MATLAB is a software package that makes it easy to manipulate
matrices. Vectors can also be easilyy handled as a special
p
case.
A matrix in MATLAB is just a 2-D array of numbers. Example:
>> A =[1 2; 3 5; 2 4; 6 7]
A=
1
2
3
5
2
4
6 7
The dimensions of a matrix is (NxM) where N is the number of rows
and N is the number of columns. The matrix A above has
dimensions (4x2)
We can add or subtract matrices by simply adding or subtracting each
element in the arrays located at the same position. Example:
>> A = [ 1 2; 3 4]
A=
1
2
3
4
>> B = [3 3; 5 5]
B=
3
3
5
5
>> C=A+B
C A B
To add matrices they must be of the same size
C=
4
5
8
9
we can also multiply matrices together in the following fashion. Let Amn
be the element in an array,
y, A,, of dimensions (MxN)
(
) at the mth row
and nth column position. Similarly for a matrix B of dimensions (NxK)
we have Bnk is the element at the nth row and kth column position .
Then the matrix product of A and B, C=A*B, is defined for each
element of C as :
N
Cmk = ∑ Amn Bnk
n =1
(MxK)
(MxN) (NxK)
Note: for this to be meaningful,
meaningful we must have the number of
columns of A be equal to the number of rows of B but the
other dimensions can be different
Example:
>> A = [ 2 3; 1 5];
>> B =[2 2; 6 3];
>> C=A*B
C=
22
13
32
17
which comes from our multiplication rule
C11 = A11 B11 + A12 B21 = (2)(2) + (3)(6) = 22
C12 = A11 B12 + A12 B22 = (2)(2) + (3)(3) = 13
C21 = A21 B11 + A22 B21 = (1)(2) + (5)(6) = 32
C22 = A21 B12 + A22 B22 = (1)(2) + (5)(3) = 17
We can consider vectors to be just examples of matrices where one of
the dimensions =1. We can either have row or column vectors:
>> v1 =[ 1 2 3 5]
(1x4) (row) vector
v1 =
1
2
3
5
>> v2 =[1;2;3;5]
v2 =
1
2
3
5
(4x1) (column) vector
we can multiply vectors A and B in the same fashion as other matrices as long
as we adhere to the previous rule:
we must have the number of columns of A be equal to the
number of rows of B in the product A*B
Consider the two vectors:
>> v1 =[ 1 2 3 4];
>> v2 =[1;2;3;4];
[
]
>> d=v1*v2
(1x4)
d=
30
(4x1)
(1x1) scalar
This is just the dot product ( or the square of the magnitude in this case)!!
Now consider reversing the order:
>> v1 =[ 1 2 3 4];
>> v2 =[1;2;3;4];
>> p= v2*v1
(4x1)
(1x4)
p=
(4x4)
Here
1
2
3
4
2
4
6
8
3
6
9
12
4
8
12
16
pmn = v 2m v1n
We can also multiply matrices with vectors if we follow the same rule:
>> M=[1 2; 3 4]
M=
1
2
3
4
>> x= [5;6]
x=
5
(2x1)
6
>> b= M*x
(2x2)
(2x1)
b=
17
39
This is equivalent to the linear system of equations:
M 11 x1 + M 12 x2 = b1
M 21 x1 + M 22 x2 = b2
another way to write these equations is
⎡ M 11
⎢M
⎣ 21
M 12 ⎤ ⎡ x1 ⎤ ⎡ b1 ⎤
=⎢ ⎥
⎥
⎢
⎥
M 22 ⎦ ⎣ x2 ⎦ ⎣b2 ⎦
We also could set
>> x= [5 6]
x=
5
6
>> M=[1 2; 3 4]
M=
((1x2))
1
2
3
4
>> b=x
b=x*M
M
b=
23
34
(1x2)
((2x2))
Note that this b vector is different from
b f
before
even though
th
h th
the x and
dMh
have
the same elements as.
If we want to get the same result for b as before but now in the form of a
row vector we need to multiply
p y x by
y the transpose
p
of M,, MT, where
if
M 12 ⎤
⎡M
M = ⎢ 11
⎥
⎣ M 21 M 22 ⎦
M 21 ⎤
⎡M
M T = ⎢ 11
⎥
⎣ M 12 M 22 ⎦
(interchange rows and columns)
This works since for any two matrices or vectors A, B, where the
product is defined we have
(AB)T =BTAT
so if we take the transpose of
⎡ M 11
⎢M
⎣ 21
M 12 ⎤ ⎡ x1 ⎤ ⎡ b1 ⎤
=⎢ ⎥
⎥
⎢
⎥
M 22 ⎦ ⎣ x2 ⎦ ⎣b2 ⎦
we get
[ x1
⎡ M 11
x2 ] ⎢
⎣ M 12
M 21 ⎤
= [b1 b2 ]
M 22 ⎥⎦
In MATLAB, the transpose is MT =M'
>> b=x*M'
b=
17
39
same b as obtained originally
but now in terms of a row vector
For vector analysis, MATLAB has built-in dot and cross product functions:
>> v1 = [ 1 2 6];
>> v2 = [ 3 5 -2];
2];
>> d =dot(v1, v2)
d=
1
>> c =cross(v1, v2)
c=
-34
20
here the vectors must be 3-D vectors
i (1
i.e.
(1x3)
3) or (3
(3x1)
1) iin di
dimensions
i
-1
we saw before we could also get the dot product by calculating v1*v2T
>> v1
v1*v2'
v2
ans =
1
In solving Statics equilibrium problems, we will typically have
to solve a set of linear equations for a set of unknown forces or
moments. We can do this easily in MATLAB:
Example Consider the following system of two equations for
the two unknowns x1 and x2
⎡ C11 C12 ⎤ ⎡ x1 ⎤ ⎡ b1 ⎤
=⎢ ⎥
⎢C
⎥
⎢
⎥
⎣ 21 C22 ⎦ ⎣ x2 ⎦ ⎣b2 ⎦
If we define the C matrix of coefficients and the vector b in MATLAB
then the solution is given by using a backslash operator:
>> C = [ 1 2; 3 7];
>> b=[1;1];
>> x =C\b
x=
5.0000
-2.0000