Mumbai University > Mechanical Engineering > Sem 6 > Finite Element Analysis

Marks: 5M

Year: Dec 2015

Gauss Elimination Method

This method involves converting the given matrix into a unit matrix by making suitable row operations. Then make the same operations ona unit matrix of the same order. This matrix will give the inverse of the given matrix. This will be illustrated clearly following example.

1) Find the inverse of the matrix by Gauss elimination method.

$\begin{bmatrix} \ 2 & 1 & 0 & 3 \\ \ -1 & 0 & 4 & 1 \\ \ 2 & 1 & 3 & 7 \\ \ 1 & 4 & 2 & 1 \\ \end{bmatrix}$

Ans:

1) Write a unit matrix by the side of the given matrix.

$\begin{bmatrix} \ 2 & 1 & 0 & 3 \\ \ -1 & 0 & 4 & 1 \\ \ 2 & 1 & 3 & 7 \\ \ 1 & 4 & 2 & 1 \\ \end{bmatrix} \begin{bmatrix} \ 1 & 0 & 0 & 0 \\ \ 0 & 1 & 0 & 0 \\ \ 0 & 0 & 1 & 0 \\ \ 0 & 0 & 0 & 1 \\ \end{bmatrix}$

2) Make 1st element of 1st row 1 by carrying out suitable row operation

$R'_1 = R_1 + R_2$

$\begin{bmatrix} \ 1 & 1 & 4 & 4 \\ \ -1 & 0 & 4 & 1 \\ \ 2 & 1 & 3 & 7 \\ \ 1 & 4 & 2 & 1 \\ \end{bmatrix} \begin{bmatrix} \ 1 & 1 & 0 & 0 \\ \ 0 & 1 & 0 & 0 \\ \ 0 & 0 & 1 & 0 \\ \ 0 & 0 & 0 & 1 \\ \end{bmatrix}$

3) Now make successive row operation such that all elements below 1 in the 1st column becomes zero.

$R'_2 = R_1 + R_2; R'_3 = R_3 - 2R_1; R'_4 = R_4 - R_1;$

$\begin{bmatrix} \ 1 & 1 & 4 & 4 \\ \ 0 & 1 & 8 & 5 \\ \ 0 & -1 & -5 & -1 \\ \ 0 & 3 & -2 & -3 \\ \end{bmatrix} \begin{bmatrix} \ 1 & 1 & 0 & 0 \\ \ 1 & 2 & 0 & 0 \\ \ -2 & -2 & 1 & 0 \\ \ -1 & -1 & 0 & 1 \\ \end{bmatrix}$

4) Make 2nd element of row 2 as 1. In this case it is already 1. Hence we proceed further.

5) Make of all elements below this element 1 as zero.

$R'_3 = R_3 + R_2; R'_4 = R_4 - 3R_2;$

$\begin{bmatrix} \ 1 & 1 & 4 & 4 \\ \ 0 & 1 & 8 & 5 \\ \ 0 & 0 & 3 & 4 \\ \ 0 & 0 & -26 & -18 \\ \end{bmatrix} \begin{bmatrix} \ 1 & 1 & 0 & 0 \\ \ 1 & 2 & 0 & 0 \\ \ -1 & 0 & 1 & 0 \\ \ -4 & -7 & 0 & 1 \\ \end{bmatrix}$

6) Make 3rd element of row 3 as 1 and elements below it zero.

$R'_3 = 9R_3 + R_4; R'_4 = R_4 + 26R_3$

$\begin{bmatrix} \ 1 & 1 & 4 & 4 \\ \ 0 & 1 & 8 & 5 \\ \ 0 & 0 & 1 & 18 \\ \ 0 & 0 & 0 & 450 \\ \end{bmatrix} \begin{bmatrix} \ 1 & 1 & 0 & 0 \\ \ 1 & 2 & 0 & 0 \\ \ -13 & 7 & 9 & 1 \\ \ -342 & -189 & 234 & 27 \\ \end{bmatrix}$

7) Make 4th element of row 4 as 1.

$R'_4 = \frac{R_4}{450}$

$\begin{bmatrix} \ 1 & 1 & 4 & 4 \\ \ 0 & 1 & 8 & 5 \\ \ 0 & 0 & 1 & 18 \\ \ 0 & 0 & 0 & 1 \\ \end{bmatrix} \begin{bmatrix} \ 1 & 1 & 0 & 0 \\ \ 1 & 2 & 0 & 0 \\ \ -13 & 7 & 9 & 1 \\ \ -0.76 & -0.42 & 0.52 & 0.06 \\ \end{bmatrix}$

8) Carry out further row operations to convert elements above the leading diagonal of LHS matrix to zero.

$R'_1 = R_1 - 4R_4 ; R'_2 = R_2 - 5 R_4; R'_3 = R_3 - 18 R_4;$

$\begin{bmatrix} \ 1 & 1 & 4 & 0 \\ \ 0 & 1 & 8 & 0 \\ \ 0 & 0 & 1 & 0 \\ \ 0 & 0 & 0 & 1 \\ \end{bmatrix} \begin{bmatrix} \ 4.04 & 2.68 & -2.08 & -0.24 \\ \ 4.8 & 4.1 & -2.6 & -0.3 \\ \ 0.68 & 0.56 & -0.36 & -0.08 \\ \ -0.76 & -0.42 & 0.52 & 0.06 \\ \end{bmatrix}$

$R'_1 = R_1 - 4R_3; R'_2 = R_2 - 8R_3;$

$\begin{bmatrix} \ 1 & 1 & 0 & 0 \\ \ 0 & 1 & 0 & 0 \\ \ 0 & 0 & 1 & 0 \\ \ 0 & 0 & 0 & 1 \\ \end{bmatrix} \begin{bmatrix} \ 1.32 & 0.44 & -0.64 & 0.08 \\ \ -0.64 & -0.38 & 0.28 & 0.34 \\ \ 0.68 & 0.56 & -0.36 & -0.08 \\ \ -0.76 & -0.42 & 0.52 & 0.06 \\ \end{bmatrix}$

$R'_1 = R_1 R_2;$

$\begin{bmatrix} \ 1 & 1 & 0 & 0 \\ \ 0 & 1 & 0 & 0 \\ \ 0 & 0 & 1 & 0 \\ \ 0 & 0 & 0 & 1 \\ \end{bmatrix} \begin{bmatrix} \ 1.96 & 0.82 & -0.92 & -0.26 \\ \ -0.64 & -0.38 & 0.28 & 0.34 \\ \ 0.68 & 0.56 & -0.36 & -0.08 \\ \ -0.76 & -0.42 & 0.52 & 0.06 \\ \end{bmatrix}$

Thus, the matrix on R.H.S. is the inverse of our original matrix. Readers can check that their product gives unit matrix.