# 9.7 Solving systems with inverses  (Page 4/8)

 Page 4 / 8

The only difference between a solving a linear equation and a system of equations written in matrix form is that finding the inverse of a matrix is more complicated, and matrix multiplication is a longer process. However, the goal is the same—to isolate the variable.

We will investigate this idea in detail, but it is helpful to begin with a $\text{\hspace{0.17em}}2\text{\hspace{0.17em}}×\text{\hspace{0.17em}}2\text{\hspace{0.17em}}$ system and then move on to a $\text{\hspace{0.17em}}3\text{\hspace{0.17em}}×\text{\hspace{0.17em}}3\text{\hspace{0.17em}}$ system.

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

Given a system of equations, write the coefficient matrix $\text{\hspace{0.17em}}A,\text{\hspace{0.17em}}$ the variable matrix $\text{\hspace{0.17em}}X,\text{\hspace{0.17em}}$ and the constant matrix $\text{\hspace{0.17em}}B.\text{\hspace{0.17em}}$ Then

$AX=B$

Multiply both sides by the inverse of $\text{\hspace{0.17em}}A\text{\hspace{0.17em}}$ to obtain the solution.

$\begin{array}{r}\hfill \left({A}^{-1}\right)AX=\left({A}^{-1}\right)B\\ \hfill \left[\left({A}^{-1}\right)A\right]X=\left({A}^{-1}\right)B\\ \hfill IX=\left({A}^{-1}\right)B\\ \hfill X=\left({A}^{-1}\right)B\end{array}$

If the coefficient matrix does not have an inverse, does that mean the system has no solution?

No, if the coefficient matrix is not invertible, the system could be inconsistent and have no solution, or be dependent and have infinitely many solutions.

## Solving a 2 × 2 system using the inverse of a matrix

Solve the given system of equations using the inverse of a matrix.

$\begin{array}{r}\hfill 3x+8y=5\\ \hfill 4x+11y=7\end{array}$

Write the system in terms of a coefficient matrix, a variable matrix, and a constant matrix.

$A=\left[\begin{array}{cc}3& 8\\ 4& 11\end{array}\right],X=\left[\begin{array}{c}x\\ y\end{array}\right],B=\left[\begin{array}{c}5\\ 7\end{array}\right]$

Then

First, we need to calculate $\text{\hspace{0.17em}}{A}^{-1}.\text{\hspace{0.17em}}$ Using the formula to calculate the inverse of a 2 by 2 matrix, we have:

So,

${A}^{-1}=\left[\begin{array}{cc}11& -8\\ -4& \text{​}\text{​}\text{\hspace{0.17em}}\text{\hspace{0.17em}}3\end{array}\right]$

Now we are ready to solve. Multiply both sides of the equation by $\text{\hspace{0.17em}}{A}^{-1}.$

The solution is $\text{\hspace{0.17em}}\left(-1,1\right).$

Can we solve for $\text{\hspace{0.17em}}X\text{\hspace{0.17em}}$ by finding the product $\text{\hspace{0.17em}}B{A}^{-1}?$

No, recall that matrix multiplication is not commutative, so $\text{\hspace{0.17em}}{A}^{-1}B\ne B{A}^{-1}.\text{\hspace{0.17em}}$ Consider our steps for solving the matrix equation.

$\begin{array}{r}\hfill \left({A}^{-1}\right)AX=\left({A}^{-1}\right)B\\ \hfill \left[\left({A}^{-1}\right)A\right]X=\left({A}^{-1}\right)B\\ \hfill IX=\left({A}^{-1}\right)B\\ \hfill X=\left({A}^{-1}\right)B\end{array}$

Notice in the first step we multiplied both sides of the equation by $\text{\hspace{0.17em}}{A}^{-1},\text{\hspace{0.17em}}$ but the $\text{\hspace{0.17em}}{A}^{-1}\text{\hspace{0.17em}}$ was to the left of $\text{\hspace{0.17em}}A\text{\hspace{0.17em}}$ on the left side and to the left of $\text{\hspace{0.17em}}B\text{\hspace{0.17em}}$ on the right side. Because matrix multiplication is not commutative, order matters.

## Solving a 3 × 3 system using the inverse of a matrix

Solve the following system using the inverse of a matrix.

$\begin{array}{r}\hfill 5x+15y+56z=35\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\\ \hfill -4x-11y-41z=-26\\ \hfill -x-3y-11z=-7\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\end{array}$

Write the equation $\text{\hspace{0.17em}}AX=B.\text{\hspace{0.17em}}$

First, we will find the inverse of $\text{\hspace{0.17em}}A\text{\hspace{0.17em}}$ by augmenting with the identity.

$\left[\begin{array}{rrr}\hfill 5& \hfill 15& \hfill 56\\ \hfill -4& \hfill -11& \hfill -41\\ \hfill -1& \hfill -3& \hfill -11\end{array}\text{\hspace{0.17em}}\text{\hspace{0.17em}}|\text{\hspace{0.17em}}\text{\hspace{0.17em}}\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right]$

Multiply row 1 by $\text{\hspace{0.17em}}\frac{1}{5}.$

$\left[\begin{array}{ccc}1& 3& \frac{56}{5}\\ -4& -11& -41\\ -1& -3& -11\end{array}\text{\hspace{0.17em}}\text{\hspace{0.17em}}|\text{\hspace{0.17em}}\text{\hspace{0.17em}}\begin{array}{ccc}\frac{1}{5}& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right]$

Multiply row 1 by 4 and add to row 2.

$\left[\begin{array}{ccc}1& 3& \frac{56}{5}\\ 0& 1& \frac{19}{5}\\ -1& -3& -11\end{array}\text{\hspace{0.17em}}\text{\hspace{0.17em}}|\text{\hspace{0.17em}}\text{\hspace{0.17em}}\begin{array}{ccc}\frac{1}{5}& 0& 0\\ \frac{4}{5}& 1& 0\\ 0& 0& 1\end{array}\right]$

Add row 1 to row 3.

$\left[\begin{array}{ccc}1& 3& \frac{56}{5}\\ 0& 1& \frac{19}{5}\\ 0& 0& \frac{1}{5}\end{array}\text{\hspace{0.17em}}\text{\hspace{0.17em}}|\text{\hspace{0.17em}}\text{\hspace{0.17em}}\begin{array}{ccc}\frac{1}{5}& 0& 0\\ \frac{4}{5}& 1& 0\\ \frac{1}{5}& 0& 1\end{array}\right]$

Multiply row 2 by −3 and add to row 1.

$\left[\begin{array}{ccc}1& 0& -\frac{1}{5}\\ 0& 1& \frac{19}{5}\\ 0& 0& \frac{1}{5}\end{array}\text{\hspace{0.17em}}\text{\hspace{0.17em}}|\text{\hspace{0.17em}}\text{\hspace{0.17em}}\begin{array}{ccc}-\frac{11}{5}& -3& 0\\ \frac{4}{5}& 1& 0\\ \frac{1}{5}& 0& 1\end{array}\right]$

Multiply row 3 by 5.

$\left[\begin{array}{ccc}1& 0& -\frac{1}{5}\\ 0& 1& \frac{19}{5}\\ 0& 0& 1\end{array}\text{\hspace{0.17em}}\text{\hspace{0.17em}}|\text{\hspace{0.17em}}\text{\hspace{0.17em}}\begin{array}{ccc}-\frac{11}{5}& -3& 0\\ \frac{4}{5}& 1& 0\\ 1& 0& 5\end{array}\right]$

Multiply row 3 by $\text{\hspace{0.17em}}\frac{1}{5}\text{\hspace{0.17em}}$ and add to row 1.

$\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& \frac{19}{5}\\ 0& 0& 1\end{array}\text{\hspace{0.17em}}\text{\hspace{0.17em}}|\text{\hspace{0.17em}}\text{\hspace{0.17em}}\begin{array}{ccc}-2& -3& 1\\ \frac{4}{5}& 1& 0\\ 1& 0& 5\end{array}\right]$

Multiply row 3 by $\text{\hspace{0.17em}}-\frac{19}{5}\text{\hspace{0.17em}}$ and add to row 2.

$\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\text{\hspace{0.17em}}\text{\hspace{0.17em}}|\text{\hspace{0.17em}}\text{\hspace{0.17em}}\begin{array}{ccc}-2& -3& 1\\ -3& 1& -19\\ 1& 0& 5\end{array}\right]$

So,

${A}^{-1}=\left[\begin{array}{ccc}-2& -3& 1\\ -3& 1& -19\\ 1& 0& 5\end{array}\right]$

Multiply both sides of the equation by $\text{\hspace{0.17em}}{A}^{-1}.\text{\hspace{0.17em}}$ We want $\text{\hspace{0.17em}}{A}^{-1}AX={A}^{-1}B:$

Thus,

${A}^{-1}B=\left[\begin{array}{r}\hfill -70+78-7\\ \hfill -105-26+133\\ \hfill 35+0-35\end{array}\right]=\left[\begin{array}{c}1\\ 2\\ 0\end{array}\right]$

The solution is $\text{\hspace{0.17em}}\left(1,2,0\right).$

So a horizontal compression by factor of 1/2 is the same as a horizontal stretch by a factor of 2, right?
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
I had just woken up when i got this message
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
For Plan A to reach $27/month to surpass Plan B's$26.50 monthly payment, you'll need 3,000 texts which will cost an additional \$10.00. So, for the amount of texts you need to send would need to range between 1-100 texts for the 100th increment, times that by 3 for the additional amount of texts...
Gilbert
...for one text payment for 300 for Plan A. So, that means Plan A; in my opinion is for people with text messaging abilities that their fingers burn the monitor for the cell phone. While Plan B would be for loners that doesn't need their fingers to due the talking; but those texts mean more then...
Gilbert
can I see the picture
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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?