In numerous applications where linear systems appear one needs to solve Ax = b for numerous differentvectors b. For illustration a structure must be tested under several different loads, not just one. As in the example of a truss the loading in such a problem is usually represented by the vector b.
Gaussian eradication with pivoting is the most efficient and accurate way to solve a linear system.Most of the work in this process is spent on the matrix Aitself. If we necessitate to solve several differentsystems with the same A and A is big then we would like to evade repeating the steps of Gaussianelimination on A for every different b. This is able to be accomplished by the LU decomposition whichin effect records the steps of Gaussian elimination.
The major idea of the LU decomposition is to record the steps used in Gaussian elimination on A inthe places where the zero is produced. Determine the matrix:
The first step of Gaussian eradication is to subtract 2 times the first row from the second row. Inorder to record what we have completed we will put the multiplier 2 into the place it was used to makea zero that is the second row first column. In order to create it clear that it is a record of the step andnot an element of A we will put it in parentheses. This leads to:
There is before now a zero in the lower left corner therefore we don’t need to eliminate anything there. Werecord this fact with a (0). To eradicate the third row second column we require subtracting −2times the second row from the third row. Footage the −2 in the spot it was utilized we have:
Let U be the upper triangular matrix produced as well as let L be the lower triangular matrix with therecords and ones on the diagonal that is:
Then we have the subsequent mysterious coincidence:
Therefore we see that A is actually the product of L and U. Here L is lower triangular as well as U isupper triangular. When a matrix be able to be written as a product of simpler matrices we call that adecomposition of A as well as this one we call the LU decomposition. Using LU to solve equations:
If we as well include pivoting then an LU decomposition for A consists of three matrices P, L and U such that:
PA = LU.
The pivot matrix P is the identity matrix with the alike rows switched as the rows of Aare switched in the pivoting for illustration
Would be the pivot matrix if the second moreover third rows of A are switched by pivoting. Mat lab will produce an LU decomposition with pivoting for a matrix A with the subsequent command
> [L U P] = lu(A)
Where P is the pivot matrix to utilize this information to solve Ax = b we first pivot both sides by multiplying by the pivot matrix
PAx = Pb ≡ d.
Substituting LU for PA we obtain
LUx = d.
Then we need merely to solve two back substitution problems
Ly = d
Ux = y.
In Mat lab this would work as follows
> A = rand(5,5)> [L U P] = lu(A)> b = rand(5,1)> d = P*b> y = L\d> x = U\y>rnorm = norm(A*x - b) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .Check the result.
We are able to then solve for any other b without redoing the LU step. Replicate the sequence for a new right hand side c = randn(5,1) you can start at the third line. While this mayn’t seem like a big savings it would be if A were a large matrix from a genuine application.
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