Math 51 Stanford Homework Solutions
Spring 2000 Math 51 pdf files
Number  Due Date  Assignment 

1  April 6 

2  April 13 

3  April 25 

4  May 4 

5  May 11 

6  May 23 

7  not turned in 

6.1 The augmented matrix is
1 0 1 0 1 2 3 4 1 2 1 2
∣
∣
∣
∣
∣
∣
5 13 5
Its reduced row echelon form is
1 0 0 −1 0 1 0 1 0 0 1 1
∣
∣
∣
∣
∣
∣
1 0 4
There is no inconsistency, and z is a free variable, so there exist infinitely many solu tions.
6.2 The augmented matrix is
[
1 2 3 2 1 −2
∣
∣
∣
∣
1 1
]
Its reduced row echelon form is
[
1 2 3 0 1 8/3
∣
∣
∣
∣
1 1/3
]
There is no inconsistency, and z is a free variable, so
there exist infinitely many solutions.
6.3 The augmented matrix is
1 −2 1 2 2 −1 3 1 2 1 −2 3
∣
∣
∣
∣
∣
∣
∣
∣
0 8 0 −7
Its reduced row echelon form is
1 0 0 0 1 0 0 0 1 0 0 0
∣
∣
∣
∣
∣
∣
∣
∣
0 0 0 1
Thus the final equation of the reduced system is 0 = 1, which means there are no solutions.
6.7 (a) After switching rows 1 and 2, the reduced row echelon form is
−1 4 0 0 11 1 0 0 0
∣
∣
∣
∣
∣
∣
0 0 0
Here z is a free variable and the solutions are (−4z/11,−z/11, z).
3