Consider the system of equations Ax = b where the coecient matrix A has the following
form. We used this example in class while trying to learn the row reduction algorithm.
A =
2
6
6
6
6
4
123
456
789
3
7
7
7
7
5
1. Compute the null space of the matrix A. Carefully write which variables are the pivot
variables and which are the free variables. Remember the pivot variables are the ones
for which there is a pivot in the corresponding column. For example if there is a pivot
in the first two columns then x1 and x2 are pivot variables. The free variable should
not have a pivot in its column.
2. Now let us consider a particular right hand side. The b vector of constants are given
as follows
b =
2
6
6
6
6
4
3
3
3
3
7
7
7
7
5
Find a particular solution to this system of equations xp = (x1, x2, x3).
3. Using the null space above write the general solutions for this system of equations.
We learned in class that the general solution will have the form
x = xp + y
where y 2 N(A). You will notice that it has a very interesting expression. Suppose it
has the form (I am not giving you the answer, this is a pseudo example)
y =
2
6
6
6
6
4
c
c
c
3
7
7
7
7
5
and the particular solution is xp = (x⇤
1, x⇤
2, x⇤
3). Then the general solution is
x =
2
6
6
6
6
4
c + x⇤
1
c + x⇤
2
c + x⇤
3
3
7
7
7
7
5