is$$\frac{1}{43}\begin{pmatrix} 7&3 \ -5&4\end{pmatrix} = \begin{pmatrix} \frac{7}{43}&\frac{3}{43} $$

(b) We can write the system of equations as $\left(\begin{array}{cc}4& -3\\ 5& 7\end{array}\right)\left(\begin{array}{c}x\\ y\end{array}\right)=\left(\begin{array}{c}-2\\ -3\end{array}\right)$. Multiplying both sides of this equation by the inverse we found in part (a) , we have $$\begin{pmatrix} \frac{7}{43}&\frac{3}{43} $$1ex] -\frac{5}{43}&\frac{4}{43}\end{pmatrix} \begin{pmatrix} 4&-3 \ 5&7\end{pmatrix} \begin{pmatrix} x \ y\end{pmatrix} =\begin{pmatrix} \frac{7}{43}&\frac{3}{43} $$1ex] -\frac{5}{43}&\frac{4}{43}\end{pmatrix}\begin{pmatrix} -2 \ -3\end{pmatrix} =\begin{pmatrix} -\frac{23}{43} $$\text{1ex] -frac{2}{43}end{pmatrix}.}$$ The product of a matrix and its inverse is the identity, so we have $$\begin{pmatrix} x \ y\end{pmatrix} = \begin{pmatrix} -\frac{23}{43} $$\text{1ex] -frac{2}{43}end{pmatrix}}$$.

(c) The determinant of the matrix is $(-0.1)(30)-(5)(-0.4)=-1$, so the inverse of the matrix is$$\frac{1}{-1}\begin{pmatrix} 30&-5 \ 0.4&-0.1\end{pmatrix} = \begin{pmatrix} -30&5 \ -0.4&0.1\end{pmatrix}.$$

(d) We write the system as $\left(\begin{array}{cc}-0.1& 5\\ -0.4& 30\end{array}\right)\left(\begin{array}{c}t\\ u\end{array}\right)=\left(\begin{array}{c}-3\\ 7\end{array}\right)$. Multiplying both sides by the inverse we found in part (c) , we have $$\begin{align*}\begin{pmatrix} -30&5 \ -0.4&0.1\end{pmatrix}\begin{pmatrix} -0.1&5 \ -0.4&30\end{pmatrix}\begin{pmatrix} t \ u\end{pmatrix}&= \begin{pmatrix} -30&5 \ -0.4&0.1\end{pmatrix}\begin{pmatrix} -3 \ 7\end{pmatrix} \&= \begin{pmatrix} 125 \ 1.9\end{pmatrix}.\end{align*}$$As before, the product of a matrix and its inverse is the identity, so we have $\left(\begin{array}{c}t\\ u\end{array}\right)=\left(\begin{array}{c}125\\ 1.9\end{array}\right)$$.

□

Problem 10.39

[!question]
If $\mathbf{\text{A}}=\left(\begin{array}{cc}4& -3\\ 2& 1\end{array}\right)$ and $\mathbf{\text{B}}=\left(\begin{array}{cc}-5& 3\\ -8& 2\end{array}\right)$, then find all matrices $\mathbf{\text{M}}$ such that $\mathbf{\text{A}}\mathbf{\text{M}}=\mathbf{\text{B}}$.

Solution for Problem 10.39: We can’t divide by $\mathbf{\text{A}}$, but because its determinant is nonzero, we can multiply by its inverse. The determinant of the matrix is 10, so$$ \textbf{A}^{-1}=\dfrac{1}{10}\begin{pmatrix} 1&3 \ -2&4\end{pmatrix} = \begin{pmatrix} 0.1&0.3 \ -0.2&0.4\end{pmatrix}. $$What’s wrong with this next step:

[! note] Bogus Solution:
Multiplying both sides of the given equation by ${\mathbf{\text{A}}}^{-1}$ gives us$$\textbf{I} \textbf{M} = \textbf{B}\textbf{A}^{-1}.$$

The problem here is that we started with $\mathbf{\text{A}}\mathbf{\text{M}}=\mathbf{\text{B}}$ and multiplied $\mathbf{\text{A}}\mathbf{\text{M}}$ on the left by ${\mathbf{\text{A}}}^{-1}$ to get $\mathbf{\text{I}}\mathbf{\text{M}}$, but we multiplied $\mathbf{\text{B}}$ on the right by ${\mathbf{\text{A}}}^{-1}$. While we can go from $t=s$ to $rt=sr$ with real numbers, we can’t go from $\mathbf{\text{T}}=\mathbf{\text{S}}$ to $\mathbf{\text{R}}\mathbf{\text{T}}=\mathbf{\text{S}}\mathbf{\text{R}}$ with matrices, since matrix multiplication is not commutative.

We multiply both sides of the equation $\mathbf{\text{A}}\mathbf{\text{M}}=\mathbf{\text{B}}$ on the left by ${\mathbf{\text{A}}}^{-1}$ to produce ${\mathbf{\text{A}}}^{-1}\mathbf{\text{A}}\mathbf{\text{M}}={\mathbf{\text{A}}}^{-1}\mathbf{\text{B}}$. The left side then becomes $\mathbf{\text{I}}\mathbf{\text{M}}$, which equals $\mathbf{\text{M}}$, so we have$$\textbf{M} =\textbf{A}^{-1}\textbf{B}= \begin{pmatrix} 0.1&0.3 \ -0.2&0.4\end{pmatrix} \begin{pmatrix} -5&3 \ -8&2\end{pmatrix} =\begin{pmatrix} -2.9&0.9 \ -2.2&0.2\end{pmatrix}.$$So, we see that there is a unique solution to the equation $\mathbf{\text{A}}\mathbf{\text{M}}=\mathbf{\text{B}}$. Moreover, our solution shows us that whenever $\mathbf{\text{A}}$ is invertible, there is a unique solution for $\mathbf{\text{M}}$ to the equation $\mathbf{\text{A}}\mathbf{\text{M}}=\mathbf{\text{B}}$, which is $\mathbf{\text{M}}={\mathbf{\text{A}}}^{-1}\mathbf{\text{B}}$. □

Problem 10.40

[!question]
Earlier we saw that $\mathbf{\text{A}}\mathbf{\text{B}}$ is not necessarily equal to $\mathbf{\text{B}}\mathbf{\text{A}}$. Must $\mathbf{\text{A}}{\mathbf{\text{A}}}^{-1}$ equal ${\mathbf{\text{A}}}^{-1}\mathbf{\text{A}}$?

Solution for Problem 10.40: We have defined ${\mathbf{\text{A}}}^{-1}$ as the matrix such that ${\mathbf{\text{A}}}^{-1}\mathbf{\text{A}}=\mathbf{\text{I}}$, so to check if $\mathbf{\text{A}}{\mathbf{\text{A}}}^{-1}={\mathbf{\text{A}}}^{-1}\mathbf{\text{A}}$, we need only compute $\mathbf{\text{A}}{\mathbf{\text{A}}}^{-1}$. Letting $\mathbf{\text{A}}=\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)$, we have $$\begin{align*}\textbf{A}\textbf{A}^{-1} &=\begin{pmatrix} a&b \ c&d\end{pmatrix}\left(\frac{1}{\det({\textbf{A}})} \begin{pmatrix} d&-b \ -c&a\end{pmatrix}\right) \&=\frac{1}{\det(\textbf{A})} \begin{pmatrix} a&b \ c&d\end{pmatrix}\begin{pmatrix} d&-b \ -c&a\end{pmatrix} \&=\frac{1}{\det(\textbf{A})} \begin{pmatrix} ad-bc&0 \ 0&ad-bc\end{pmatrix} \&= \textbf{I}.\end{align*}$$Therefore, $\mathbf{\text{A}}{\mathbf{\text{A}}}^{-1}={\mathbf{\text{A}}}^{-1}\mathbf{\text{A}}=\mathbf{\text{I}}$. □

We therefore see that if $\mathbf{\text{A}}$ is invertible, then the inverse of ${\mathbf{\text{A}}}^{-1}$ is $\mathbf{\text{A}}$.

[! Danger] Important:
If $\mathbf{\text{A}}$ is invertible, then we have $\mathbf{\text{A}}{\mathbf{\text{A}}}^{-1}={\mathbf{\text{A}}}^{-1}\mathbf{\text{A}}=\mathbf{\text{I}}$.

Problem 10.41

[!question]
Suppose $\mathbf{\text{A}}$ and $\mathbf{\text{B}}$ are invertible. Prove that $(\mathbf{\text{A}}\mathbf{\text{B}}{)}^{-1}={\mathbf{\text{B}}}^{-1}{\mathbf{\text{A}}}^{-1}$.

Solution for Problem 10.41: If $\mathbf{\text{M}}$ is the inverse of $\mathbf{\text{A}}\mathbf{\text{B}}$, then we must have $\mathbf{\text{M}}(\mathbf{\text{A}}\mathbf{\text{B}})=\mathbf{\text{I}}$. So, we test $\mathbf{\text{M}}={\mathbf{\text{B}}}^{-1}{\mathbf{\text{A}}}^{-1}$, and using the associative property, we find $$\begin{align*}(\textbf{B}^{-1}\textbf{A}^{-1})(\textbf{A}\textbf{B})&= \textbf{B}^{-1}(\textbf{A}^{-1}(\textbf{A}\textbf{B})) \&= \textbf{B}^{-1}((\textbf{A}^{-1}\textbf{A})\textbf{B}) \&= \textbf{B}^{-1}\textbf{I}\textbf{B} \&= \textbf{B}^{-1}\textbf{B} \&= \textbf{I}.\end{align*}$$Therefore, the inverse of $\mathbf{\text{A}}\mathbf{\text{B}}$ is ${\mathbf{\text{B}}}^{-1}{\mathbf{\text{A}}}^{-1}$. □

Exercises

10 .6.1:

[!question]
What is the inverse of $\mathbf{\text{I}}$?

10 .6.2:

[!question]
For each of the following matrices, find the inverse or show that the matrix does not have an inverse.
(a) $\left(\begin{array}{cc}6& 8\\ -2& 4\end{array}\right)$

(b) $\left(\begin{array}{cc}2& -1\\ -0.9& 0.5\end{array}\right)$

(c) $\left(\begin{array}{cc}-40& 30\\ 8& -6\end{array}\right)$

10 .6.3:

[!question]
Suppose we are given matrix $\mathbf{\text{A}}$ and vector $\mathbf{\text{v}}$.

(a) If $\mathbf{\text{A}}$ is nonsingular, must there be a vector $\mathbf{\text{w}}$ such that $\mathbf{\text{A}}\mathbf{\text{w}}=\mathbf{\text{v}}$?

(b) If $\mathbf{\text{A}}$ is singular, can we conclude that there is no vector such that $\mathbf{\text{A}}\mathbf{\text{w}}=\mathbf{\text{v}}$?

10 .6.4:

[!question]
All of the entries of $\mathbf{\text{M}}$ and ${\mathbf{\text{M}}}^{-1}$ are integers. Find all possible values of $det(\mathbf{\text{M}})$.

10 .6.5:

[!question]
In ## Problem 10.41, we showed that if $\mathbf{\text{A}}$ and $\mathbf{\text{B}}$ are invertible, then $(\mathbf{\text{A}}\mathbf{\text{B}}{)}^{-1}={\mathbf{\text{B}}}^{-1}{\mathbf{\text{A}}}^{-1}$. If $\mathbf{\text{A}}$ and $\mathbf{\text{B}}$ are invertible, then must we have $(\mathbf{\text{A}}+\mathbf{\text{B}}{)}^{-1}={\mathbf{\text{B}}}^{-1}+{\mathbf{\text{A}}}^{-1}$?

10 .6.6:

[!question]
In ## Problem 10.18, we found that multiplying $\left(\begin{array}{c}x\\ y\end{array}\right)$ by the matrix $\left(\begin{array}{cc}a& -b\\ b& a\end{array}\right)$ corresponds to multiplying the complex number $x+yi$ by $a+bi$. Find the matrix $\mathbf{\text{M}}$ such that multiplying $\left(\begin{array}{c}x\\ y\end{array}\right)$ by $\mathbf{\text{M}}$ corresponds to dividing $x+yi$ by $a+bi$ (assuming $a$ and $b$ are not both 0).