Download Homework # 2 Solutions

Homework # 2 Solutions
Math 152, Fall 2014
Instructor: Dr. Doreen De Leon
p. 48: 26, 38
26. Suppose A is the 3 × 3 zero matrix (with all zero entries). Describe the solution set of
the equation Ax = 0.
Solution: First, look at Ax:

 
 
 
 
0 0 0 x1
0
0
0









Ax = 0 0 0 x2 = x1 0 + x2 0 + x3 0
0 0 0 x3
0
0
0
 
0

= 0 = 0.
0
This is true for any x ∈ R3 , so the solution set of the equation is R3 .
38. Let A be an m × n matrix and let w be a vector in Rn that satisfies the equation Ax = 0.
Show that for any scalar c, the vector cw also satisfies Ax = 0.
Solution: We need to show that A(cw) = 0.
A(cw) = c(Aw)
= c(0)
= 0.
Therefore, cw solves the equation Ax = 0.
p. 61: 12, 36
12. Find the value(s) of h for which the vectors are linearly dependent. Justify each answer.
     
3
−6
9
−6 ,  4 , h
1
3
3
1
Solution: If we denote the vectors as v1 , v2 , and v3 , respectively, then the vectors are
linearly dependent if there is a nontrivial solution of
x1 v1 + x2 v2 + x3 v3 = 0.
The solution of the vector equation is the same as the solution of the system
augmented matrix is





3 −6 9 | 0 r → 1 r
1 −2 3 | 0
1 −2
3 |
1
r2 →r2 +6r1
3 1
−6




4 h | 0 −−−−→ −6
4 h | 0 −−−−−−→ 0 −8 h + 18 |
r3 →r3 −r1
1 −3 3 | 0
1 −3 3 | 0
0 −1
0 |




1 −2
3 | 0
1 −2
3 | 0
r2 ↔r3
r3 →r3 −8r2



0 | 0 −−−−−−→ 0 −1
0 | 0
−−−→ 0 −1
0 −8 h + 18 | 0
0
0 h + 18 | 0
whose

0
0
0
This system has a nontrivial solution if h + 18 = 0. Therefore, the vectors are linearly
dependent if h = −18 .
36. If v1 , v2 , v3 are in R3 and v3 is not a linear combination of v1 and v2 , then {v1 , v2 , v3 }
is linearly independent.
Solution: This is a false statement. Let
 
 
 
1
−1
1
v1 = 2 , v2 = −2 , v3 = 0 .
3
−3
0
Then v3 is not a linear combination of v1 and v2 . We can see this by solving
c 1 v1 + c 2 v2 = v3 ,
which has the augmented matrix


1 −1 | 1
2 −2 | 0 .
3 −3 | 0
The echelon form of this matrix is


1 −1 | 1
0 0 | 2 .
0 0 | 0
Therefore, there is no solution. We also have
1v1 + 1v2 + 0v3 = 0,
a linear dependence relation. Therefore, the vectors are linearly dependent.
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p. 69-70: 24, 36
24. An affine transformation T : Rn → Rn has the form T (x) = Ax + b, with A an m × n
matrix and b ∈ Rm . Show that T is not a linear transformation when b 6= 0. (Affine
transformations are important in computer graphics.)
Solution: For any vectors u, v ∈ Rn ,
T (u + v) = A(u + v) + b
T (u) + T (v) = Au + b + Av + b
= A(u + v) + 2b 6= T (u + v) unless b = 0.
Therefore, T is not a linear transformation when b 6= 0.
Alternate solution: T (0) = A(0) + b = b 6= 0, so T is not a linear transformation
when b 6= 0.
36. Let T : Rn → Rm be a linear transformation. Suppose {u, v} is a linearly independent
set, but {T (u), T (v)} is a linearly dependent set. Show that T (x) = 0 has a nontrivial
solution.
Solution: Since {T (u), T (v)} is linearly dependent, there exist constants c1 and c2 , not
both zero, such that
c1 T (u) + c2 T (u) = 0.
Since T is linear, c1 T (u) + c2 T (u) = T (c1 u1 + c2 u2 ). So, x = c1 u + c2 v is such that
T (x) = 0. Since {u, v} is linearly independent, we know that x 6= 0. (Otherwise, there
would be a nontrivial linear combination of u and v that is zero, which is impossible.)
Therefore, the equation T (x) = 0 has a nontrivial solution.
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