# 7.7 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

what is math number
Tric Reply
x-2y+3z=-3 2x-y+z=7 -x+3y-z=6
Sidiki Reply
Need help solving this problem (2/7)^-2
Simone Reply
x+2y-z=7
Sidiki
what is the coefficient of -4×
Mehri Reply
-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
Alfred Reply
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?
Kala Reply
lim x to infinity e^1-e^-1/log(1+x)
given eccentricity and a point find the equiation
Moses Reply
12, 17, 22.... 25th term
Alexandra Reply
12, 17, 22.... 25th term
Akash
College algebra is really hard?
Shirleen Reply
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
Adu
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
Jerwin Reply
I know this work
salma
The given of f(x=x-2. then what is the value of this f(3) 5f(x+1)
virgelyn Reply
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
kinnecy Reply
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.
Jeffrey Reply
sure. what is your question?
ninjadapaul
20/(×-6^2)
Salomon
okay, so you have 6 raised to the power of 2. what is that part of your answer
ninjadapaul
I don't understand what the A with approx sign and the boxed x mean
ninjadapaul
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
ninjadapaul
oops. ignore that.
ninjadapaul
so you not have an equal sign anywhere in the original equation?
ninjadapaul
hmm
Abhi
is it a question of log
Abhi
🤔.
Abhi
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salma
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salma
Commplementary angles
Idrissa Reply
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

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