I bet it can be done, but I'll be darned if I can!
The identity matrix is a square one with ones (1s) down its main diagonal and zeroes (0s) elsewhere. That is, it must have the same number of rows as columns, and where the row number is the same as the column number, the entry must be 1, elsewhere, it must be 0.
It will either be a 1*23 row matrix or a 23*1 column matrix.
No.
Using the method derived from the usual definition using the minors, the complexity is O(n!). But it seems that one could just do the Gaussian elimination on the matrix, turning the matrix into a triangular one while keeping track of any neccessary row swaps, and then just multiply the values of the diagona. This method would get the complexity of O(n*n).
Reduced matrix is a matrix where the elements of the matrix is reduced by eliminating the elements in the row which its aim is to make an identity matrix.
It is a matrix with 1 row and two columns: something like (x, y).
Starting with the square matrix A, create the augmented matrix AI = [A:I] which represents the columns of A followed by the columns of I, the identity matrix.Using elementary row operations only (no column operations), convert the left half of the matrix to the identity matrix. The right half, which started off as I, will now be the inverse of A.Starting with the square matrix A, create the augmented matrix AI = [A:I] which represents the columns of A followed by the columns of I, the identity matrix.Using elementary row operations only (no column operations), convert the left half of the matrix to the identity matrix. The right half, which started off as I, will now be the inverse of A.Starting with the square matrix A, create the augmented matrix AI = [A:I] which represents the columns of A followed by the columns of I, the identity matrix.Using elementary row operations only (no column operations), convert the left half of the matrix to the identity matrix. The right half, which started off as I, will now be the inverse of A.Starting with the square matrix A, create the augmented matrix AI = [A:I] which represents the columns of A followed by the columns of I, the identity matrix.Using elementary row operations only (no column operations), convert the left half of the matrix to the identity matrix. The right half, which started off as I, will now be the inverse of A.
Since the columns of AT equal the rows of A by definition, they also span the same space, so yes, they are equivalent.
#include<iostream> #include<iomanip> #include<vector> class matrix { private: // a vector of vectors std::vector< std::vector< int >> m_vect; public: // default constructor matrix(unsigned rows, unsigned columns) { // rows and columns must be non-zero if (!rows !columns) { throw; } m_vect.resize (rows); for (unsigned row=0; row<rows; ++row) { m_vect[row].resize (columns); for (unsigned column=0; column<columns; ++column) { m_vect[row][column] = 0; } } } // copy constructor matrix(const matrix& copy) { m_vect.resize (copy.rows()); for (unsigned row=0; row<copy.rows(); ++row) { m_vect[row] = copy.m_vect[row]; } } // assignment operator (uses copy/swap paradigm) matrix operator= (const matrix copy) { // note that copy was passed by value and was therefore copy-constructed // so no need to test for self-references (which should be rare anyway) m_vect.clear(); m_vect.resize (copy.rows()); for (unsigned row=0; row<copy.m_vect.size(); ++row) m_vect[row] = copy.m_vect[row]; } // allows vector to be used just as you would a 2D array (const and non-const versions) const std::vector< int >& operator[] (unsigned row) const { return m_vect[row]; } std::vector< int >& operator[] (unsigned row) { return m_vect[row]; } // product operator overload matrix operator* (const matrix& rhs) const; // read-only accessors to return dimensions const unsigned rows() const { return m_vect.size(); } const unsigned columns() const { return m_vect[0].size(); } }; // implementation of product operator overload matrix matrix::operator* (const matrix& rhs) const { // ensure columns and rows match if (columns() != rhs.rows()) { throw; } // instantiate matrix of required size matrix product (rows(), rhs.columns()); // calculate elements using dot product for (unsigned x=0; x<product.rows(); ++x) { for (unsigned y=0; y<product.columns(); ++y) { for (unsigned z=0; z<columns(); ++z) { product[x][y] += (*this)[x][z] * rhs[z][y]; } } } return product; } // output stream insertion operator overload std::ostream& operator<< (std::ostream& os, matrix& mx) { for (unsigned row=0; row<mx.rows(); ++row) { for (unsigned column=0; column<mx.columns(); ++column) { os << std::setw (10) << mx[row][column]; } os << std::endl; } return os; } int main() { matrix A(2,3); matrix B(3,4); int value=0, row, column; // initialise matrix A (incremental values) for (row=0; row<A.rows(); ++row) { for (column=0; column<A.columns(); ++column) { A[row][column] = ++value; } } std::cout << "Matrix A:\n\n" << A << std::endl; // initialise matrix B (incremental values) for (row=0; row<B.rows(); ++row) { for (column=0; column<B.columns(); ++column) { B[row][column] = ++value; } } std::cout << "Matrix B:\n\n" << B << std::endl; // calculate product of matrices matrix product = A * B; std::cout << "Product (A x B):\n\n" << product << std::endl; }
Transposing a matrix is O(n*m) where m and n are the number of rows and columns. For an n-row square matrix, this would be quadratic time-complexity.
The minor is the determinant of the matrix constructed by removing the row and column of a particular element. Thus, the minor of a34 is the determinant of the matrix which has all the same rows and columns, except for the 3rd row and 4th column.
#include<stdio.h> #include<unistd.h> #define SIZE 3000; void main() { int a[5][5],row,columns,i,j; printf("Enter the order of the matrix(5*5)"); scanf("%d %d",&row,&columns); printf("Enter the element of the matrix\n"); for(i=0;i<row;i++) for(j=0;j<columns;j++) { scanf("%d", &a[i][j]); } printf("3-tuple representation"); for(i=0;i<row;i++) for(j=0;j<columns;j++) { if(a[i][j]!=0) { printf("%d %d %d", (i+1),(j+1),a[i][j]); } } getchar(); }
Gaussian elimination as well as Gauss Jordan elimination are used to solve systems of linear equations. If, using elementary row operations, the augmented matrix is reduced to row echelon form, then the process is called Gaussian elimination. If the matrix is reduced to reduced row echelon form, the process is called Gauss Jordan elimination. In the case of Gaussian elimination, assuming that the system is consistent, the solution set can be obtained by back substitution whereas, if the matrix is in reduced row echelon form, the solution set can usually be obtained directly from the final matrix or at most by a few additional simple steps.
And your question is......................?
A fixed point for pivoting in linear algebra refers to a scenario where the pivot element in a matrix remains constant during row operations. In other words, the pivot element does not change its position in the matrix as row operations are performed. This is important for maintaining the consistency and accuracy of solutions when using techniques like Gaussian elimination for solving systems of linear equations.
I bet it can be done, but I'll be darned if I can!