# 11.7 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

exercise 1.2 solution b....isnt it lacking
I dnt get dis work well
what is one-to-one function
what is the procedure in solving quadratic equetion at least 6?
Almighty formula or by factorization...or by graphical analysis
Damian
I need to learn this trigonometry from A level.. can anyone help here?
yes am hia
Miiro
tanh2x =2tanhx/1+tanh^2x
cos(a+b)+cos(a-b)/sin(a+b)-sin(a-b)=cotb ... pls some one should help me with this..thanks in anticipation
f(x)=x/x+2 given g(x)=1+2x/1-x show that gf(x)=1+2x/3
proof
AUSTINE
sebd me some questions about anything ill solve for yall
cos(a+b)+cos(a-b)/sin(a+b)-sin(a-b)= cotb
favour
how to solve x²=2x+8 factorization?
x=2x+8 x-2x=2x+8-2x x-2x=8 -x=8 -x/-1=8/-1 x=-8 prove: if x=-8 -8=2(-8)+8 -8=-16+8 -8=-8 (PROVEN)
Manifoldee
x=2x+8
Manifoldee
×=2x-8 minus both sides by 2x
Manifoldee
so, x-2x=2x+8-2x
Manifoldee
then cancel out 2x and -2x, cuz 2x-2x is obviously zero
Manifoldee
so it would be like this: x-2x=8
Manifoldee
then we all know that beside the variable is a number (1): (1)x-2x=8
Manifoldee
so we will going to minus that 1-2=-1
Manifoldee
so it would be -x=8
Manifoldee
so next step is to cancel out negative number beside x so we get positive x
Manifoldee
so by doing it you need to divide both side by -1 so it would be like this: (-1x/-1)=(8/-1)
Manifoldee
so -1/-1=1
Manifoldee
so x=-8
Manifoldee
Manifoldee
so we should prove it
Manifoldee
x=2x+8 x-2x=8 -x=8 x=-8 by mantu from India
mantu
lol i just saw its x²
Manifoldee
x²=2x-8 x²-2x=8 -x²=8 x²=-8 square root(x²)=square root(-8) x=sq. root(-8)
Manifoldee
I mean x²=2x+8 by factorization method
Kristof
I think x=-2 or x=4
Kristof
x= 2x+8 ×=8-2x - 2x + x = 8 - x = 8 both sides divided - 1 -×/-1 = 8/-1 × = - 8 //// from somalia
Mohamed
hii
Amit
how are you
Dorbor
well
Biswajit
can u tell me concepts
Gaurav
Find the possible value of 8.5 using moivre's theorem
which of these functions is not uniformly cintinuous on (0, 1)? sinx
helo
Akash
hlo
Akash
Hello
Hudheifa
which of these functions is not uniformly continuous on 0,1
solve this equation by completing the square 3x-4x-7=0
X=7
Muustapha
=7
mantu
x=7
mantu
3x-4x-7=0 -x=7 x=-7
Kr
x=-7
mantu
9x-16x-49=0 -7x=49 -x=7 x=7
mantu
what's the formula
Modress
-x=7
Modress
new member
siame