Solution The procedure of LU decomposition runs similar to the process of Gaussian Elimination. Firstly A is reduced to upper triangular form, which is U, using just the third elementary row operation, namely: add to one row of matrix a scalar time another row of that same matrix. Those scalar used during this process are co-efficient in the L matrix.
The final result will look like this: .
Here are the solution for this problem:
According to Gaussian Elimination, the first number in row 2 must be zero-out by adding the first row of matrix a scalar times second row. This scalar, fortunately, is .
The idea of using LU decomposition to solve systems of simultaneous linear equations Ax=b is rewriting the systems as L(Ux)=b. To solve x, we first solve the systems Ly=b for y, and then, once y is determined, we solve the systems: Ux=y for x. Both systems are easy to solve, the first by forward substitution and the second by backward substitution.