Math 211 Fall 2007: Solutions: HW #11
Instructor: S. Cautis
1. section 7.4, #15
a) There are 3 equations and 6 unknowns so there must be at least 3 free
variables.
b) The system could have more free variables if the equations are linearly
dependent. Adding the first to the second and thrice the first to the third
gives
−x1 + x2 + x3 − x4 − x6 = 0
−3x2 + 3x3 + 3x4 − 3x5 − 3x6 = 0
0=0
so there are actually 4 free variables.
2. section 7.5, #4
We try to solve

   

1
0
−3
a1 −4 + a2 −2 =  2 
4
1
7
This means
a1 = −3
−4a1 − 2a2 = 2
4a1 + a2 = 7
But then the second equation means a2 = −2 while the third implies
a2 = 19. So there is no solution and hence w is not in the span of v1 and
v2 .
3. section 7.5, #12
We find v1 − v2 + v3 = 0.
4. section 7.5, #30
We reduce the matrix to row-echelon form as follows. Multiply first row
by 2 and subtract three times 3rd row from 1st row and twice 3rd row
from second to get


0 1 −1
A= 0 0 0 
−2 5 −7
We look for a vector v = (x1 , x2 , x3 )T such that Av = 0. This gives
x2 − x3 = 0
−2x1 + 5x2 − 7x3 = 0
so that if we take x2 = t then x3 = t and x1 = −t so the null space looks
like t(−1, 1, 1)T .
5. section 7.6, #16
We need to solve
x1 + x2 + x3 = 0
x2 + x3 = 0


0
So if x3 = t then x2 = −t and x1 = 0. So the set of solutions is t −1
1
and thus A is singular.
6. section 7.6, #18
Adding the appropriate multiple of the second row to the other rows gives


0
0
−6
6
−1 −4
2
−3 


 0 −11 11 −11
0
−9
9
−9
which gives


0
0 −1 1
−1 −4 2 −3


 0 −1 1 −1
0
0
0
0
So we need to solve
−x3 + x4 = 0
−x1 − 4x2 + 2x3 − 3x4 = 0
−x2 + x3 − x4 = 0
If we let x4 
= t then
 x3 = t and x2 = 0 and x1 = −t so that the null space
−1
0

looks like t 
 1 . Thus A is again singular.
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