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Linear systems
of equations and
their solution
Video 12
Daniel J. Bodony
Department of Aerospace Engineering
University of Illinois at Urbana-Champaign
In this video you will learn. . .
1
What is a linear system of equations?
2
How to write one in matrix form.
3
4
When is there a solution?
How to solve A~x = ~b in Matlab.
5
Example: fitting a quadratic polynomial to three points
D. J. Bodony (UIUC)
AE199 IAC
Video 12
2 / 20
What is a linear system of equations?
Possibly a long time ago you learned how to solve equations of the type
x +y +z =2
2x + 8y = −1
y − 3z = 0.
Notice that we have 3 equations with 3 unknowns, x, y , and z. This is a
system of equations. Using the rules of matrix-vector multiplication, we
can write the same system as

   
x
2
1 1 1
2 8 0  y  = −1
z
0
0 1 −3
{z
} |{z} | {z }
|
A
~
x
~
b
A~x = ~b
D. J. Bodony (UIUC)
← This is in matrix form.
AE199 IAC
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Matrix form of linear system of equations
The generic equation
A~x = ~b
is the matrix form of a linear system of equations. We call
A the matrix of coefficients, or coefficient matrix
~x the vector of unknowns, or just the unknowns
~b the right-hand-side vector, or just the right-hand-side (rhs).
Matlab has a large set of tools useful for solving A~x = ~b. Generically we
write the solution as
~x = A−1 ~b;
however, A−1 , called the inverse of A, may not exist.
D. J. Bodony (UIUC)
AE199 IAC
Video 12
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When is there a solution?
Let’s see what conditions on A are important for solving A~x = ~b by way of
a 2 × 2 example. Suppose that our equations are
ax + by = r1
cx + dy = r2
and we are to solve for ~x = [x y ]T in terms of ~b = [r1 r2 ]T . In matrix form
we write this is
a b x
r
= 1
r2
c d y
and the solution is
−1 x
a b
r1
=
,
y
c d
r2
provided the inverse exists.
D. J. Bodony (UIUC)
AE199 IAC
Video 12
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When is there a solution?
It is easy to find the inverse of a 2 × 2 matrix; it is
a b
c d
−1
1 d −b
=
∆ −c a
where
∆ = ad − bc
is the determinant of A.
From the expression on the right we see clearly that if ∆ = 0 then the
inverse does not exist!
When ∆ = 0 the matrix A is called singular!1
1
When ∆ = 0 a solution may exist but we will not discuss this in class.
D. J. Bodony (UIUC)
AE199 IAC
Video 12
6 / 20
How to solve A~x = ~b in Matlab
Assuming the determinant of A, ∆, is not zero, Matlab will very easily
solve A~x = ~b in several ways.
The most natural way for you will be to use
x = inv(A) * b; ← ~x = A−1 b
where inv is the matrix inverse command.
How does this command work?
1
The inverse of A is explicitly computed
2
A matrix-vector multiplication between inv(A) and b happens
3
The result is stored in x
D. J. Bodony (UIUC)
AE199 IAC
Video 12
7 / 20
How to solve A~x = ~b in Matlab
Matlab provides another way to solve A~x = ~b. It is
x = A\b;
Notice the use of the backslash (\) command. The (\) is a very special
command that Matlab uses to solve a large variety of linear system
problems. It works roughly like this:
1 If A is square, then
1
2
2
Look at the structure of A (e.g., symmetric, triangular, ...) and pick a
good solution method
Store the solution in x
Else, use a least squares method if A is rectangular and store the
solution in x.
D. J. Bodony (UIUC)
AE199 IAC
Video 12
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Choosing inv(A)*b or A\b
How do you choose? Well, the answer is trivial:
ALWAYS CHOOSE A\b
Why? Let me count the ways
1
A\b is faster because it doesn’t compute the inverse of A directly
(you’ll learn more about this in AE370)
2
A\b works even if A is singular (we won’t discuss this)
3
A\b works even if A is rectangular (we won’t discuss this)
D. J. Bodony (UIUC)
AE199 IAC
Video 12
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Example: quadratic polynomial fit
To illustrate things, let’s do a simple example of finding the coefficients of
a quadratic polynomial
p(x) = ax 2 + bx + c
that exactly go through the three points in the x-y plane (x1 , y1 ), (x2 , y2 )
and (x3 , y3 ).
What do we mean by ‘go through the three points’ ? We mean the value
of the polynomial p(x) at the points xi is exactly equal to yi , for
i = 1, 2, 3. That is we want to find the {a, b, c} such that
p(x1 ) = y1
p(x2 ) = y2
p(x3 ) = y3
D. J. Bodony (UIUC)
AE199 IAC
Video 12
10 / 20
Example: quadratic polynomial fit
Using the definition of the polynomial p(x) we find three equations in the
three unknowns {a, b, c}:
ax12 + bx1 + c = y1
ax22 + bx2 + c = y2
ax32 + bx3 + c = y3
or, in matrix form,
 2
   
x1 x1 1
a
y1
x22 x2 1 b  = y2 
x32 x3 1
c
y3
or
A~x = ~b.
D. J. Bodony (UIUC)
AE199 IAC
Video 12
11 / 20
Example: quadratic polynomial fit
For real numbers, let’s find the polynomial that passes through the three
red points that lie on the cosine: (2, cos 2), (π, cos π), (4, cos 4).
1.5
1
0.5
0
−0.5
−1
−1.5
0
D. J. Bodony (UIUC)
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Example: quadratic polynomial fit
In this example we have
(x1 , y1 ) = (2, cos 2)
(x2 , y2 ) = (π, cos π)
(x3 , y3 ) = (4, cos 4)
so that our matrix A and vector ~b are




4 2 1
cos 2
A = π 2 π 1 and ~b = cos π 
16 4 1
cos 4
It is easy to check that det(A) = -1.9599 and so we expect a solution.
D. J. Bodony (UIUC)
AE199 IAC
Video 12
13 / 20
Example: quadratic polynomial fit
Once A and b are entered into Matlab, we can find the answer using \:
>> format long e
>> x = A\b
x =
4.574622948019237e-01
-2.863522160969777e+00
3.481048306184716e+00
So, according to Matlab, the polynomial we are after is
p(x) = (4.574622948019237 × 10−1 )x 2
+ (−2.863522160969777 × 100 )x + (3.481048306184716 × 100 )
D. J. Bodony (UIUC)
AE199 IAC
Video 12
14 / 20
Example: quadratic polynomial fit
A plot of the polynomial fit (in black) shows that it performs as desired.
1.5
1
0.5
0
−0.5
−1
−1.5
0
D. J. Bodony (UIUC)
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Example: quadratic polynomial fit
We can check our answer by using polyfit in Matlab:
>> polyfit([2 pi 4],cos([2 pi 4]),2).’
ans =
4.574622948019231e-01
-2.863522160969772e+00
3.481048306184707e+00
Which is the same to 14 decimal places. (Matlab uses a different
algorithm within polyfit so the coefficients are different to within
numerical roundoff.)
D. J. Bodony (UIUC)
AE199 IAC
Video 12
16 / 20
Another example: bunch of linear springs
Let’s start simple by having one spring only, as shown in the figure:
A force balance says that
k1 x = W
so that
x = k1−1 W .
(Notice the way I’ve written this.)
D. J. Bodony (UIUC)
AE199 IAC
Video 12
17 / 20
Another example: bunch of linear springs
Add another spring so that our picture looks like
Our goal: find x given W .
D. J. Bodony (UIUC)
AE199 IAC
Video 12
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We start by identifying the equilibrium lengths of the springs as x1e and x2e
when no weight is on the platform:
Undeformed
Deformed
We perform a force balance on the platform (top red box) to find that
k1 (x1 − x1e ) − W = 0
and a force balance on the inter-spring connection (bottom red box) to get
k2 (x2 − x2e ) − k1 (x1 − x1e ) = 0.
D. J. Bodony (UIUC)
AE199 IAC
Video 12
19 / 20
Another example: bunch of linear springs
If we define the change in length of the springs as
δ1 = x1 − x1e
δ2 = x2 − x2e
then our linear system can be written in matrix form as
k1
0 δ1
W
=
−k1 k2 δ2
0
which has solution
W k2
1 k2 0 W
δ1
=
=
0
δ2
k1 k2 k1 k1
k1 k2 k1
such that the total displacement of platform is
x = δ1 + δ2 = W
D. J. Bodony (UIUC)
AE199 IAC
k1 + k2
k1 k2
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