Module 4 : Solving Linear Algebraic Equations
Section 3 : Direct Solution Techniques
 

which implies MATH and, if you consider the fact that exchanging two rows of a matrix in succession will bring back the original matrix, then this result is obvious. Coming back to the construction of the transformation matrix for the linear system of equations under consideration, matrix $\QTR{bf}{T}$ is now constructed as follows MATH

This approach of constructing the transformation matrix $\QTR{bf}{T,}$ demonstrated using a $3\times 3$ matrix $\QTR{bf}{A,}$ can be easily generalized for a system of equations involving an $n\times n$ matrix $\QTR{bf}{A}$. For example, elementary matrix $\QTR{bf}{E}_{i1},$ which reduces element $(i,1)$ in matrix $\QTR{bf}{A}$ to zero, can be constructed by simply inserting $-e_{i1}=$ $-(a_{i1}/a_{11})$ at $(i,1)^{\prime }th$ location in an $n\times n$ identity matrix, i.e. MATH--------(21)while the permutation matrix $\QTR{bf}{P}_{ij},$ which interchanges i'th and j'th rows, can be created by interchanging i'th and j'th rows of the $n\times n$ identity matrix. The transformation matrix $\QTR{bf}{T}$ for a general $n\times n$ matrix $\QTR{bf}{A}$ can then be constructed by multiplying the elementary matrices, $\QTR{bf}{E}_{ij},$ and the permutation matrices, $\QTR{bf}{P}_{ij},$ in appropriate order such that $\QTR{bf}{TA=U.}$

It is important to note that the explanation presented in this subsection provides insights into the internal working of the Gaussian elimination process. While performing the Gaussian elimination on a particular matrix through a computer program, neither matrices (MATH nor matrix $\QTR{bf}{T}$ is constructed explicitly. For example, reducing elements $(i,1)$ in the first column of matrix $\QTR{bf}{A}$ to zero is achieved by performing the following set of computations MATHwhere $i=1,2,...n.$ Performing these elimination calculations, which are carried out row wise and may require row exchanges, is equivalent to constructing matrices (MATH and effectively matrix $\QTR{bf}{T,}$ which is an invertible matrix. Once we have reduced MATH to MATH form, such that u$_{ii}\neq 0$ for all i, then it is easy to recover the solution $\QTR{bf}{x}$ using the back substitution method. Invertibility of matrix $\QTR{bf}{T}$ guarantees that the solution of the transformed problem is identical to that of the original problem.