Math 436.  September 25, 2002. Midterm 1.   Name__________________________________
5 problems, 10 pts each.

1. Solve for x,y
ax + y = 1, x + ay = a2
where a is a given number.

the augmented matrix is
a   1  1
a  a2
Adding the second row to the fist one with coeeficient  -a we obtain
0   1- a2    1 - a3
1      a         a2
Switching  the rows, we obtain
1      a         a2
0   1- a2    1 - a3
If   1- a2    is not 0,     we multiply the second row by 1/(1- a2 ) and obtain
1      a         a2
0    1    (1 +a + a2) /(1+ a).
1      0     (-1 -a + a3) /(1+ a)
0     1       (1 +a + a2) /(1+ a).
So when  a is not 1 or -1, the answer is
x =  (-1 -a + a3) /(1+ a), y = (1 +a + a2) /(1+ a).
When a = 1, the answer is        x =1 -y.
When a = -1, the answer is       0 = 1.

2. Solve the system
x1 + 2x2+ x3= 3,  -x1+ 3x2 + x4 = -2.

x3= 3  - x1 - 2x2 ,
x4 = -2  +x- 3x2 .

3. Find the rank of the matrix    A=
1   2   3   4
5   6   7   8
9 10 11 12 .
Row and column operations do not change the rank.
By two row addition operations we obtain
1     2     3        4
0   -4   -8   -12
0   -8  -16  -24 .
By two row multiplication operations we obtain
1     2    3     4
0     1    2     3
0     1    2     3 .
By  a row addition operation, we obtain
1     2    3     4
0     1    2     3
0     0    0     0 .
By  a row addition operation, we obtain
1     0    1     1
0     1    2     3
0     0    0     0 .
By 4 column addition operations we obtain
1     0    0     0
0     1    0     0
0     0    0     0 .
So the rank is 2.

4. Find a basis for the column space of  A.
By row operations we obtain
1     0    1     1
0     1    2     3
0     0    0     0 .
So the first two columns are linearly independent and they span the column space.
Answer: the first two columns of A form a basis for the column space.
There are other correct answers. For example, any two distinct columns of A form a basis.

5. Find a basis for the row space of  A.
By row operations we obtain
1     0    1     1
0     1    2     3
0     0    0     0 .
Row operations do not change the row space. So the rows
[1     0    1     1] and  [0     1    2     3] form a basis for the row space.
There are other correct answers and other ways to find a basis..
For example, any two distinct rows of A form a basis.