# 9.7 Solving systems with inverses  (Page 3/8)

 Page 3 / 8

Use the formula to find the inverse of matrix $\text{\hspace{0.17em}}A.\text{\hspace{0.17em}}$ Verify your answer by augmenting with the identity matrix.

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

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

## Finding the inverse of the matrix, if it exists

Find the inverse, if it exists, of the given matrix.

$A=\left[\begin{array}{cc}3& 6\\ 1& 2\end{array}\right]$

We will use the method of augmenting with the identity.

$\left[\begin{array}{cc}3& 6\\ 1& 3\end{array}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}|\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]$
1. Switch row 1 and row 2.
$\left[\begin{array}{cc}1& 3\\ 3& 6\text{\hspace{0.17em}}\text{​}\end{array}\text{​}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{​}\text{​}|\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\begin{array}{cc}0& 1\\ 1& 0\end{array}\right]$
2. Multiply row 1 by −3 and add it to row 2.
$\left[\begin{array}{cc}1& 2\\ 0& 0\end{array}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}|\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\begin{array}{cc}1& 0\\ -3& 1\end{array}\right]$
3. There is nothing further we can do. The zeros in row 2 indicate that this matrix has no inverse.

## Finding the multiplicative inverse of 3×3 matrices

Unfortunately, we do not have a formula similar to the one for a $\text{\hspace{0.17em}}2\text{}×\text{}2\text{\hspace{0.17em}}$ matrix to find the inverse of a $\text{\hspace{0.17em}}3\text{}×\text{}3\text{\hspace{0.17em}}$ matrix. Instead, we will augment the original matrix with the identity matrix and use row operations    to obtain the inverse.

Given a $\text{\hspace{0.17em}}3\text{}×\text{}3\text{\hspace{0.17em}}$ matrix

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

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

To begin, we write the augmented matrix    with the identity on the right and $\text{\hspace{0.17em}}A\text{\hspace{0.17em}}$ on the left. Performing elementary row operations    so that the identity matrix    appears on the left, we will obtain the inverse matrix on the right. We will find the inverse of this matrix in the next example.

Given a $\text{\hspace{0.17em}}3\text{\hspace{0.17em}}×\text{\hspace{0.17em}}3\text{\hspace{0.17em}}$ matrix, find the inverse

1. Write the original matrix augmented with the identity matrix on the right.
2. Use elementary row operations so that the identity appears on the left.
3. What is obtained on the right is the inverse of the original matrix.
4. Use matrix multiplication to show that $\text{\hspace{0.17em}}A{A}^{-1}=I\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}{A}^{-1}A=I.$

## Finding the inverse of a 3 × 3 matrix

Given the $\text{\hspace{0.17em}}3\text{\hspace{0.17em}}×\text{\hspace{0.17em}}3\text{\hspace{0.17em}}$ matrix $\text{\hspace{0.17em}}A,\text{\hspace{0.17em}}$ find the inverse.

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

Augment $\text{\hspace{0.17em}}A\text{\hspace{0.17em}}$ with the identity matrix, and then begin row operations until the identity matrix replaces $\text{\hspace{0.17em}}A.\text{\hspace{0.17em}}$ The matrix on the right will be the inverse of $\text{\hspace{0.17em}}A.\text{\hspace{0.17em}}$

$-{R}_{2}+{R}_{1}={R}_{1}\to \left[\begin{array}{ccc}1& 0& 0\\ 2& 3& 1\\ 2& 4& 1\end{array}\text{\hspace{0.17em}}\text{\hspace{0.17em}}|\text{\hspace{0.17em}}\text{\hspace{0.17em}}\begin{array}{rrr}\hfill -1& \hfill 1& \hfill 0\\ \hfill 1& \hfill 0& \hfill 0\\ \hfill 0& \hfill 0& \hfill 1\end{array}\right]$
$-{R}_{2}+{R}_{3}={R}_{3}\to \left[\begin{array}{ccc}1& 0& 0\\ 2& 3& 1\\ 0& 1& 0\end{array}\text{\hspace{0.17em}}\text{\hspace{0.17em}}|\text{\hspace{0.17em}}\text{\hspace{0.17em}}\begin{array}{rrr}\hfill -1& \hfill 1& \hfill 0\\ \hfill 1& \hfill 0& \hfill 0\\ \hfill -1& \hfill 0& \hfill 1\end{array}\right]$
$-2{R}_{1}+{R}_{3}={R}_{3}\to \left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 3& 1\end{array}\text{\hspace{0.17em}}\text{\hspace{0.17em}}|\text{\hspace{0.17em}}\text{\hspace{0.17em}}\begin{array}{rrr}\hfill -1& \hfill 1& \hfill 0\\ \hfill -1& \hfill 0& \hfill 1\\ \hfill 3& \hfill -2& \hfill 0\end{array}\right]$
$-3{R}_{2}+{R}_{3}={R}_{3}\to \left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\text{\hspace{0.17em}}\text{\hspace{0.17em}}|\text{\hspace{0.17em}}\text{\hspace{0.17em}}\begin{array}{rrr}\hfill -1& \hfill 1& \hfill 0\\ \hfill -1& \hfill 0& \hfill 1\\ \hfill 6& \hfill -2& \hfill -3\end{array}\right]$

Thus,

${A}^{-1}=B=\left[\begin{array}{ccc}-1& \text{\hspace{0.17em}}1& \text{\hspace{0.17em}}0\\ -1& \text{\hspace{0.17em}}\text{\hspace{0.17em}}0& \text{\hspace{0.17em}}\text{\hspace{0.17em}}1\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}6& -2& -3\end{array}\text{\hspace{0.17em}}\right]$

Find the inverse of the $\text{\hspace{0.17em}}3×3\text{\hspace{0.17em}}$ matrix.

$A=\left[\begin{array}{ccc}\text{\hspace{0.17em}}\text{\hspace{0.17em}}2& -17& 11\\ -1& \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}11& -7\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}0& \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}3& -2\end{array}\right]$

${A}^{-1}=\left[\begin{array}{ccc}1& 1& \text{\hspace{0.17em}}\text{\hspace{0.17em}}2\\ 2& 4& -3\\ 3& 6& -5\end{array}\right]$

## Solving a system of linear equations using the inverse of a matrix

Solving a system of linear equations using the inverse of a matrix requires the definition of two new matrices: $\text{\hspace{0.17em}}X\text{\hspace{0.17em}}$ is the matrix representing the variables of the system, and $\text{\hspace{0.17em}}B\text{\hspace{0.17em}}$ is the matrix representing the constants. Using matrix multiplication , we may define a system of equations with the same number of equations as variables as

$AX=B$

To solve a system of linear equations using an inverse matrix , let $\text{\hspace{0.17em}}A\text{\hspace{0.17em}}$ be the coefficient matrix    , let $\text{\hspace{0.17em}}X\text{\hspace{0.17em}}$ be the variable matrix, and let $\text{\hspace{0.17em}}B\text{\hspace{0.17em}}$ be the constant matrix. Thus, we want to solve a system $\text{\hspace{0.17em}}AX=B.\text{\hspace{0.17em}}$ For example, look at the following system of equations.

$\begin{array}{c}{a}_{1}x+{b}_{1}y={c}_{1}\\ {a}_{2}x+{b}_{2}y={c}_{2}\end{array}$

From this system, the coefficient matrix is

$A=\left[\begin{array}{cc}{a}_{1}& {b}_{1}\\ {a}_{2}& {b}_{2}\end{array}\right]$

The variable matrix is

$X=\left[\begin{array}{c}x\\ y\end{array}\right]$

And the constant matrix is

$B=\left[\begin{array}{c}{c}_{1}\\ {c}_{2}\end{array}\right]$

Then $\text{\hspace{0.17em}}AX=B\text{\hspace{0.17em}}$ looks like

Recall the discussion earlier in this section regarding multiplying a real number by its inverse, $\text{\hspace{0.17em}}\left({2}^{-1}\right)\text{\hspace{0.17em}}2=\left(\frac{1}{2}\right)\text{\hspace{0.17em}}2=1.\text{\hspace{0.17em}}$ To solve a single linear equation $\text{\hspace{0.17em}}ax=b\text{\hspace{0.17em}}$ for $\text{\hspace{0.17em}}x,\text{\hspace{0.17em}}$ we would simply multiply both sides of the equation by the multiplicative inverse (reciprocal) of $\text{\hspace{0.17em}}a.\text{\hspace{0.17em}}$ Thus,

#### Questions & Answers

The average annual population increase of a pack of wolves is 25.
Brittany Reply
how do you find the period of a sine graph
Imani Reply
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
Jhon Reply
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
Baptiste Reply
the sum of any two linear polynomial is what
Esther Reply
divide simplify each answer 3/2÷5/4
Momo Reply
divide simplify each answer 25/3÷5/12
Momo
how can are find the domain and range of a relations
austin Reply
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?
Diddy Reply
6000
Robert
more than 6000
Robert
can I see the picture
Zairen Reply
How would you find if a radical function is one to one?
Peighton Reply
how to understand calculus?
Jenica Reply
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.
rachel Reply
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?
Reena Reply
what is foci?
Reena Reply
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
how to determine the vertex,focus,directrix and axis of symmetry of the parabola by equations
Bryssen Reply

### Read also:

#### Get the best Precalculus course in your pocket!

Source:  OpenStax, Precalculus. OpenStax CNX. Jan 19, 2016 Download for free at https://legacy.cnx.org/content/col11667/1.6
Google Play and the Google Play logo are trademarks of Google Inc.

Notification Switch

Would you like to follow the 'Precalculus' conversation and receive update notifications?

 By By By Subramanian Divya By By