>> A=[1 3;2 8]
A =
1 3
2 8
>> size(A)
ans =
2 2
>> V=input('enter matrix')
enter matrix[4 5 6;1 8 2]
V =
4 5 6
1 8 2
>> c=ones(4,2)
c =
1 1
1 1
1 1
1 1
>> D=eye(3)
D =
1 0 0
0 1 0
0 0 1
>> E=zeros(3,4)
E =
0 0 0 0
0 0 0 0
0 0 0 0
>> X=rand(2,3)
X =
0.9501 0.6068 0.8913
0.2311 0.4860 0.7621
>> B=[1 3 5 4 1;2 4 7 3 2]
B =
1 3 5 4 1
2 4 7 3 2
>> mean(B)
ans =
1.5000 3.5000 6.0000 3.5000 1.5000
>> C=B'
C =
1 2
3 4
5 7
4 3
1 2
>> mean(C)
ans =
2.8000 3.6000
>> sum(C(:,2))
ans =
18
>> sum(C([2:4],:))
ans =
12 14
>> %adds the number of males from .2 to .8, and the number of females between .2 and .8
>> D=rand(3,4)
D =
0.4565 0.4447 0.9218 0.4057
0.0185 0.6154 0.7382 0.9355
0.8214 0.7919 0.1763 0.9169
>> P=[10;15;20;10]
P =
10
15
20
10
>> D*P
ans =
33.7285
33.5354
32.7875
>> %took a 3 row by 4 column matrix and multiplied by a vector column to get delays from weather, plane problems, and traffic
>> P2=[10 20; 15 0; 20 30; 10 10]
P2 =
10 20
15 0
20 30
10 10
>> D*P2
ans =
33.7285 40.8408
33.5354 31.8710
32.7875 30.8852
>> %gives the delays in the first month and the delays in the second month
>> %in matrix multiplication, the number of rows in the second matrix must equal the number of columns in the first matrix
>> A=[2 5 4 3 7];
>> sumo=0
sumo =
0
>> for ii = [1:5]
sumo=sumo+A(ii)
end
sumo =
2
sumo =
7
sumo =
11
sumo =
14
sumo =
21
>> %repeats the process of addition from the first to the fifth number (eg. first+sumo=2, second+sumo=7)
>> x=input('enter the number you are looking for')
enter the number you are looking for5
x =
5
>> for ii=[1:5]
if A(ii)==x
fprintf('I found %f at index %f',x,ii)
end
end
I found 5.000000 at index 2.000000>>
>> %searches for the x value within the set of numbers until it finds it
>> for n=[1:10]
fprintf('%f\n',1/n)
end
1.000000
0.500000
0.333333
0.250000
0.200000
0.166667
0.142857
0.125000
0.111111
0.100000