Write a C program to compute Matrix Multiplication of two matrices. Use one dimensional array to store each matrix, where each row is stored after another. Hence, the size of the array will be a product of number of rows times number of columns of that matrix. Get number of row and column from user and use variable length array to initialize the size of the two matrices as well as the resultant matrix. Check whether the two matrices can be multiplied or not. Write a getMatrix() function to generate the array elements randomly. Write a printMatrix() function to print the 1D array elements in 2D Matrix format. Also, write another function product(), which multiplies the two matrices and stores in the resultant matrix. With SEED 5, the following output is generated.
Sample Output
Enter the rows and columns of Matrix A with space in between: 3 5
Enter the rows and columns of Matrix B with space in between: 5 4
Matrix A:
8 6 4 1 6
2 9 7 7 5
1 3 1 1 2
Matrix B:
9 5 4 5
9 9 8 1
4 4 3 5
2 6 2 1
4 5 2 4
Product AxB:
168 146 106 91
161 186 125 81
50 52 37 22
In conclusion, the answer is 5x1
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Answer:
Step-by-step explanation:
We are asked the equation of the graph which passes through the points at (0,5) and (-3,-4).
Now, we know that the equation of a straight line passing through the points (
) and (
) is given by
.
So, in our case () ≡ (0,5) and () ≡ (-3,-4)
Therefore, the equation of the straight line will be
⇒ 
⇒ (Answer)
Answer:
A) 
B) 
C) for n = 2
= 1
for n = 3
= 3
for n = 4
= 8
for n = 5
= 19
Step-by-step explanation:
A) A recurrence relation for the number of bit strings of length n that contain a pair of consecutive Os can be represented below
if a string (n ) ends with 00 for n-2 positions there are a pair of consecutive Os therefore there will be : 
strings
therefore for n ≥ 2
The recurrence relation for the number of bit strings of length 'n' that contains consecutive Os
b ) The initial conditions
The initial conditions are :
C) The number of bit strings of length seven containing two consecutive 0s
here we apply the re occurrence relation and the initial conditions
for n = 2
= 1
for n = 3
= 3
for n = 4
= 8
for n = 5
= 19
What is it that I am supposed to be doing?
Answer:
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