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Solving Linear Systems of Equations - Inverse Matrix
• Consider the following
system of equations ...
a1x  b1 y  c1
• Let the matrix A
represent the coefficients ...
 a1
A
a 2
a 2 x  b2 y  c2
• Let matrix B hold the
constants ...
b1 

b2 
 c1 
B 
c 2 
x 
X 
 y
• Finally, let matrix X
represent the variables ...
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Solving Linear Systems of Equations - Inverse Matrix
• Now notice what the result is when we work out the
following matrix equation ...
 a1
AX  B  
a 2
b1  x   c1 
 



b 2   y c 2 
 a1x b1 y   c1 

 

a 2 x b 2 y  c 2 

a1x  b1 y  c1
a 2 x  b2 y  c2
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Slide 2
Solving Linear Systems of Equations - Inverse Matrix
• Thus, AX = B represents the system of equations.
This matrix equation can be solved for X as follows ...
• Recall that matrix multiplication
is not commutative, so each side of
the equation must be multiplied on
the left by A-1
• Matrix multiplication is
associative.
AX  B
1
1
A (AX)  A B
1
1
( A A) X  A B
1
( I) X  A B
1
XA B
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Slide 3
Solving Linear Systems of Equations - Inverse Matrix
• Method of solution:
(1) Given a system of equations, form matrices
A, X, and B.
A
Coefficients
X
Variables (vertical matrix)
B
Constants (vertical matrix)
(2) Find A-1.
(3) Find the solution by multiplying A-1 times B.
X = A-1 B
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Slide 4
Solving Linear Systems of Equations - Inverse Matrix
• Example:
Use an inverse matrix to solve the
system at the right.
3  2
A

1  1
x 
X 
 y
3x  2 y  8
x  y3
8
B 
3
• Using the methods of finding an inverse, A-1 is ...
1  2
A 

1

3


1
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Slide 5
Solving Linear Systems of Equations - Inverse Matrix
• Now find X ...
1  2 8
2
XA B 
 



1  3 3
 1
1
• The solution is (2, -1), or
x=2
y = -1
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Slide 6
Solving Linear Systems of Equations - Inverse Matrix
• This same method can be used on any size system of
equations as long as the coefficient matrix is square and
the solution is unique.
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Slide 7
Solving Linear Systems of Equations - Inverse Matrix
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