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

Questions & Answers

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
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
how do they get the third part x = (32)5/4
kinnecy Reply
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
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
Idrissa Reply
hello
Sherica
im all ears I need to learn
Sherica
right! what he said ⤴⤴⤴
Tamia
hii
Uday
hi
salma
what is a good calculator for all algebra; would a Casio fx 260 work with all algebra equations? please name the cheapest, thanks.
Kevin Reply
a perfect square v²+2v+_
Dearan Reply
kkk nice
Abdirahman Reply
algebra 2 Inequalities:If equation 2 = 0 it is an open set?
Kim Reply
or infinite solutions?
Kim
The answer is neither. The function, 2 = 0 cannot exist. Hence, the function is undefined.
Al
y=10×
Embra Reply
if |A| not equal to 0 and order of A is n prove that adj (adj A = |A|
Nancy Reply
rolling four fair dice and getting an even number an all four dice
ramon Reply
Kristine 2*2*2=8
Bridget Reply

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Source:  OpenStax, College algebra. OpenStax CNX. Feb 06, 2015 Download for free at https://legacy.cnx.org/content/col11759/1.3
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