Question

Problem 2: Matrix Vector Operations

a) Create a program named my_dotProduct which USING LOOPS calculates the dot product of two vectors (of dimensions \(1 \times n\) ) from the keyboard. The program should check the input from the user making sure that the vectors are the right dimensions. If the vectors are not then an error message should be outputted to the user. If the vectors are of the right dimensions then the answer should be outputted to the user. Using your code, complete the following table:

b) Create a program named my_arrayCount which inputs an \(n \times m\) array and determines (USING LOOPS) how many elements in the array are greater than zero and how many are less than zero. The program should then output the answers. Using your code, complete the following table:

Answer 1

Below are the code in MATLAB for the above questions:

(my_dotProduct.m)

clear,clc;
%taking input vector u,v from user.
u = input('Enter the elements of vector u: ');
v = input('Enter the elements of vector v: ');

%storing length of both vectors in n and m respectively.
n=length(u);
m=length(v);

%checking for dimension mismatch.
if(n==m)
%variable result that stores dot product of vector u and v.
result = 0;
%finding dot product.
for i = 1 : n
result = result + u(i)*v(i);
end
%displaying result.
fprintf('u.v = %d\n',result);
else
%displays error if vector mismatch.
disp('Error:vector dimension mismatch');
end

(my_arrayCount.m)

clear,clc;
%taking nxm array input from the user.
arr = input('enter the size of 2D array: ');
%storing dimension of array arr.
[n, m] = size(arr);
%variables which stores count of negative and
%positve numbers in an array arr.
neg = 0;
pos = 0;

%iterating in the array.
for i = 1:n
for j = 1:m
%array is less than zero then increment neg.
if(arr(i,j)<0)
neg = neg + 1;
else
%else we increment pos.
pos = pos + 1;
end
end
end
%displaying count of pos and neg in the array arr.
fprintf('n(>0): %d\nn(<0): %d\n',pos,neg);

Refer to the screenshot attached below to better understand the code and indentation:

(my_dotProduct.m)

1- my_dotProduct. m x my_arrayCount. m x + clear, clc; ftaking input vector u,v from user. u = input('Enter the elements of vector u: '); v = input('Enter the elements of vector v: '); storing length of both vectors in n and m respectively. n=length(u); m=length (v); 15 - 9 checking for dimension mismatch. if(n==m) *variable result that stores dot product of vector u and v. result = 0; $finding dot product. for i = 1:n result = result + u(i) *v(i); end $displaying result. fprintf('u.v = $d', result); else %displays error if vector mismatch. disp('Error:vector dimension mismatch'); end

Output:

Enter the elements of vector u: (3 -24) Enter the elements of vector v: [2 2 -3] u.v = -10



Enter the elements of vector u: 0 0 -1] Enter the elements of vector v: [1 0 0] u.v = 0



Enter the elements of vector u: [2 1 2] Enter the elements of vector v: (-1 2 5] u.v = 10

a)

--- -------------------- IV ----- I u.V 31-2j+4k | 2i+2j-3k | -10 0i+0j-1k | 11-6j+ek o 2i+1j+2k -1i+2j+5k | 10



(my_arrayCount.m)

my_dotProduct.m x my_arrayCount. m x + clear, clc; Staking nxm array input from the user. arr = input('enter the size of 2D array: '); storing dimension of array arr. [n, m] = size (arr); $variables which stores count of negative and $positve numbers in an array arr. neg = 0; pos = 0; TL LLLLLLL fiterating in the array. for i = 1:n for j = 1:m farray is less than zero then increment neg. if(arr(i,j)<0) neg = neg + 1; else felse we increment pos. pos = pos + 1; end end end $displaying count of pos and neg in the array arr. fprintf('n>0: %d\nn<0: d\n', pos, neg);

Output:

enter the size of 2D array: (3,2,-1,5,0,-2,-3] n>0: 4 n<0: 3



enter the size of 2D array: n>0: 2 n<0: 2



enter the size of 2D array: [1 -2 3; 2 10 -10; -pi 5 3; -5 17 20] n>0: 8 n<0: 4


b)

Array n(>0) n(<0) [3,2,-1,5,0-2-3] [H1 2 3 2 -10 -12 - 53 5 17 20]


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