Download How do you solve a matrix equation using the

Question 2: How do you solve a matrix equation using the matrix inverse?
In the previous question, we wrote systems of equations as a matrix equation AX  B .
In this format, the matrix A contains the coefficients on the variables, matrix X contains
the variables, and matrix B contains the constants. Solving the system of equations
means that we need to solve for the variable matrix X. This is accomplished by
multiplying both sides of the matrix equation by the inverse of the coefficient matrix, A1 .
A1 A X  A1 B
The product of a matrix A and its inverse is the identity matrix I. This means we can
simplify the matrix equation to
IX  A1 B
The product of an identity matrix with X is simply X, so the solution to the matrix
equation is
X  A 1 B How To Solve a System of Linear Equations with
Inverses
1. Make sure the system is in proper format with variable
terms on the left side of the equation and constants on the
right side. The variable terms should be listed in the same
order in each equation. Identify the coefficient matrix A,
the variable matrix X, and the constant matrix B.
3. Compute the inverse A1 . If the matrix is not invertible, it
is not possible to solve the system with inverses.
4. Compute the product A1 B .
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5. The solution to the matrix equation AX  B is X  A1 B .
The corresponding solution for the system of linear
equations is found by matching the variables in X with the
corresponding entries in the product A1 B .
Example 3
Solve a Linear System with the Inverse
Solve the system of linear equations using the inverse of the coefficient
matrix.
x  y  5
3x  4 y  1
Solution In 0, we wrote this system as the matrix equation AX  B ,
where
1 1 
A
,
3 4 
 x
 5
X    , and B   
 y
1
In section 3.3, we found the inverse of A,
 4 1
A1  

 3 1 
We can use the inverse to compute the solution to the system.
The solution is found by multiplying the inverse of A times the constant
matrix B,
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X
A1
B
 4 1  5

 
 3 1   1 
Insert the matrices. The product of a 2 x 2
matrix and a 2 x 1 matrix is a 2 x 1 matrix
 4  5    11 


  3 5   1 1 
Multiply the row entries in the inverse times
the corresponding column entries in B
 21


 16 
Since the individual entries in X correspond to the variables x and y, this
tells us that x  21 and y  16 . We can check these values in the
original system to make sure they solve the system:
 21 
Example 4
16  5
TRUE
3  21  4 16   1
TRUE
Solve a Linear System with the Inverse
Solve the system of linear equations with the inverse of the coefficient
matrix.
x1  x2  x3  2
2 x1  2 x2  3 x3  1
 x1
 2 x3  7
Solution This system of equations is equivalent to the matrix equation
AX  B where
 1 1 1
A   2 2 3 ,
 1 0 2 
 x
2


X   y  , and B   1
 z 
 7 
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The variable matrix may be solved for using the inverse of the
coefficient matrix, X  A1 B . We found the inverse of this coefficient
matrix in Example 5 of section 3.3.
X
A1
B
 4 2 1  2 
  1 1 1   1
 2 1 0   7 
A1 was computed in Example 5 of
section 3.3
 4  2   2  1   1  7 


  1 2  1 1  1  7 
 2  2   1 1  0  7 
Multiply the entries in the rows of A
the entries in the column of B
1
by
3
  4 
 5 
Equating the entries in the variable matrix with the entries in this
product, we observe that x  3 , y  4 , and z  5 .
For systems of linear equations with unique solutions, we can use inverses to solve for
the variables. Many of the applications in Chapter 2 may be solved using this strategy.
Example 5
Mixing Ethanol Blends
In Example 12 of section 2.2 we created a system of equations to
describe a mix of E10 and E85 ethanol,
E10 
E85  10
0.10 E10  0.85 E85  2
where E10 is the amount of 10% ethanol pumped in gallons and E85 is
the amount of 85% ethanol pumped in gallons. The first equation
describes the total amount in the mixture, 10 gallons. The second
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equation describes the total amount of ethanol in the mixture, 20% of 10
gallons or 2 gallons.
a. Solve this system of equations by finding the inverse of the
coefficient matrix.
Solution This system of equations was solved in Chapter 2 using the
Substitution Method. In this section we’ll solve the same system by
writing the system as a matrix equation.
We can write this system of equations as the matrix equation AX  B
by identifying the matrices,
1 
 1
A
,
0.10 0.85
 E10 
10 
X 
, and B   

 E85 
2
The solution to the matrix equation is X  A1 B and requires the inverse
of the matrix A.
We begin the process of finding the inverse of A by placing the
coefficient matrix in a new matrix alongside a 2 x 2 identity matrix.
1
1 0
 1
0.10 0.85 0 1 


Since the entry in the first row, first column is already a 1, we’ll make
the rest of the column into zeros using row operations. To do this,
multiply the first row by -0.10 and add it to the second row. Place the
sum in the second row:
0.10 R1 :
 R2 :
0.10 0.10 0.10 0
0.10 0.85
0
1
0
0.75
0.10 1
0.10 R1  R2
becomes
R2
1
1
0
1
 0 0.75 0.10 1 


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With the first column transformed, proceed to the second column and
use row operations to create a 1 in the second row, second column.
We put a 1 in the second row, second column by multiplying the second
row by
1
3
4
. Since 0.75  , this is the same as multiplying by :
4
0.75
3
0 0.75 .10 1
1

0.75
0
1

2
15
1
0.75
R2
becomes
R2
4
3

1 1 1

0 1  2
15


0

4
3 
Now multiply the second row by -1 and add it to the first row. Place the
result in the first row:
1R2 :
 R1 :
0 1
2
15
1
1
1
1
0
17
15

4
3
0

4
3
1R2  R1
becomes
R1
17
4

1
0


15
3


4 
0 1  2

15 3 
4
 17
 

15
3
The inverse matrix is the right hand side of this matrix, A1  

4 
 2
 15 3 
The solution to the original system of equations is
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X
 17
 15

 2
 15
A1
B
4
 
3 10 

4   2 
3 
Multiply the entries in the rows of A
the entries in the column of B
17
 4 
15 10    3   2 

 

2
4


  15 10  3  2 
1
by
Combine the terms in each entry and
simplify all fractions
 26 
3
 
4
 3 
 E10 
Since the variable matrix X represents 
 , this means that the
 E85
amount of E10 needed is
26
gallons and the amount of E85 needed is
3
4
gallons.
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b. If the number of gallons in the mixture should be 12 gallons, how
much 10% ethanol and 85% ethanol must be pumped?
Solution If the total number of gallons in the mixture is increased to 12
gallons, the total amount of ethanol in the mixture is .20 12 or 2.4
gallons. The system of equations becomes
E10 
E85  12
0.10 E10  0.85 E85  2.4
The only change to the system of equations is in the constants, not the
coefficients.
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This means that we may use the existing coefficient matrix A and the
corresponding inverse
A 1
to find the solution of this system,
E10 
E85  12
0.10 E10  0.85 E85  2.4
The constant matrix for this system is
 12 
B 
 2.4 
and the solution is
X
A1
B
4
 17

 15
3   12 


4   2.4 
 2
 15 3 
17

 4
 15 12    3   2.4 




2
4


  15 12  3  2.4 
 52 
5
 
8
 5 
Use the same inverse as in part a
Multiply the entries in the rows of A
by the entries in the column of B
1
Combine the terms in each entry and
simplify all fractions
Since the coefficients in the system have not changed, there is no need
to recompute the inverse. We may use the inverse from part a to
calculate the product. We must mix
52
8
gallons of E10 and gallons of
5
5
E85.
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This example illustrated the power of using inverses to solve a system of linear
equations. As long as the coefficient matrix does not change, the same inverse may be
used to solve several different problems with different constant matrices. The solutions
are simply the product of the inverse of the coefficient matrices and the different
10 
constants. For part a, the constant was B    and for part b the constant was
2
 12 
B    . There was no need to recomputed the inverse for the different parts.
 2.4 
16