please explain the working of 2nd ,3rd and 4th statements of this coding portion:
 for i = 1:n
 j = 1:n;
 j(i) = [];
 b = abs(a(i,j));
 check(i) = abs(a(i,i)) - sum(b); 
 if check(i) < 0
 fprintf('the matrix is not strictly diagonally dominant at row %2i

',i)
 end
end
hello, 

this code checks to see if the diagonal elements of a given matrix a (assuming n x n) are larger in magnitude than the sum of the magnitude of the non-diagonal elements in its row.

line by line explanation:

the first line loops through all the rows of a.

 for i = 1:n

the second line is defining j = [1 2 3 ... n], so a vector containing 1..n,
 
 j = 1:n;

the third line is removing the diagonal element, aka, the ith row and this removes the ith column, so the i,i th element,
 
 j(i) = [];

the fourth line takes the absolute value of all the elements in the ith row of a, excluding the diagonal element: a(i,j) = a(ith row, elements 1,2,...,i-1,i+1,...,n) since we removed the ith element from j previously.
 
 b = abs(a(i,j));

the fifth line computes the absolute value of the diagonal element minus the sum of the absolute value of all the other elements in the row. if check is positive, we know the magnitude of the diagonal element was greater than the sum of all the magnitudes of the other elements in its row.
 
 check(i) = abs(a(i,i)) - sum(b);

the sixth line checks if it's less than zero, in which case the diagonal elements magnitude is not greater than or equal to the sum of the magnitudes of the other elements.
 
 if check(i) < 0

if this is the case, the matrix is not strictly diagonally dominant, which the seventh line prints.
 
 fprintf('the matrix is not strictly diagonally dominant at row %2i

',i)

the eighth and ninth lines end the cycle.
 
 end
 end

sorry if this was too much detail,

hope this helps!
a - ii, b - iv, c - i, d - iii let i = `(1)/(sinx*(3 + 2cosx))*dx` = `int sinx/(sin^2x*(3 + 2cosx))*dx` = `int sinx/((1 - cos^2x)(3 + 2cosx))*dx` = `int sinx/((1 - cosx)(1 + cosx)(3 + 2cosx))*dx` put cos x = t∴ – sinx.dx = dt∴ sinx.dx = – dt ∴ i = `int (1)/((1 - t)(1 + t)(3 + 2t))*(-dt)` = `int (-1)/((1 - t)(1 + t)(3 + 2t))*dt` let `(-1)/((1 - t)(1 + t)(3 + 2t)) = "a"/(1 - t) + "b"/(1 + t) + "c"/(3 + 2t)` ∴ – 1 = a(1 + t)(3 + 2t) + b(1 - t)(3 + 2t) + c(1 - t)(1 + t)put 1 – t = 0, i.e. t = 1, we get– 1 = a(2)(5) + b(0)(5) + c(0)(2) ∴ – 1 = 10a ∴ a = `(-1)/(10)`put 1 + t = 0, i.e. t = – 1, we get– 1 = a(0)(1) + b(2)(1) + c(2)(0) ∴ – 1 = 2b ∴ b = `-(1)/(2)` put 3 + 2t = 0, i.e. t = `-(3)/(2)`, we get – 1 = `"a"(-1/2)(0) + "b"(5/2)(0) + "c"(5/2)(-1/2)` ∴ – 1 = `-(5)/(4)"c"` ∴ c = `(4)/(5)` ∴ `(-1)/((1 - t)(1 + t)(3 + 2t)) = (((-1)/(10)))/(1 - t) + ((-1/2))/(1 + t) + ((4/5))/(3 + 2t)` ∴ i = `int [(((-1)/10))/(1 - t) + ((-1/2))/(1 + t) + ((4/5))/(3 + 2t)]*dt` = `-(1)/(10) int 1/(1 - t)*dt - (1)/(2) int 1/(1 + t)*dt + (4)/(5) int 1/(3 + 2t)*dt` = `-(1)/(10) (log|1 - t|)/(-1) - (1)/(2) log | 1 + t| + 4/5 (log|3 + 2t|)/(2) + c` = `(1)/(10)log|1 - cosx| - (1)/(2)log|1 + cosx| + (2)/(5)log|3 + 2cos| + c`. Notice that the matrix A is upper-triangle, so its determinant is a  
product   
of diagonal elements:  
  
det(A) = 1*2*3 = 6.  
  
Moreover, det(A) is   
non-zero.  
Therefore the columns of A are linearly independent and   
therefore  
they constitutive a basis for R^3, so R^3 = Col(A).  
  
Hence   
any base of R^3 fits to be a base for Col(A),   
for instance the columns of   
A:  
  
(1,0,0), (2,2,0), (3,3,3)  
  
or   
  
(1,0,0), (0,1,0),   
(0,0,1)  
  
are the base.  
  
The dimension of Col(A) is the dimension of   
R^3 and therefore is equal  
to 3. a, b and c if response.keys: # if the list is not empty

border.opacity = 1 # needs to be reset on each trial

if response.keys[0] == ('f'):

border.pos = [-0.3, 0]

elif response.keys[0] == ('j'):

border.pos = [0.3, 0]

else:

border.opacity = 0

else:

continueroutine = fa… finding the correct answer here is important, but perhaps equally important is finding it using the method of matrix reduction, since those are the instructions.the augmented matrix is4 -2 -5 -514 -1 -2 -364 -2 4 -24you'll have to forgive the poor formatting. this answer box doesn't have a proper math symbol editor.using elementary row operations, the reduced row echelon form of the augmented matrix is1 0 0 -60 1 0 60 0 1 3from which we conclude that the system has a unique solution (-6,6,3). [image] nirgrahamuk:

observe({

fv <- req(input$filter\_var)

td <- req(tabledata())

updateselectizeinput(inputid = "filter\_vals",

choices = pull(td,fv))

})

observe({

fv <- as.character(req(input$filter\_var))

td <- req(table… Hi Valdeth. You should use input and output file streams if you want to write and read from a file. Basically you open the file for writing and once you are done close the file. Later you can open the file for reading. The example below reads numbers from the file and just prints them out. If you want to read the file line by line you should look at the getline() function. I have just used your initial array to come up with a result per array which gives the product of all negative numbers in the array. Note that there **HAS** to be at least 1 negative number per array.



```
const int m=4, n=5;
	int i,j;
	int num;
	int result;
	int A[m][n]={{5,3,-4,8,2},
				 {1,5,2,7,-8},
				 {-3,9,5,2,4},
				 {4,-1,3,6,8}};

	ofstream outfile;
	ifstream infile; 

	outfile.open("c:/matrix/matrix.txt");

	for (i=0;i<m;i++)
	{
		result = 1;

		for (j=0;j<n;j++)
		{
			outfile << setw(3)<< A[i][j];

                            // Multiply negative numbers in each array (has to be at least 1 negative number in each array)
			if (A[i][j] < 0)              				                      result *= A[i][j];  // result = result x A[i][j]
		}
		cout << "Result of array " << i << " is " << result << endl;
		outfile << "\n";
	}

	outfile.close(); // Close for writing

	infile.open("c:/matrix/matrix.txt"); // opens the file
    
	infile >> num;
	while ( !infile.eof() ) 
	{ // keep reading until end-of-file
		cout << "The next number is " << num << endl;
		infile >> num; // sets EOF flag if no value found
	}
	infile.close();  //Close for reading
``` if statements always check if an expression evaluates to the value of true. since you’re checking whether the value true  value true then it’ll run every time. as for the “purpose” of the if statement, it’s here to create a “block scope” within the “function scope”.

interestingly, if you remove … 4x3 – 12x2 + 11x – 5 is divided by 2x – 1remainder theorem states that if p(x) is any polynomial and a is any real number and if p(x) is divided by the linear polynomial (x – a) , then the remainder is p(a).let p(x) = 4x3 – 12x2 + 11x – 5 and we have (2x – 1)the zero of (2x – 1) is now using remainder theorem,p(x) = 4x3 – 12x2 + 11x – 5 is divided by 2x – 1 then, is the remainderremainder = –2