Here's an ideal solution to HW1. Note the following
format compact to minimize empty lines and thus save electrons and trees. help and so on. My homework set prints in just three pages, if I show just one way of solving each problem!
Please be sure to make your homework solutions as compact and readable as possible, using format compact and editing out extraneous material!
% John F. Gibson
% 2014-09-12
% Math 445 HW1
format compact
% Problem 1: Given two numeric variables x and y, write a Matlab expression that
% evaluates to true (1) if and have opposite signs and false (0) otherwise.
% Test the expression by evaluating it with the following pairs of numbers
% (x,y) = (-5, 4), (5,4), (5,-4), (0,-2), and (3,0).
% There are many ways to answer this question, ranging from the most obvious and
% literal to the most readable and efficiently executed. Here's the obvious and
% literal way
x=-5; y=4; ((x<0) && (y>0)) || ((x>0) && (y<0))
ans =
1
x=5; y=4; ((x<0) && (y>0)) || ((x>0) && (y<0))
ans =
0
x=5; y=-4; ((x<0) && (y>0)) || ((x>0) && (y<0))
ans =
1
x=0; y=-2; ((x<0) && (y>0)) || ((x>0) && (y<0))
ans =
0
x=3; y=0; ((x<0) && (y>0)) || ((x>0) && (y<0))
ans =
0
% Here's a less complicated expression that accomplishes the same
x=-5; y=4; x*y < 0
ans =
1
% and another
x=-5; y=4; sign(x)*sign(y) == -1
ans =
1
% Problem 2: Find a matlab expression to computer the combined resistance RT of
% three resistors R1,R2, R3.
% The simple, straightforward way
R1=5; R2=3; R3=4;
RT = 1/(1/R1 + 1/R2 + 1/R3)
RT =
1.2766
% Here's a slicker way to do the same
R = [5 3 4];
RT = 1/sum(R.^(-1))
RT =
1.2766
% or even better
RT = 1/sum([5 3 4].^(-1))
RT =
1.2766
% Problem 3:
% (a) Create a row vector x whose elements are the numbers 5, 7, 10, 1.
x = [5 7 10 1]
x =
5 7 10 1
% (b) Create a column vector x whose elements are the numbers 5, 7, 10, 1.
x = [5 ; 7 ; 10 ; 1]
x =
5
7
10
1
x = [5 7 10 1]'
x =
5
7
10
1
% (c) Use colon syntax to create a row vector x whose elements start at 0, end at 1,
% and increase in steps of 0.1.
x = 0:0.1:1
x =
Columns 1 through 10
0 0.1000 0.2000 0.3000 0.4000 0.5000 0.6000 0.7000 0.8000 0.9000
Column 11
1.0000
% (d) Determine the dimension of x from (c) and assign the value to the variable d
% (using Matlab, not by counting!).
d = length(x)
d =
11
d = size(x,2)
d =
11
% (e) Use the linspace function to create a 10-dimensional vector of numbers evenly spaced between 0 and 1.
x = linspace(0,1,10)
x =
0 0.1111 0.2222 0.3333 0.4444 0.5556 0.6667 0.7778 0.8889 1.0000
% Problem 4
% (a) Create the given matrix
A = [ 3 9 2 ; -1 4 6 ; 5 -2 0]
A =
3 9 2
-1 4 6
5 -2 0
% (b) Change A(2,3) to 7
A(2,3) = 7
A =
3 9 2
-1 4 7
5 -2 0
% (c) Assign the third column of A to the variable v
v = A(:,3)
v =
2
7
0
% (d) Change the first row of A to 8,1,4
A(1,:) = [8 1 4]
A =
8 1 4
-1 4 7
5 -2 0
% Problem 5. Create the given A matrix and b vector and solve Ax=b
A = [4 2 ; -1 5]
A =
4 2
-1 5
b = [3 4]'
b =
3
4
x = A\b
x =
0.3182
0.8636
% verify solution satisfies Ax=b
A*x
ans =
3
4
A*x-b
ans =
0
0
norm(A*x-b)
ans =
0
% Problem 6: Use Matlab to solve the problem. Nilanjana has 40 coins worth $6.40.
% They're all quarters and nickels. How many nickels and how many quarters does
% she have? Verify that your answer solves the problem.
A = [1 1 ; 0.10 0.25]
A =
1.0000 1.0000
0.1000 0.2500
b = [40 ; 6.40]
b =
40.0000
6.4000
x = A\b
x =
24.0000
16.0000
% She has 24 dimes and 16 quarters. That sums to 40 coins, and 24 * $0.10 + 16 * $0.25 = $6.40
A*x
ans =
40.0000
6.4000
% Problem 7: Suhasini has 44 coins worth $7.50. They're all quarter, dimes, and nickels.
% She has twice as many dimes as nickels. How many of each type of coin does she have?
% Find the answer, and then verify that the solution satisfies the problem. The equations
% are
%
% n + d + q = 44
% 0.05 n + 0.10 d + 0.25 q = 7.50
% d - 2n = 0
A = [1 1 1 ; 0.05 0.10 0.25 ; -2 1 0]
A =
1.0000 1.0000 1.0000
0.0500 0.1000 0.2500
-2.0000 1.0000 0
b = [44 7.50 0]'
b =
44.0000
7.5000
0
x = A\b
x =
7
14
23
% She has 7 nickels, 14 dimes, and 23 quarters. Verifying...
n = x(1); d = x(2); q = x(3);
n + d + q
ans =
44
0.05*n + 0.10*d + 0.25*q
ans =
7.5000
d - 2*n
ans =
0
norm(A*x-b)
ans =
0