Download Solutions to Assignment 3

Solutions to Assignment 3
Math 217, Fall 2002
1.9.14 Let T : R2 → R2 be a linear transformation with standard matrix A = [a1 a2 ],
where
a1 and a2 are shown in the figure. Using the figure, draw the image of
−1
under the transformation T .
3
xO 2
>
~~
~
~~
~~
~
~~a
~~ 1
~
~~
~V~V
VVVV
VVVV
VVVV
V
a2 VVVVVV
VVVV
+
/ x1
−1
−1
−1
The key to this problem is realizing that T
=A
= [a1 a2 ]
=
3
3
3
−a1 + 3a2 . This is what it means for A to be the standard matrix of T . So we
need to draw the vector −a1 + 3a2 . That is simply:
1
@ •K
xO 2
a1 >•
}
}
}}
}
}
}
}
}} a1

}

}
}
−1
}

T 
}}
3
•
@
a
@•
1
• kVVVVVV
a1
VVVVV −a
VVVV2V
VVVVV
VVVVV V VV
VVVVV
VVVVV
VVVVV
VVVVV
a2
VVVVV
+
/ x1
•


x1 − 2x2
x1
1.9.22 Let T : R2 → R3 be a linear transformation such that T
= −x1 + 3x2 .
x2
3x1 − 2x2
 
−1
Find x such that T (x) =  4 .
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This problem isn’t hard if we write down the standard matrix of T . So,


 
  

x1 − 2x2
1
−2
1 −2 x1
x
T
= −x1 + 3x2  = x1 −1 + x2  3  = −1 3  1 ,
x2
x2
3x1 − 2x2
3
−2
3 −2
and it is clear that the standard matrix of T is


1 −2
A = −1 3  .
3 −2
So now we put the augmented matrix


1 −2 −1
4
A = −1 3
3 −2 9
2
is row reduced echelon form. After the dust clears, we get the matrix


1 0 5
0 1 3 .
0 0 0
5
5
The solution to this linear system is x =
, that is,
is an x such that
3
3
 
−1

T (x) = 4 .
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1.9.40 Let

9
 14

A=
−8
−5
13
13
15
−9
−6
14

5
6 −1
−7 −6 4 

12 −5 −9

−8 9
8
15 2 11
be the standard matrix of T : R5 → R5 . Is T onto?
According to Theorem 12 page 89, this map is onto if and only if the columns
of A span R5 . In turn, by Theorem 4, page 43, we know that A spans R5 if and
only if A has 5 pivot positions. So we put the monster in reduced row echelon
form (row echelon form is enough, but my calculator gives the answer reduced).
We find the matrix


1 0 0 0 5
0 1 0 0 −4


0 0 1 0 0  .


0 0 0 1 1 
0 0 0 0 0
As this fails to have 5 pivot positions, we conclude that T is not onto.
1.10.2 One serving (28 g) of Kellogg’s Cracklin’ Oat Bran supplies 110 calories, 3 g of
protein, 21 g of carbohydrate, and 3 g of fat. One serving of Kellogg’s Crispix
supplies 110 calories, 2 g of protein, 25 g of carbohydrate, and .4 g of fat.
(a) Set up a matrix B and a vector u such that Bu gives the amounts of
calories, protein, carbohydrate, and fat contained in a mixture of three
servings of Cracklin’ Oat Bran and two servings of Crispix.
Let u1 be the number of servings of Cracklin’ Oat Bran and u2 be the
number of servings of Crispix in our mixture. Then there are 110u1 +
110u2 calories in any mixture, 3u1 + 2u2 g of protein, 21u1 +
 25u2 g of
110
 3 

carbohydrate, and 3u1 + .4u2 g of fat. This corresponds to 
 21  u1 +
3
3
 
110
 2 
  u2 , that is,
 25 
.4


110 110  3
2 
 u1

 21 25  u2 .
3
.4
So the matrix is


110 110
 3
2 

B=
 21 25  .
3
.4
For 3 servings of Cracklin’ Oat Bran
and 2 servings of Crispix, we would
3
multiply B by the vector u =
.
2
(b) Suppose that you want a cereal with more protein than Crispix but less
fat than Cracklin’ Oat Bran. It is possible for a mixture of the two cereals
to supply 110 calories, 2.25 g of protein, 24 g of carbohydrate, and 1 g of
fat? If so, what is the mixture?


110
2.25

Solve the system Bu = 
 24 . The row reduced echelon form of the
1
augmented matrix


110 110 110
 3
2 2.25


 21 25 24 
3
.4
1
is

1
0

0
0
0
1
0
0

0
0
.
1
0
We see that the system is inconsistent, so there is no mixture that works.
2.1.4 Compute A − 5I3 and (5I3 )A, when


9 −1 3
A = −8 7 −6 .
−4 1
8
Well,

 
 

9 −1 3
5 0 0
4 −1 3
A − 5I3 = −8 7 −6 − 0 5 0 = −8 2 −6 .
−4 1
8
0 0 5
−4 1
3
4
Also,


 

5 0 0
9 −1 3
45 −5 15
(5I3 )A = 0 5 0 −8 7 −6 = −40 35 −30 .
0 0 5 −4 1
8
−20 5
40
2.1.18 Suppose the first two columns, b1 and b2 , of B are equal. What can you say
about the columns of AB (if AB is defined)? Why?
Well, we can write B = [b1 b1 b2 · · · bn ], and then AB = [Ab1 Ab1 Ab2 · · · Abn ].
One can see that the first two columns of AB are equal.
2.1.24 Suppose AD = Im (the m × m identity matrix). Show for any b in Rm , the
equation Ax = b has a solution. Explain why A cannot have more columns
then rows.
Let b be any vector in Rm . Then Db is a vector as well, and A(Db) = (AD)b =
Im b = b, so Ax = b always has a solution. By Theorem 4, page 43, this means
that A has a pivot position in every row. If there were more rows then columns,
then not each row could have a pivot (because there is at most one pivot per
column). So there must be at least as many columns as rows.
2.2.14 Suppose AB = AC, where B and C are n × p matrices and A is invertible.
Show that B = C. Is this true, in general, when A is not invertible?
We may multiply both sides of the equality above by A−1 . The result is
A−1 AB = A−1 AC. Simplifying both sides we
obtain
B = C. This is not
1 0
0 0
true in general, however. Suppose that A =
, B =
, and C =
0 0
1 0
0 0
. Note that A is not invertible (see Theorem 4 of this chapter). Then
0 1 0 0
AB =
= AC, but B 6= C.
0 0
2.2.16 Suppose A and B are n × n, B is invertible, and AB is invertible. Show that A
is invertible.
We know that B −1 and (AB)−1 exist. Let C = AB. Then CB −1 = A, after
a little painless algebra. But we have noted already (see page 122) that the
product of invertible things is invertible. Both C and B −1 are invertible (see
Theorem 6 for the second fact), and thus A is invertible. In this case, A−1 =
BC −1 .
2.2.22 Explain why the columns of an n × n matrix A span Rn when A is invertible.
If A is invertible, then Ax = b has a solution for all b ∈ Rn (Theorem 5). But
by Theorem 4, page 43, this occurs if and only if the columns of A span Rn .
2.3.16 Is it possible for a 5 × 5 matrix to be invertible when it columns do not span
R5 ?
No. If the matrix in question is invertible, then because it is square, it columns
must span R5 by the Invertible Matrix Theorem (look at (a) and (h) of that
theorem).
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2.3.18 If C is 6 × 6 and the equation Cx = v is consistent for every v in R6 , is it
possible that for some v, the equation Cx = v has more than one solution?
Why or why not?
By the Invertible Matrix Theorem, Cx = v is consistent for every v in R6 if
and only if C is invertible. Theorem 5, page 120, tells us that if this is so, then
the solution to Cx = v is unique (it is C −1 v).
2.3.36 Let T be a linear transformation that maps Rn onto Rn . Show that T −1 exists
and maps Rn onto Rn . Is T −1 also one-to-one?
Let A be the standard matrix of T . So x → Ax is onto, and by the Invertible
Matrix Theorem A is invertible. Then by Theorem 9, T −1 exists with standard
matrix A−1 . But by Theorem 6 page 121, A−1 is invertible, so that again by the
Invertible Matrix Theorem we know that x → A−1 x is onto (and one-to-one).
This means that T −1 is onto (and one-to-one).
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