Java Programming Lab Code

One of the application of LU matrix decomposition finds inverse of a square matrix,A. A A -1 = I The square matrix (coefficient matrix) A decomposed into LU (lower triangular and upper triangular) matrix by elementary-row-operations . A = LU after that, inverse matrix A -1 is found by applying forward and back substitution on each column of identity matrix, I and place returning solution vectors x to column of A -1 matrix. Algorithm steps for Matrix Inverse by LU Decomposition Input : a square matrix , A Output : a square matrix, A -1 read matrix A [L,U]= LUDecompose (A) [nrow,ncol]=size(A) I = Identity-matrix(nrow,ncol); iA = zero-matrix(nrow,ncol); For c =1 to ncol Y = forward-substitute (L,I(c)) X = back-substitute (U,Y) A -1 (c,:) = X // set vector X to column of A -1

One of the application of LU matrix decomposition is solving system of linear equations. Ax = b A - coefficient matrix b - non-homogeneous vector x - unknown vector (solution vector) The square matrix (coefficient matrix) A decomposed into LU (lower triangular and upper triangular) matrix by elementary-row-operations. A = LU the solution vector X of system found by forward and back substitution. y = forward-substitute L and b X = back-substitute U and y The two or more system of equations have same coefficient matrix A but have different non-homogeneous vectors (b1 ,b2,..), the solutions vectors (x1,x2,..) obtained by once applying LU decomposition over matrix A. System 1 &nbsp&nbsp Ax = b1

A square matrix is transformed into a Lower triangular matrix L or an Upper triangular matrix U by applying elementary row operation (Gaussian elimination) for solving system linear of equations. A solution vector X of system of linear equations is obtained by applying substitution method. The forward substitution method is applied to matrix L The back substitution method is applied to matrix U Algorithm steps for forward substitution to matrix L Input : a square matrix, A a non-homogeneous vector b Output : Solution vector, X read matrix A read vector b L = transform_to_L (A,b) X = substitute-forward(L); Example of forward substitution to matrix L substitute unknown variables top to bottom approach 0.4286x = 1.7143 3.6667x + 2.3334y = 10.0 2.0x+ + 4.0y + 6.0z = 18.0 find x, from equation 1