# 9.7 Solving systems with inverses  (Page 2/8)

 Page 2 / 8

Show that the following two matrices are inverses of each other.

$A=\left[\begin{array}{rrr}\hfill 1& \hfill & \hfill 4\\ \hfill -1& \hfill & \hfill -3\end{array}\right],B=\left[\begin{array}{rrr}\hfill -3& \hfill & \hfill -4\\ \hfill 1& \hfill & \hfill 1\end{array}\right]$
$\begin{array}{l}AB=\left[\begin{array}{rrr}\hfill 1& \hfill & \hfill 4\\ \hfill -1& \hfill & \hfill -3\end{array}\right]\begin{array}{r}\hfill \end{array}\left[\begin{array}{rrr}\hfill -3& \hfill & \hfill -4\\ \hfill 1& \hfill & \hfill 1\end{array}\right]=\left[\begin{array}{rrr}\hfill 1\left(-3\right)+4\left(1\right)& \hfill & \hfill 1\left(-4\right)+4\left(1\right)\\ \hfill -1\left(-3\right)+-3\left(1\right)& \hfill & \hfill -1\left(-4\right)+-3\left(1\right)\end{array}\right]=\left[\begin{array}{rrr}\hfill 1& \hfill & \hfill 0\\ \hfill 0& \hfill & \hfill 1\end{array}\right]\hfill \\ BA=\left[\begin{array}{rrr}\hfill -3& \hfill & \hfill -4\\ \hfill 1& \hfill & \hfill 1\end{array}\right]\begin{array}{r}\hfill \end{array}\left[\begin{array}{rrr}\hfill 1& \hfill & \hfill 4\\ \hfill -1& \hfill & \hfill -3\end{array}\right]=\left[\begin{array}{rrr}\hfill -3\left(1\right)+-4\left(-1\right)& \hfill & \hfill -3\left(4\right)+-4\left(-3\right)\\ \hfill 1\left(1\right)+1\left(-1\right)& \hfill & \hfill 1\left(4\right)+1\left(-3\right)\end{array}\right]=\left[\begin{array}{rrr}\hfill 1& \hfill & \hfill 0\\ \hfill 0& \hfill & \hfill 1\end{array}\right]\hfill \end{array}$

## Finding the multiplicative inverse using matrix multiplication

We can now determine whether two matrices are inverses, but how would we find the inverse of a given matrix? Since we know that the product of a matrix and its inverse is the identity matrix, we can find the inverse of a matrix by setting up an equation using matrix multiplication .

## Finding the multiplicative inverse using matrix multiplication

Use matrix multiplication to find the inverse of the given matrix.

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

For this method, we multiply $\text{\hspace{0.17em}}A\text{\hspace{0.17em}}$ by a matrix containing unknown constants and set it equal to the identity.

Find the product of the two matrices on the left side of the equal sign.

Next, set up a system of equations with the entry in row 1, column 1 of the new matrix equal to the first entry of the identity, 1. Set the entry in row 2, column 1 of the new matrix equal to the corresponding entry of the identity, which is 0.

Using row operations, multiply and add as follows: $\text{\hspace{0.17em}}\left(-2\right){R}_{1}+{R}_{2}\to {R}_{2}.\text{\hspace{0.17em}}$ Add the equations, and solve for $\text{\hspace{0.17em}}c.$

$\begin{array}{r}\hfill 1a-2c=1\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\\ \hfill 0+1c=-2\\ \hfill c=-2\end{array}$

Back-substitute to solve for $\text{\hspace{0.17em}}a.$

$\begin{array}{r}\hfill a-2\left(-2\right)=1\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\\ \hfill a+4=1\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\\ \hfill a=-3\end{array}$

Write another system of equations setting the entry in row 1, column 2 of the new matrix equal to the corresponding entry of the identity, 0. Set the entry in row 2, column 2 equal to the corresponding entry of the identity.

$\begin{array}{rr}\hfill 1b-2d=0& \hfill {R}_{1}\\ \hfill 2b-3d=1& \hfill {R}_{2}\end{array}$

Using row operations, multiply and add as follows: $\text{\hspace{0.17em}}\left(-2\right){R}_{1}+{R}_{2}={R}_{2}.\text{\hspace{0.17em}}$ Add the two equations and solve for $\text{\hspace{0.17em}}d.$

$\begin{array}{r}\hfill 1b-2d=0\\ \hfill \frac{0+1d=1}{\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}d=1}\\ \hfill \end{array}$

Once more, back-substitute and solve for $\text{\hspace{0.17em}}b.$

$\begin{array}{r}\hfill b-2\left(1\right)=0\\ \hfill b-2=0\\ \hfill b=2\end{array}$
${A}^{-1}=\left[\begin{array}{rrr}\hfill -3& \hfill & \hfill 2\\ \hfill -2& \hfill & \hfill 1\end{array}\right]$

## Finding the multiplicative inverse by augmenting with the identity

Another way to find the multiplicative inverse is by augmenting with the identity. When matrix $\text{\hspace{0.17em}}A\text{\hspace{0.17em}}$ is transformed into $\text{\hspace{0.17em}}I,\text{\hspace{0.17em}}$ the augmented matrix $\text{\hspace{0.17em}}I\text{\hspace{0.17em}}$ transforms into $\text{\hspace{0.17em}}{A}^{-1}.$

For example, given

$A=\left[\begin{array}{rrr}\hfill 2& \hfill & \hfill 1\\ \hfill 5& \hfill & \hfill 3\end{array}\right]$

augment $\text{\hspace{0.17em}}A\text{\hspace{0.17em}}$ with the identity

Perform row operations    with the goal of turning $\text{\hspace{0.17em}}A\text{\hspace{0.17em}}$ into the identity.

1. Switch row 1 and row 2.
2. Multiply row 2 by $\text{\hspace{0.17em}}-2\text{\hspace{0.17em}}$ and add to row 1.
3. Multiply row 1 by $\text{\hspace{0.17em}}-2\text{\hspace{0.17em}}$ and add to row 2.
4. Add row 2 to row 1.
5. Multiply row 2 by $\text{\hspace{0.17em}}-1.$

The matrix we have found is $\text{\hspace{0.17em}}{A}^{-1}.$

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

## Finding the multiplicative inverse of 2×2 matrices using a formula

When we need to find the multiplicative inverse of a $\text{\hspace{0.17em}}2\text{\hspace{0.17em}}×\text{\hspace{0.17em}}2\text{\hspace{0.17em}}$ matrix, we can use a special formula instead of using matrix multiplication or augmenting with the identity.

If $\text{\hspace{0.17em}}A\text{\hspace{0.17em}}$ is a $\text{\hspace{0.17em}}2×2\text{\hspace{0.17em}}$ matrix, such as

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

the multiplicative inverse of $\text{\hspace{0.17em}}A\text{\hspace{0.17em}}$ is given by the formula

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

where $\text{\hspace{0.17em}}ad-bc\ne 0.\text{\hspace{0.17em}}$ If $\text{\hspace{0.17em}}ad-bc=0,\text{\hspace{0.17em}}$ then $\text{\hspace{0.17em}}A\text{\hspace{0.17em}}$ has no inverse.

## Using the formula to find the multiplicative inverse of matrix A

Use the formula to find the multiplicative inverse of

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

Using the formula, we have

$\begin{array}{l}{A}^{-1}=\frac{1}{\left(1\right)\left(-3\right)-\left(-2\right)\left(2\right)}\left[\begin{array}{cc}-3& 2\\ -2& 1\end{array}\right]\hfill \\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\frac{1}{-3+4}\left[\begin{array}{cc}-3& 2\\ -2& 1\end{array}\right]\hfill \\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\left[\begin{array}{cc}-3& 2\\ -2& 1\end{array}\right]\hfill \end{array}$

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
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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
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?
what is foci?
This term is plural for a focus, it is used for conic sections. For more detail or other math questions. I recommend researching on "Khan academy" or watching "The Organic Chemistry Tutor" YouTube channel.
Chris