Up: 1.53 (a) Previous: 1.53 (a), §1

1.53 (a), §2 Solution

We need the null space of the matrix:

Forward elimination:

Continue to row canonical:

Back substitution:

From (3''), nothing; from (2'), z=-3s-t; from (1'), x=-3y+7s+3t. Variables y, s, and t cannot be determined: space is 3D.

Vector form;

The three vectors in the right hand side form a basis for the solution space.


Up: 1.53 (a) Previous: 1.53 (a), §1
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