1. Solve for x,y
ax + y = 1, x + ay = a2
where a is a given number.
the augmented matrix is
a 1 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).
A row addition operation gives
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 +x1 - 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.