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Answers to Even-Numbered Practice Problems
Section 1.1)
#38 (a) x 1 3 , x 2 5 , x 3 2 , x 4 0
(b) x 4
0 , x 3
2 , x 2
1 , x 1
Section 1.2)
#18) Only (b) and (d) are in RREF. In (a), the third column contains two leading ones and in (c),
the row of zeros must be at the bottom of the matrix.
#22) Including the zero matrix, there are 7 types of 2 3 matrices in RREF (discussed in class).
#36) Show that
reduces to
and so the unique
solution is x 1 2 , x 2 3 , x 3 4.
Section 1.3)
#20) (a)
(b)
#22) The augmented matrix is a 3 4 matrix with three leading ones and no "forbidden" rows.
So the rank of the 3 3 matrix of corfficients is 3 and the RREF of the coefficient matrix
is necessarily
#28) If A is a 5 3 matrix of rank 3, then RREF( A ) is
#36) Interpreting the given products as linear combinations of the columns of A , it follows that
A
#50) From the given information, the last column of the augmented matrix must contain a leading 1.
Thus the system is inconsistent and there are no solutions.
#52) C
b
c
is a linear combination of
and
if and only if b 9 and c 6.
Section 2.1)
#2) This is a linear transformation.
#4) A
#6) Yes, T is linear and the matrix for this transformation is A
#8) x 1
20 y 1
7 y 2
and x 2
3 y 1
y 2
; inverse matrix is
#32) Because B is 2 3, B x = 0 has infinitely many solutions, and hence there is a nonzero vector
v such that B v = 0. But then A B v = A ( B v ) = 0. So if A B I 3 , then
A B v = 0 I 3
v = 0 v = 0 , which is a contradiction! So the answer to the question is NO.
Section 3.1)
#30) An easy example is the matrix A
because a linear combination of the columns of this
matrix will simply be a scalar multiple of
. This is the matrix for the linear transformation
T : R
1 R
2 defined by T x x , 5 x . A more "substantial" example would be to define
A as any 2 n matrix all of whose columns are scalar multiples of
, for instance
A
#32) This is similar to Problem 30. A simple example is A
. As before, we could use any
3 n matrix all of whose columns are scalar multiples of
#40) If ker A im B , then A B x A B x 0 for all x Applying this to each of the vectors e
i
we see that each column of A B is the zero vector and so A B is the n m zero matrix.
#50) In general we know that ker( A
3 ) ker( A
4 ) for any square matrix A. To show the reverse inclusion in this problem,
suppose that x is in ker( A
4 ), so A
4
x 0 A
3
A x A x is in ker A
3 . Now we are
given that ker( A
3 ) = ker( A
2
), so we now know that A x is in ker A
2 A
2
A x 0 A
3
x 0
x is in ker( A
3 ). So we have shown that ker( A
4 ) ker( A
3 ) and can conclude that in this problem,
ker( A
3 ) = ker( A
4 )
Section 3.2)
. At this point it is clear that the matrix has rank 3
and hence A x b is consistent for every b in R
3 ; in other words the image of A is R
3
. Another way to see this is to note
that the rank computation shows that the columns of the matrix are linearly independent and a set of three linearly indepen-
dent vectorrs in a vector space of dimension three is necessarily a basis for that space. So the image of A is all of R
3 .
#34) If the columns of A are v 1
, v 2
, v 3
, v 4
, then
is in ker A if and only if A
if and only if v 1 2 v 2 3 v 3 4 v 4 0 or v 4
1
4
v 1
1
2
v 2
3
4
v 3.
#40) To show that the columns of A B are linearly independent, it is enough to show (using our equivalent statements for
linear independence) that ker( A B ) = { 0 }. Applying the same line of reasoning as in Problem 51 of Section 3.1, we see that
if x is in ker( A B ), then A B x 0 A B x 0 B x
is in the kernel of A. Because we are given that the columns of A are linearly independent, it follows that
ker( A ) = { 0 } and so B x 0. This means that x is in ker( B ), and because the columns of B are linearly independent (i.e.
ker ( B ) = { 0 } ), it follows that x = 0. Thus the only vector in ker( A B ) is the zero
vector, which implies that the columns of A B are linearly independent.
Section 3.4)
#56) Here is one approach: if S is the matrix whose columns are v 1 and v 2 , then the given information
can be expressed as S
, so S
1
So the desired basis is v 1
and v 2
#72) The previous problem shows that ker( A ) and ker( B ) have the same dimension, in other words
nullity( A ) = nullity( B ). Now the Dimension Theorem implies that
rank( A ) + nullity( A ) = n = rank( B ) + nullity( B ) ( A and B are n n matrices).
Hence rank( A ) = rank( B ) when A and B are similar n n matrices.
#48) (a) If W 1
is the subspace of all even polynomials in P 4
, then p t is in P 4
if and only if
p t a b t
2 c t
4 , so a basis for W 1 is 1 , t
2 , t
4 and dim W 1 3.
(b) If W 2
is the subspace of all odd polynomials in P 4
, then p t is in P 4
if and only if
p t a t b t
3 , so a basis for W 2
is t , t
3 and dim W 2
Section 4.2)
# 2) T M
1
M
2
7 M
1
M
2
7 M
1
7 M
2
T M
1
T M
2
and
T k M 7 k M k 7 M k T M , so T is a linear transformation from R
2 2 to R
2 2 .
Furthermore T is invertible and T
1 M
1
7
M , so that it follows that T is an isomorphism.
Another way to see this is to note that T M O 7 M O M O ,
so ker T contains only the "zero element" which in this case is the 2×2 zero matrix.
#22) Properties of the definite integral show that the given transformation is linear. However because
dim( P 2
) = 3 and dim( R ) = 1, the transformation is not an isomorphism.
#60) im T R
2 and so rank T 2
ker T span p t where p t t 7 t 11 t
2 18 t 77 and so nullity T 1
Note: rank T nullity T 2 1 3 dim P 2
#72) It is routine to show that Zn p t Pn : p 0 0 is a subspace of Pn. (This is done just like on Problem 4 of
Assignment # 4.) A basis for Zn is t , t
2 ,... , t
n and so dim Zn n.
#34) I
n
A A
1 A
1 det A
adj A I n
A adj A det A I n
det A 0. 0
0 det A. 0
0 0. det A
Similarly, In A
1 A
1
det A
adj A A adj A A det A In
det A 0. 0
0 det A. 0
0 0. det A
#38) For an invertible matrix A, A
1
1 det A
adj A adj A det A A
1 .
Now if A and B are invertible matrices of the same size, then A B is also inverible.
Then adj A B det A B A B
1 det A det B B
1 A
1 det B ) B
1 det A A
1
adj B adj A .
Section 7.1)
#4) If v is an eigenvector for A with associated eigenvalue , then 7 A v 7 A v 7 v 7 v.
So v is an eigenvector for 7 A with associated eigenvalue 7 .
#6) Suppose that v is an eigenvector for A with associated eigenvalue and that v is an eigenvector for B
with associated eigenvalue . Then A B v A B v A v A v v v.
So v is an eigenvector for A B with associated eigenvalue .
Section 7.2)
#22) By properties of transposition, A I
T A
T I for any n n matrix A. (Here of course
I is the n n identity matrix.) So det A
T I det A I
T det A I . The last step
uses the fact that a square matrix and its transpose have the same determinant.
So it follows that A and A
T have the same characteristic polynomial and hence have exactly the
same eigenvalues (with the same corresponding algebraic multiplicities).
#38) The trace of A is the sum of the eigenvalues and the determinant of A is the product of the
eigenvalues. So the characteristic equation for the 2 2 matrix A is
2
5 14 0 or
2 5 14 0 7 2 0 and so the
eigenvalues of A are 7 and 2.
#48) If v 1 and v 2 are the columns of S , then A S A v 1 v 2 A v 1 A v 2 and
S D v 1
v 2
(2 v 1
3 v 2
). Now because S is invertible , we know that
v 1
and v 2
must be nonzero vectors such that A v 1
2 v 1
and A v 2
3 v 2
In other words, A is a 2 2 matrix whose eigenvalues are 2 and 3. Contrast this with
Problem 47 in which you are asked to find those matrices A for which there exists a
nonzero matrix M such that A M M D. This means that at least one of the columns of
M is nonzero. Then reasoning as above, we see that such a matrix M exists when A is a
matrix for whch either 2 o r 3 is an eigenvalue (possibly both).
Section 7.3)
#28) Clearly, J n
k has only one eigenvalue, namely k , and the algebraic multiplicity of this eigenvalue
is n. Now J n
k k I
and it is clear that rank J n
k k I n 1
nullity J n
k k I 1 dim E k
1 geometric multiplicity of the eigenvalue k is 1.
Section 7.4)
#38) Let A
and B
. Then det A det B tr A tr B 4. It is clear that A
is diagonalizable. ( A is itself a diagonal matrix so what would be an appropriate matrix S ?)
However, for B , dim E 2
1 and so B is not diagonalizable. So A and B are not similar.
#54) Using the hint, A
2 O but B
2
O. Now the only matrix that is similar to the
zero matrix O is O itself. So A
2 and B
2 are not similar. However, if A and B were similar,
properties of similarlity would show that A
2 and B
2 would necessarily be similar. So we can
conclude that A and B are not similar.
