7.7 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

what is the coefficient of -4×
-1
Shedrak
the operation * is x * y =x + y/ 1+(x × y) show if the operation is commutative if x × y is not equal to -1
An investment account was opened with an initial deposit of \$9,600 and earns 7.4% interest, compounded continuously. How much will the account be worth after 15 years?
lim x to infinity e^1-e^-1/log(1+x)
given eccentricity and a point find the equiation
12, 17, 22.... 25th term
12, 17, 22.... 25th term
Akash
College algebra is really hard?
Absolutely, for me. My problems with math started in First grade...involving a nun Sister Anastasia, bad vision, talking & getting expelled from Catholic school. When it comes to math I just can't focus and all I can hear is our family silverware banging and clanging on the pink Formica table.
Carole
I'm 13 and I understand it great
AJ
I am 1 year old but I can do it! 1+1=2 proof very hard for me though.
Atone
hi
Not really they are just easy concepts which can be understood if you have great basics. I am 14 I understood them easily.
Vedant
find the 15th term of the geometric sequince whose first is 18 and last term of 387
I know this work
salma
The given of f(x=x-2. then what is the value of this f(3) 5f(x+1)
hmm well what is the answer
Abhi
If f(x) = x-2 then, f(3) when 5f(x+1) 5((3-2)+1) 5(1+1) 5(2) 10
Augustine
how do they get the third part x = (32)5/4
make 5/4 into a mixed number, make that a decimal, and then multiply 32 by the decimal 5/4 turns out to be
AJ
how
Sheref
can someone help me with some logarithmic and exponential equations.
20/(×-6^2)
Salomon
okay, so you have 6 raised to the power of 2. what is that part of your answer
I don't understand what the A with approx sign and the boxed x mean
it think it's written 20/(X-6)^2 so it's 20 divided by X-6 squared
Salomon
I'm not sure why it wrote it the other way
Salomon
I got X =-6
Salomon
ok. so take the square root of both sides, now you have plus or minus the square root of 20= x-6
oops. ignore that.
so you not have an equal sign anywhere in the original equation?
hmm
Abhi
is it a question of log
Abhi
🤔.
Abhi
I rally confuse this number And equations too I need exactly help
salma
But this is not salma it's Faiza live in lousvile Ky I garbage this so I am going collage with JCTC that the of the collage thank you my friends
salma
Commplementary angles
hello
Sherica
im all ears I need to learn
Sherica
right! what he said ⤴⤴⤴
Tamia
hii
Uday
hi
salma
hi
Ayuba
Hello
opoku
hi
Ali
greetings from Iran
Ali
salut. from Algeria
Bach
hi
Nharnhar
what is a good calculator for all algebra; would a Casio fx 260 work with all algebra equations? please name the cheapest, thanks.
a perfect square v²+2v+_
kkk nice