Question

Let

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

1. Find $A^{-1}$. [Hint: Use Exercise 19.]
2. Find $A^3$.
3. Find ${\left(A^{-1}\right)}^3$.

Use your answers to (b) and (c) to show that ${\left(A^{-1}\right)}^3$is the inverse of $A^3$.

Short answer

1. $\left[\begin{array}{cc}-\frac{3}{5}& \frac{2}{5}\\ \frac{1}{5}& \frac{1}{5}\end{array}\right]$.
2. $\left[\begin{array}{cc}1& 18\\ 9& 37\end{array}\right]$.
3. $\left[\begin{array}{cc}-\frac{37}{125}& \frac{8}{125}\\ \frac{9}{125}& \frac{1}{125}\end{array}\right]$.

d.${\left(A^3\right)}^{-1}={\left(A^{-1}\right)}^3$ .

Step-by-step solution

Step 1

Given:

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

a. By the previous exercise, we know that the inverse of

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

In this case, $a=-1,b=2,c=1,d=3$:

$\begin{array}{r}{A}^{-1}=\left[\begin{array}{cc}\frac{d}{ad-bc}& -\frac{b}{ad-bc}\\ -\frac{c}{ad-bc}& \frac{a}{ad-bc}\end{array}\right]\\ =\left[\begin{array}{ll}\frac{3}{(-1)\left(3\right)-\left(2\right)\left(1\right)}& \frac{-2}{(-1)\left(3\right)-\left(2\right)\left(1\right)}\\ \frac{-1}{(-1)\left(3\right)-\left(2\right)\left(1\right)}& \frac{-1}{(-1)\left(3\right)-\left(2\right)\left(1\right)}\end{array}\right]\\ =\left[\begin{array}{cc}-\frac{3}{5}& \frac{2}{5}\\ \frac{1}{5}& \frac{1}{5}\end{array}\right]\end{array}$.

Step 2

b.$A^2$is the product of A and A.

localid="1668444840049" $\begin{array}{r}{A}^2=A\cdot A\\ =\left[\begin{array}{rr}-1& 2\\ 1& 3\end{array}\right]\cdot \left[\begin{array}{rr}-1& 2\\ 1& 3\end{array}\right]\\ =\left[\begin{array}{ll}(-1)(-1)+\left(2\right)\left(1\right)& (-1)\left(2\right)+\left(2\right)+\left(2\right)\left(3\right)\\ \left(2\right)(-1)+\left(11\right)\left(1\right)& \left(2\right)\left(2\right)+\left(11\right)\left(3\right)\end{array}\right]\\ =\left[\begin{array}{ll}1& 18\\ 9& 37\end{array}\right].\end{array}$

Step 3

c. By part (a):

$A^d=\left[\begin{array}{cc}-\frac{3}{5}& \frac{2}{5}\\ \frac{1}{5}& \frac{1}{5}\end{array}\right]$

role="math" localid="1668444524238" ${\left(A^{-1}\right)}^2$is the product of $A^{-1}$and $A^{-1}$.

$\begin{array}{r}{\left(A^{-1}\right)}^2=A^{-1}\cdot A^{-1}\\ =\left[\begin{array}{cc}-\frac{3}{5}& \frac{2}{5}\\ \frac{1}{5}& \frac{1}{5}\end{array}\right]\cdot \left[\begin{array}{cc}-\frac{3}{5}& \frac{2}{5}\\ \frac{1}{5}& \frac{1}{5}\end{array}\right]\\ =\left[\begin{array}{cc}-\frac{3}{5}\cdot \left(-\frac{3}{5}\right)+\frac{2}{5}\cdot \frac{1}{5}& -\frac{3}{5}\cdot \frac{2}{5}+\frac{2}{5}\cdot \frac{1}{5}\\ \frac{1}{5}\cdot \left(-\frac{3}{5}\right)+\frac{1}{5}\cdot \frac{1}{5}& \frac{1}{5}\cdot \frac{2}{5}+\frac{1}{5}\cdot \frac{1}{5}\end{array}\right]\\ =\left[\begin{array}{ll}\frac{11}{25}& -\frac{4}{25}\\ -\frac{2}{25}& \frac{3}{25}\end{array}\right]\end{array}$

$A^3$is the product of $A^2$and A.

$\begin{array}{r}{\left(A^{-1}\right)}^3={\left(A^{-1}\right)}^2\cdot A^{-1}\\ =\left[\begin{array}{lr}\frac{11}{25}& -\frac{4}{25}\\ -\frac{2}{25}& \frac{3}{25}\end{array}\right]\cdot \left[\begin{array}{rr}-\frac{3}{5}& \frac{2}{5}\\ \frac{1}{5}& \frac{1}{5}\end{array}\right]\\ =\left[\begin{array}{ll}\frac{11}{25}\cdot \left(-\frac{3}{5}\right)-\frac{4}{25}\cdot \frac{1}{5}& \frac{11}{25}\cdot \frac{2}{5}-\frac{4}{25}\cdot \frac{1}{5}\\ -\frac{2}{25}\cdot \left(-\frac{3}{25}\right)+\frac{3}{25}\cdot \frac{1}{5}& -\frac{2}{25}\cdot \left(-\frac{3}{5}\right)+\frac{3}{25}\cdot \frac{1}{5}\end{array}\right]\\ =\left[\begin{array}{cc}-\frac{37}{125}& \frac{18}{125}\\ \frac{9}{125}& -\frac{1}{125}\end{array}\right]\end{array}$.

Step 4

d. By part (b):

$A^3=\left[\begin{array}{cc}1& 18\\ 9& 37\end{array}\right]$

By the previous exercise, we know that the inverse of

$A=\left[\begin{array}{cc}a& b\\ c& d\end{array}\right]$

is

role="math" localid="1668444753345" $A^{-1}=\left[\begin{array}{cc}\frac{d}{ad-bc}& \frac{b}{dd--c}\\ -\frac{c}{dd-c}& \frac{a}{d-bc}\end{array}\right]$

In this case, $a=1,b=18,c=9,d=37$:

$\begin{array}{r}{\left(A^3\right)}^{-1}=\left[\begin{array}{cc}\frac{d}{ad-bc}& -\frac{b}{ad-bc}\\ -\frac{c}{ad-bc}& \frac{a}{ad-bc}\end{array}\right]\\ =\left[\begin{array}{cc}\frac{37}{\left(1\right)\left(37\right)-\left(18\right)\left(9\right)}& \frac{-18}{\left(1\right)\left(37\right)-\left(18\right)\left(9\right)}\\ \frac{-9}{\left(1\right)\left(37\right)-\left(18\right)\left(9\right)}& \frac{-(-1)}{\left(1\right)\left(37\right)-\left(18\right)\left(9\right)}\end{array}\right]\\ =\left[\begin{array}{cc}-\frac{37}{125}& \frac{18}{125}\\ \frac{9}{125}& -\frac{1}{125}\end{array}\right]\end{array}$

We now note that ${\left(A^3\right)}^{-1}={\left(A^{-1}\right)}^3$.