# 9.7 Solving systems with inverses  (Page 6/8)

 Page 6 / 8

## Algebraic

In the following exercises, show that matrix $\text{\hspace{0.17em}}A\text{\hspace{0.17em}}$ is the inverse of matrix $\text{\hspace{0.17em}}B.$

$A=\left[\begin{array}{cc}1& 0\\ -1& 1\end{array}\right],\text{\hspace{0.17em}}B=\left[\begin{array}{cc}1& 0\\ 1& 1\end{array}\right]$

$A=\left[\begin{array}{cc}1& 2\\ 3& 4\end{array}\right],\text{\hspace{0.17em}}B=\left[\begin{array}{cc}-2& 1\\ \frac{3}{2}& -\frac{1}{2}\end{array}\right]$

$AB=BA=\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]=I$

$A=\left[\begin{array}{cc}4& 5\\ 7& 0\end{array}\right],\text{\hspace{0.17em}}B=\left[\begin{array}{cc}0& \frac{1}{7}\\ \frac{1}{5}& -\frac{4}{35}\end{array}\right]$

$A=\left[\begin{array}{cc}-2& \frac{1}{2}\\ 3& -1\end{array}\right],\text{\hspace{0.17em}}B=\left[\begin{array}{cc}-2& -1\\ -6& -4\end{array}\right]$

$AB=BA=\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]=I$

$A=\left[\begin{array}{ccc}1& 0& 1\\ 0& 1& -1\\ 0& 1& 1\end{array}\right],\text{\hspace{0.17em}}B=\frac{1}{2}\left[\begin{array}{ccc}2& 1& -1\\ 0& 1& 1\\ 0& -1& 1\end{array}\right]$

$A=\left[\begin{array}{ccc}1& 2& 3\\ 4& 0& 2\\ 1& 6& 9\end{array}\right],\text{\hspace{0.17em}}B=\frac{1}{4}\left[\begin{array}{ccc}6& 0& -2\\ 17& -3& -5\\ -12& 2& 4\end{array}\right]$

$AB=BA=\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right]=I$

$A=\left[\begin{array}{ccc}3& 8& 2\\ 1& 1& 1\\ 5& 6& 12\end{array}\right],\text{\hspace{0.17em}}B=\frac{1}{36}\left[\begin{array}{ccc}-6& 84& -6\\ 7& -26& 1\\ -1& -22& 5\end{array}\right]$

For the following exercises, find the multiplicative inverse of each matrix, if it exists.

$\left[\begin{array}{cc}3& -2\\ 1& 9\end{array}\right]$

$\frac{1}{29}\left[\begin{array}{cc}9& 2\\ -1& 3\end{array}\right]$

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

$\left[\begin{array}{cc}-3& 7\\ 9& 2\end{array}\right]$

$\frac{1}{69}\left[\begin{array}{cc}-2& 7\\ 9& 3\end{array}\right]$

$\left[\begin{array}{cc}-4& -3\\ -5& 8\end{array}\right]$

$\left[\begin{array}{cc}1& 1\\ 2& 2\end{array}\right]$

There is no inverse

$\left[\begin{array}{cc}0& 1\\ 1& 0\end{array}\right]$

$\left[\begin{array}{cc}0.5& 1.5\\ 1& -0.5\end{array}\right]$

$\frac{4}{7}\left[\begin{array}{cc}0.5& 1.5\\ 1& -0.5\end{array}\right]$

$\left[\begin{array}{ccc}1& 0& 6\\ -2& 1& 7\\ 3& 0& 2\end{array}\right]$

$\left[\begin{array}{ccc}0& 1& -3\\ 4& 1& 0\\ 1& 0& 5\end{array}\right]$

$\frac{1}{17}\left[\begin{array}{ccc}-5& 5& -3\\ 20& -3& 12\\ 1& -1& 4\end{array}\right]$

$\left[\begin{array}{ccc}1& 2& -1\\ -3& 4& 1\\ -2& -4& -5\end{array}\right]$

$\left[\begin{array}{ccc}1& 9& -3\\ 2& 5& 6\\ 4& -2& 7\end{array}\right]$

$\frac{1}{209}\left[\begin{array}{ccc}47& -57& 69\\ 10& 19& -12\\ -24& 38& -13\end{array}\right]$

$\left[\begin{array}{ccc}1& -2& 3\\ -4& 8& -12\\ 1& 4& 2\end{array}\right]$

$\left[\begin{array}{ccc}\frac{1}{2}& \frac{1}{2}& \frac{1}{2}\\ \frac{1}{3}& \frac{1}{4}& \frac{1}{5}\\ \frac{1}{6}& \frac{1}{7}& \frac{1}{8}\end{array}\right]$

$\left[\begin{array}{ccc}18& 60& -168\\ -56& -140& 448\\ 40& 80& -280\end{array}\right]$

$\left[\begin{array}{ccc}1& 2& 3\\ 4& 5& 6\\ 7& 8& 9\end{array}\right]$

For the following exercises, solve the system using the inverse of a $\text{\hspace{0.17em}}2\text{\hspace{0.17em}}×\text{\hspace{0.17em}}2\text{\hspace{0.17em}}$ matrix.

$\left(-5,6\right)$

$\begin{array}{l}8x+4y=-100\\ 3x-4y=1\end{array}$

$\begin{array}{l}\text{\hspace{0.17em}}3x-2y=6\hfill \\ -x+5y=-2\hfill \end{array}$

$\left(2,0\right)$

$\begin{array}{l}5x-4y=-5\hfill \\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}4x+y=2.3\hfill \end{array}$

$\begin{array}{l}-3x-4y=9\hfill \\ \text{\hspace{0.17em}}12x+4y=-6\hfill \end{array}$

$\left(\frac{1}{3},-\frac{5}{2}\right)$

$\begin{array}{l}-2x+3y=\frac{3}{10}\hfill \\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}-x+5y=\frac{1}{2}\hfill \end{array}$

$\begin{array}{l}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\frac{8}{5}x-\frac{4}{5}y=\frac{2}{5}\hfill \\ -\frac{8}{5}x+\frac{1}{5}y=\frac{7}{10}\hfill \end{array}$

$\left(-\frac{2}{3},-\frac{11}{6}\right)$

$\begin{array}{l}\frac{1}{2}x+\frac{1}{5}y=-\frac{1}{4}\\ \frac{1}{2}x-\frac{3}{5}y=-\frac{9}{4}\end{array}$

For the following exercises, solve a system using the inverse of a $\text{\hspace{0.17em}}3\text{}×\text{}3\text{\hspace{0.17em}}$ matrix.

$\begin{array}{l}3x-2y+5z=21\hfill \\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}5x+4y=37\hfill \\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}x-2y-5z=5\hfill \end{array}$

$\left(7,\frac{1}{2},\frac{1}{5}\right)$

$\left(5,0,-1\right)$

$\begin{array}{l}6x-5y+2z=-4\hfill \\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}2x+5y-z=12\hfill \\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}2x+5y+z=12\hfill \end{array}$

$\begin{array}{l}4x-2y+3z=-12\hfill \\ 2x+2y-9z=33\hfill \\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}6y-4z=1\hfill \end{array}$

$\frac{1}{34}\left(-35,-97,-154\right)$

$\begin{array}{l}\frac{1}{10}x-\frac{1}{5}y+4z=\frac{-41}{2}\\ \frac{1}{5}x-20y+\frac{2}{5}z=-101\\ \frac{3}{10}x+4y-\frac{3}{10}z=23\end{array}$

$\begin{array}{l}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\frac{1}{2}x-\frac{1}{5}y+\frac{1}{5}z=\frac{31}{100}\hfill \\ -\frac{3}{4}x-\frac{1}{4}y+\frac{1}{2}z=\frac{7}{40}\hfill \\ -\frac{4}{5}x-\frac{1}{2}y+\frac{3}{2}z=\frac{1}{4}\hfill \end{array}$

$\frac{1}{690}\left(65,-1136,-229\right)$

$\begin{array}{l}0.1x+0.2y+0.3z=-1.4\hfill \\ 0.1x-0.2y+0.3z=0.6\hfill \\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}0.4y+0.9z=-2\hfill \end{array}$

## Technology

For the following exercises, use a calculator to solve the system of equations with matrix inverses.

$\begin{array}{l}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}2x-y=-3\hfill \\ -x+2y=2.3\hfill \end{array}$

$\left(-\frac{37}{30},\frac{8}{15}\right)$

$\begin{array}{l}-\frac{1}{2}x-\frac{3}{2}y=-\frac{43}{20}\hfill \\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\frac{5}{2}x+\frac{11}{5}y=\frac{31}{4}\hfill \end{array}$

$\begin{array}{l}12.3x-2y-2.5z=2\hfill \\ 36.9x+7y-7.5z=-7\hfill \\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}8y-5z=-10\hfill \end{array}$

$\left(\frac{10}{123},-1,\frac{2}{5}\right)$

$\begin{array}{l}0.5x-3y+6z=-0.8\hfill \\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}0.7x-2y=-0.06\hfill \\ 0.5x+4y+5z=0\hfill \end{array}$

## Extensions

For the following exercises, find the inverse of the given matrix.

$\left[\begin{array}{cccc}1& 0& 1& 0\\ 0& 1& 0& 1\\ 0& 1& 1& 0\\ 0& 0& 1& 1\end{array}\right]$

$\frac{1}{2}\left[\begin{array}{rrrr}\hfill 2& \hfill 1& \hfill -1& \hfill -1\\ \hfill 0& \hfill 1& \hfill 1& \hfill -1\\ \hfill 0& \hfill -1& \hfill 1& \hfill 1\\ \hfill 0& \hfill 1& \hfill -1& \hfill 1\end{array}\right]$

$\left[\begin{array}{rrrr}\hfill -1& \hfill 0& \hfill 2& \hfill 5\\ \hfill 0& \hfill 0& \hfill 0& \hfill 2\\ \hfill 0& \hfill 2& \hfill -1& \hfill 0\\ \hfill 1& \hfill -3& \hfill 0& \hfill 1\end{array}\right]$

$\left[\begin{array}{rrrr}\hfill 1& \hfill -2& \hfill 3& \hfill 0\\ \hfill 0& \hfill 1& \hfill 0& \hfill 2\\ \hfill 1& \hfill 4& \hfill -2& \hfill 3\\ \hfill -5& \hfill 0& \hfill 1& \hfill 1\end{array}\right]$

$\frac{1}{39}\left[\begin{array}{rrrr}\hfill 3& \hfill 2& \hfill 1& \hfill -7\\ \hfill 18& \hfill -53& \hfill 32& \hfill 10\\ \hfill 24& \hfill -36& \hfill 21& \hfill 9\\ \hfill -9& \hfill 46& \hfill -16& \hfill -5\end{array}\right]$

$\left[\begin{array}{rrrrr}\hfill 1& \hfill 2& \hfill 0& \hfill 2& \hfill 3\\ \hfill 0& \hfill 2& \hfill 1& \hfill 0& \hfill 0\\ \hfill 0& \hfill 0& \hfill 3& \hfill 0& \hfill 1\\ \hfill 0& \hfill 2& \hfill 0& \hfill 0& \hfill 1\\ \hfill 0& \hfill 0& \hfill 1& \hfill 2& \hfill 0\end{array}\right]$

$\left[\begin{array}{rrrrrr}\hfill 1& \hfill 0& \hfill 0& \hfill 0& \hfill 0& \hfill 0\\ \hfill 0& \hfill 1& \hfill 0& \hfill 0& \hfill 0& \hfill 0\\ \hfill 0& \hfill 0& \hfill 1& \hfill 0& \hfill 0& \hfill 0\\ \hfill 0& \hfill 0& \hfill 0& \hfill 1& \hfill 0& \hfill 0\\ \hfill 0& \hfill 0& \hfill 0& \hfill 0& \hfill 1& \hfill 0\\ \hfill 1& \hfill 1& \hfill 1& \hfill 1& \hfill 1& \hfill 1\end{array}\right]$

$\left[\begin{array}{rrrrrr}\hfill 1& \hfill 0& \hfill 0& \hfill 0& \hfill 0& \hfill 0\\ \hfill 0& \hfill 1& \hfill 0& \hfill 0& \hfill 0& \hfill 0\\ \hfill 0& \hfill 0& \hfill 1& \hfill 0& \hfill 0& \hfill 0\\ \hfill 0& \hfill 0& \hfill 0& \hfill 1& \hfill 0& \hfill 0\\ \hfill 0& \hfill 0& \hfill 0& \hfill 0& \hfill 1& \hfill 0\\ \hfill -1& \hfill -1& \hfill -1& \hfill -1& \hfill -1& \hfill 1\end{array}\right]$

## Real-world applications

For the following exercises, write a system of equations that represents the situation. Then, solve the system using the inverse of a matrix.

2,400 tickets were sold for a basketball game. If the prices for floor 1 and floor 2 were different, and the total amount of money brought in is $64,000, how much was the price of each ticket? In the previous exercise, if you were told there were 400 more tickets sold for floor 2 than floor 1, how much was the price of each ticket? Infinite solutions. A food drive collected two different types of canned goods, green beans and kidney beans. The total number of collected cans was 350 and the total weight of all donated food was 348 lb, 12 oz. If the green bean cans weigh 2 oz less than the kidney bean cans, how many of each can was donated? Students were asked to bring their favorite fruit to class. 95% of the fruits consisted of banana, apple, and oranges. If oranges were twice as popular as bananas, and apples were 5% less popular than bananas, what are the percentages of each individual fruit? 50% oranges, 25% bananas, 20% apples A sorority held a bake sale to raise money and sold brownies and chocolate chip cookies. They priced the brownies at$1 and the chocolate chip cookies at $0.75. They raised$700 and sold 850 items. How many brownies and how many cookies were sold?

A clothing store needs to order new inventory. It has three different types of hats for sale: straw hats, beanies, and cowboy hats. The straw hat is priced at $13.99, the beanie at$7.99, and the cowboy hat at $14.49. If 100 hats were sold this past quarter,$1,119 was taken in by sales, and the amount of beanies sold was 10 more than cowboy hats, how many of each should the clothing store order to replace those already sold?

10 straw hats, 50 beanies, 40 cowboy hats

Anna, Ashley, and Andrea weigh a combined 370 lb. If Andrea weighs 20 lb more than Ashley, and Anna weighs 1.5 times as much as Ashley, how much does each girl weigh?

Three roommates shared a package of 12 ice cream bars, but no one remembers who ate how many. If Tom ate twice as many ice cream bars as Joe, and Albert ate three less than Tom, how many ice cream bars did each roommate eat?

Tom ate 6, Joe ate 3, and Albert ate 3.

A farmer constructed a chicken coop out of chicken wire, wood, and plywood. The chicken wire cost $2 per square foot, the wood$10 per square foot, and the plywood $5 per square foot. The farmer spent a total of$51, and the total amount of materials used was He used more chicken wire than plywood. How much of each material in did the farmer use?

Jay has lemon, orange, and pomegranate trees in his backyard. An orange weighs 8 oz, a lemon 5 oz, and a pomegranate 11 oz. Jay picked 142 pieces of fruit weighing a total of 70 lb, 10 oz. He picked 15.5 times more oranges than pomegranates. How many of each fruit did Jay pick?

124 oranges, 10 lemons, 8 pomegranates

So a horizontal compression by factor of 1/2 is the same as a horizontal stretch by a factor of 2, right?
The center is at (3,4) a focus is at (3,-1), and the lenght of the major axis is 26
The center is at (3,4) a focus is at (3,-1) and the lenght of the major axis is 26 what will be the answer?
Rima
I done know
Joe
What kind of answer is that😑?
Rima
I had just woken up when i got this message
Joe
Rima
i have a question.
Abdul
how do you find the real and complex roots of a polynomial?
Abdul
@abdul with delta maybe which is b(square)-4ac=result then the 1st root -b-radical delta over 2a and the 2nd root -b+radical delta over 2a. I am not sure if this was your question but check it up
Nare
This is the actual question: Find all roots(real and complex) of the polynomial f(x)=6x^3 + x^2 - 4x + 1
Abdul
@Nare please let me know if you can solve it.
Abdul
I have a question
juweeriya
hello guys I'm new here? will you happy with me
mustapha
The average annual population increase of a pack of wolves is 25.
how do you find the period of a sine graph
Period =2π if there is a coefficient (b), just divide the coefficient by 2π to get the new period
Am
if not then how would I find it from a graph
Imani
by looking at the graph, find the distance between two consecutive maximum points (the highest points of the wave). so if the top of one wave is at point A (1,2) and the next top of the wave is at point B (6,2), then the period is 5, the difference of the x-coordinates.
Am
you could also do it with two consecutive minimum points or x-intercepts
Am
I will try that thank u
Imani
Case of Equilateral Hyperbola
ok
Zander
ok
Shella
f(x)=4x+2, find f(3)
Benetta
f(3)=4(3)+2 f(3)=14
lamoussa
14
Vedant
pre calc teacher: "Plug in Plug in...smell's good" f(x)=14
Devante
8x=40
Chris
Explain why log a x is not defined for a < 0
the sum of any two linear polynomial is what
Momo
how can are find the domain and range of a relations
the range is twice of the natural number which is the domain
Morolake
A cell phone company offers two plans for minutes. Plan A: $15 per month and$2 for every 300 texts. Plan B: $25 per month and$0.50 for every 100 texts. How many texts would you need to send per month for plan B to save you money?
6000
Robert
more than 6000
Robert
For Plan A to reach $27/month to surpass Plan B's$26.50 monthly payment, you'll need 3,000 texts which will cost an additional \$10.00. So, for the amount of texts you need to send would need to range between 1-100 texts for the 100th increment, times that by 3 for the additional amount of texts...
Gilbert
...for one text payment for 300 for Plan A. So, that means Plan A; in my opinion is for people with text messaging abilities that their fingers burn the monitor for the cell phone. While Plan B would be for loners that doesn't need their fingers to due the talking; but those texts mean more then...
Gilbert
can I see the picture
How would you find if a radical function is one to one?
how to understand calculus?
with doing calculus
SLIMANE
Thanks po.
Jenica
Hey I am new to precalculus, and wanted clarification please on what sine is as I am floored by the terms in this app? I don't mean to sound stupid but I have only completed up to college algebra.
I don't know if you are looking for a deeper answer or not, but the sine of an angle in a right triangle is the length of the opposite side to the angle in question divided by the length of the hypotenuse of said triangle.
Marco
can you give me sir tips to quickly understand precalculus. Im new too in that topic. Thanks
Jenica
if you remember sine, cosine, and tangent from geometry, all the relationships are the same but they use x y and r instead (x is adjacent, y is opposite, and r is hypotenuse).
Natalie
it is better to use unit circle than triangle .triangle is only used for acute angles but you can begin with. Download any application named"unit circle" you find in it all you need. unit circle is a circle centred at origine (0;0) with radius r= 1.
SLIMANE
What is domain
johnphilip
the standard equation of the ellipse that has vertices (0,-4)&(0,4) and foci (0, -15)&(0,15) it's standard equation is x^2 + y^2/16 =1 tell my why is it only x^2? why is there no a^2?