06.09.2021 Views

Linear Algebra, Theory And Applications, 2012a

Linear Algebra, Theory And Applications, 2012a

Linear Algebra, Theory And Applications, 2012a

SHOW MORE
SHOW LESS

You also want an ePaper? Increase the reach of your titles

YUMPU automatically turns print PDFs into web optimized ePapers that Google loves.

14.5. ITERATIVE METHODS FOR LINEAR SYSTEMS 355<br />

The iterates are defined as<br />

⎛<br />

⎞<br />

∗ 0 ··· 0<br />

⎛<br />

. 0 ∗ .. . ⎜<br />

⎝<br />

.<br />

. .. . ⎟ ⎜<br />

.. 0 ⎠ ⎝<br />

0 ··· 0 ∗<br />

⎛<br />

⎞<br />

0 ∗ ··· ∗<br />

⎛<br />

.<br />

= −<br />

∗ 0 .. . ⎜<br />

⎝<br />

.<br />

. .. . ⎟ ⎜<br />

.. ∗ ⎠ ⎝<br />

∗ ··· ∗ 0<br />

x r+1<br />

1<br />

x r+1<br />

2. .<br />

x r+1<br />

n<br />

x r 1<br />

x r 2<br />

.<br />

x r n<br />

⎞<br />

⎟<br />

⎠<br />

⎞ ⎛<br />

⎟<br />

⎠ + ⎜<br />

⎝<br />

b 1<br />

b 2<br />

. .<br />

b n<br />

⎞<br />

⎟<br />

⎠<br />

(14.13)<br />

The matrix on the left in (14.13) is obtained by retaining the main diagonal of A and<br />

setting every other entry equal to zero. The matrix on the right in (14.13) is obtained from<br />

A by setting every diagonal entry equal to zero and retaining all the other entries unchanged.<br />

Example 14.5.2 Use the Jacobi method to solve the system<br />

⎛<br />

⎞ ⎛ ⎞ ⎛ ⎞<br />

3 1 0 0 x 1 1<br />

⎜ 1 4 1 0<br />

⎟ ⎜ x 2<br />

⎟<br />

⎝ 0 2 5 1 ⎠ ⎝ x 3<br />

⎠ = ⎜ 2<br />

⎟<br />

⎝ 3 ⎠<br />

0 0 2 4 x 4 4<br />

Of course this is solved most easily using row reductions. The Jacobi method is useful<br />

when the matrix is 1000×1000 or larger. This example is just to illustrate how the method<br />

works. First lets solve it using row operations. The augmented matrix is<br />

The row reduced echelon form is<br />

⎛<br />

⎜<br />

⎝<br />

⎛<br />

⎜<br />

⎝<br />

3 1 0 0 1<br />

1 4 1 0 2<br />

0 2 5 1 3<br />

0 0 2 4 4<br />

1 0 0 0<br />

6<br />

29<br />

0 1 0 0<br />

11<br />

29<br />

0 0 1 0<br />

8<br />

29<br />

0 0 0 1<br />

25<br />

29<br />

which in terms of decimals is approximately equal to<br />

⎛<br />

1.0 0 0 0 . 206<br />

⎜ 0 1.0 0 0 . 379<br />

⎝ 0 0 1.0 0 . 275<br />

0 0 0 1.0 . 862<br />

⎞<br />

⎟<br />

⎠<br />

⎞<br />

⎟<br />

⎠<br />

⎞<br />

⎟<br />

⎠ .<br />

In terms of the matrices, the Jacobi iteration is of the form<br />

⎛<br />

⎞ ⎛<br />

3 0 0 0 x r+1 ⎞ ⎛<br />

⎞ ⎛<br />

1<br />

0 1 0 0<br />

⎜ 0 4 0 0<br />

⎟ ⎜ x r+1<br />

2 ⎟<br />

⎝ 0 0 5 0 ⎠ ⎝ x r+1 ⎠ = − ⎜ 1 0 1 0<br />

⎟ ⎜<br />

⎝<br />

3<br />

0 2 0 1 ⎠ ⎝<br />

0 0 0 4 x r+1<br />

4<br />

0 0 2 0<br />

x r 1<br />

x r 2<br />

x r 3<br />

x r 4<br />

⎞ ⎛<br />

⎟<br />

⎠ + ⎜<br />

⎝<br />

1<br />

2<br />

3<br />

4<br />

⎞<br />

⎟<br />

⎠ .

Hooray! Your file is uploaded and ready to be published.

Saved successfully!

Ooh no, something went wrong!