Download 1. Let A = 3 2 −1 1 3 2 4 5 1 . The rank of A is (a) 2 (b) 3 (c) 0 (d) 4 (e

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Transcript


3 2 −1
2. The rank of A is
1. Let A = 1 3
4 5
1
(a) 2
(b) 3
(c) 0
(d) 4
(e) 1
2. Let P2 = {a0 +a1 t+a2 t2 } where {a0 , a1 , a2 } range over all real numbers, and let T : P2 → P2
be a linear transformation dedifined by
T (a0 + a1 t + a2 t2 ) = a1 + 9a2 t
�
�
If T p(t) = λp(t) for some non-zero polynomial and some real number λ, then p(t) is called
an eigenvector corresponding to λ. The linear transformation T has an eigenvector:
(a) 100
(b) 0
(c) t2
(d) t + 9t2
(e) t
�
�
1 6
3. Let {λ1 , λ2 } be two eigenvalues of
. Then the product of two eigenvalues, λ1 λ2 , is
5 2
equal to
(b) −28
(a) 28
(c) 3
(d) 4
(e) 7
� �
2
(d)
3
� �
3
(e)
2
�
�
4 −3
5. The matrix A =
has a complex eigenvector:
3
4
� �
� �
� �
� �
3
4
1
4
(a)
(b)
(c)
(d)
4i
3i
−1
3
� �
1
(e)
i
�
�
� �
1
1
4. Let b1 =
, Let b2 =
,
−1
2
� �
5
Let x =
and B = {b1 , b2 } Find [x]B .
4
� �
� �
� �
1
1
5
(a)
(b)
(c)
2
−1
4


2
4
3
6. Let A = −4 −6 −3. The eigenvalues of A are
3
3
1
(a) 1, 2, 3
(b) 1, ±2
(c) 0, 1, 2
(d) 0, 1, −2
(e) 1, −2, −2
� �
� �
5
3
7. Let S be the parallelogram determined by the vectors b1 =
and b2 =
and let
3
2
�
�
−2 99
A=
. Compute the area of the images of S under the mapping v �→ Av.
0 1
(a) 3
(b) 99
(c) 2
(d) −2
(e) 5
(d) 4
(e) ±i
�
�
5 −5
8. Let A =
. The complex eigenvalues of A are
1
1
(a) 1 ± 3i
(b) 3 ± i
(c) ±3i
 
1
9. For what value(s) of h will y be in the subspace spanned by {v1 , v2 , v3 }, if v1 = 2,
3
 
 
 
2
3
6
v2 = 1, v3 1 and y =  4.
3
4
h
(a) 2
(b) 4
(c) 10
(d) 6
(e) 3
� �
� �
� �
� �
−9
−5
1
3
10.
Let b1 =
, b2 =
, c1 =
, c2 =
, B = {b1 , b2 } and
1
−1
−4
−5
�
�
C = {c1 , c2 }. If P = [b1 ]C , [b2 ]C is the change-of-coordinates matrix from B to C, then
find P .
�
�
�
�
�
�
�
�
�
�
1 1 −5
1 −5 3
−9 −5
6
4
1
3
(a)
(b)
(c)
(d)
(e)
9
4 1
1 −1
−5 −3
−4 −5
4 1
7


 9a − 8b 
11. Find a matrix A such that W = Col(A) where W =  a + 2b and {a, b} range over


−5a
all real numbers.


9 −8
2
(a) 1
0 −5


1 0
(b) 0 1
0 0


9 −8
2
(c)  1
−5
0
�
�
�
�
9 1 −5
0 1 0
(d)
(e)
−8 2
0
1 0 0
   
0 
 0
12. Let S = 0 ,  2 . Then the subset S


1
−1
 
 
0
0



2
(a) 0 is orthogonal to
(b) a basis of R3
1
−1
(c) spans R3
(d) a linearly dependent subset
(e) a linearly independent subset


 
1 −4
2
0



8 −9 Use Cramer’s rule to solve Ax = 1.
13. Let A = −2
−1
7
0
0
14. Let P2 = {a0 +a1 t+a2 t2 } where {a0 , a1 , a2 } range over all real numbers, and let T : P2 → P1
be a linear transformation given by T (a0 +a1 t+a2 t2 ) = a1 +2a2 t. Suppose that B = {1, t, t2 }
is a basis of P2 and C = {1, t} is a basis of P1 .
(1) Find a matrix A such that [T v]C = A[x]B .
(2) Find N ul(A) and Col(A).
�
�
1
7 2
15. Let A =
. Find lim Ak . (Hint: Find the Diagonalization D of A use the formula
−4
1
k→∞
5
Ak = P Dk P −1 .
Solutions
1.

 
 
 

3 2 −1
1 3
2
1
3
2
1 3 2
2 3 2 −1 0 −7 −7 0 1 1 There are
Reduce to echelon form: 1 3
4 5
1
4 5
1
0 −7 −7
0 0 0
two pivots so rank is 2
2.


0 1 0
With the basis B = {1, t, t2 }, the matrix for T is 0 0 9. Because the matrix is triangular,
0 0 0
 
1
3

the characteristic polynomial is −λ and hence the eigenvalue is 0 and the eigenvector is 0,
0
2
so 100 = 100 + 0t + 0t is an eigenvector with eigenvalue 0.
3.
The product of the roots, with multiplicities, of any polynomial with leading coefficient 1 is
(−1)n times the product of the roots where n is the degree: (t − λ1 )(t − λ2 ) · · · (t − λn ) =
tn + · · · + (−1)n (λ1 · · · λn ). In this case λ1 λ2 = det A so the answer is −28.
4.
�
�
1 1
We need to solve Ay = x where A =
. So reduce to reduced row echelon form
−1 2
�
� �
� �
� �
�
� �
1 1 5
1 1 5
1 1 5
1 0 2
2
Hence the solution is [x]B =
−1 2 4
0 3 9
0 1 3
0 1 3
3
5.
�
�
�4 − λ
�
−3
� = (4−λ)2 +9 = 16−8λ+λ2 +9 = λ2 −8λ+25.
The characteristic equation is det ��
3 4 − λ�
√
8 ± 64 − 100
The roots are λ =
= 4 ± 3i. The eigenspace for 4 + 3i is the null space for
2
�
�
� �
−3i −3
1
and an eigenvector is
.
3 3i
i
�
�
�2 − λ
�
�
�
4
3
�
�
�−6 − λ
�
−3
�−
−3�� = (2 − λ) det ��
The characteristic equation is det �� −4 −6 − λ
3 1 − λ�
�
�
3 1−λ
�
�
�
� 3
�−4
�
�
�
�
�
�
−3�
�−4 −6 − λ� = (2 − λ) (−6 − λ)(1 − λ) + 9 − 4(−4(1 − λ) + 9 +
4 det ��
+
3
det
� 3
3 1 − λ�
3�
�
�
3 −12 − 3(−6 − λ) and good luck to you!
OR
The sums of the five answers are all the different and the sum of the eigenvalues is the
trace of the matrix 2 + (−6) + 1 = −3. Hence the only choice for the eigenvalues in the given
answers is 1, −2, −2.
6.
7.
The area is the area of S times
� the
� absolute value of the determinant of A, which is 2. The
�5 3�
� = 10 − 9 = 1.
area of S is det [b1 b2 ] = det ��
3 2�
8.
2
2
The characteristic equation is λ −trAλ+det A = λ −6λ+10 with roots λ =
3 ± i.
9.

1 2

We are looking for the h such that the equation Ax = y with A = 2 1
3 3
reduce
the
following
matrix
to
reduced
row
echelon
form:

 
 
 
1 2 3 6
1
2
3
6
1
2
3
6
1
2 1 1 4 0 −3 −5
−8 0 −3 −5
−8 = 0
3 3 4 h
0 −3 −5 h − 18
0
0
0 h − 18 + 8
0
To have a solution, h − 10 = 0.
10.
We need to write b1
�
� �
1
3 −9
1
−4 −5
1
0
�
� �
1
3 −5
1
−4 −5 −1
0
6+
√
36 − 40
=
2

3
1 so we need to
4

2
3
6
−3 −5
−8.
0
0 h − 10
� �
� �
� �
�
�
� �
−9
1
3
1
3
−9
=
= ac1 + bc2 = a
+b
or solve
x=
.
1
−4
−5
−4 −5
1
� �
� �
�
� �
3 −9
1 3 −9
1 0
6
6
so the first column of P is
.
7 −35
0 1 −5
0 1 −5
−5
� �
� �
�
� �
3 −5
1 3 −5
1 0
4
4
so the second column of P is
.
7 −21
0 1 −3
0 1 −3
−3
�
�
6
4
Hence P =
.
−5 −3
11.

 
 


9a − 8b
9
−8
9 −8
 a + 2b = a  1 + b  2 so a matrix is  1
2
−5a
−5
0
−5
0
12.
Since there are only two vectors they can not be a basis for R3 . They can not even span
R3 . If the two are dependent, one is a multiple of the other and this is clearly not the case.
Hence they are independent. The dot product is −1 so they are not orthogonal.
13.
To use� Cramer’s rule
� we need to compute four determinants:
� 1 −4
�
�
�
�
2��
�
�−2 8�
� 1 −4�
�
�
�
�
�
� + 0 = −12 + 27 = 15
8 −9� = 2 det �
det �−2
� − (−9) det �−1
�
−1
7
7
�−1
�
7
0
�
�
�0 −4
�
�
�
2
�
�
�−4 2�
�
�
�
� + 0 = 14
8 −9� = 0 − 1 det �
det �1
�
7
0
�0
�
7
0
�
�
� 1 0
�
�
�
2
�
�
� 1 2�
�
�
�
�−0=2
det �−2 1 −9� = −0 + 1 det �
�
−1
0
�−1 0
0�
and�
�
� 1 −4 0�
�
�
�
�
� 1 −4�
� + 0 = −3
8 1�� = 0 − 1 det ��
det ��−2
−1
7�
�−1
�
7 0


14
 15 


 2 



Hence the solution vector is x = 
 15 .


 3 
−

15
14.
A is a 2 × 3 matrix and the ith column is found by working out T on the ith basis element
of B.
� �
0
Column 1: T (1) = 0 so
.
0
� �
1
Column 2: T (t) = 1 so
.
0
� �
0
2
Column 3: T (t ) = 0 so
.
2
�
�
0 1 0
Hence A =
.
0 0 2
The matrix A is already in echelon form and there are two pivot positions.
Col(A)
 Hence

1
has dimension 2 and so is all of P1 . The space N ul(A) has dimension 1, 0 is clearly a
0
non-zero vector in the null space so it is a basis for it.
15.
The characteristic equation for A is λ2 − trace(A)λ + det(A) = λ2 − 85 λ + 15
= (λ − 35 )(λ − 1).
25
3
Hence the eigenvalues are and 1.
5
� 4
2�
� �
5
5
3
2
The eigenspace for is the null space for
and this is spanned by
.
4
2
−4
5
−5 −5
� 2
2�
� �
5
5
1
The eigenspace for 1 is the null space for
and this is spanned by
4
4
−1
−5 −5
�
�
�3 �
�
�
2
1
0 0
0
k
−1
5
If P =
and D =
, then A = P DP . Note lim D =
so
−4 −1
0 1
0 1
k→∞
�
��
� �
�
�
��
�
�
� �
�
1
1
2
1 0 0 1 −1 −1
2
1 0 0
4
2
2
1
k
lim A =
=
=
=
.
−4 −1 0 1 2
4
2
−2 −1
k→∞
2 −4 −1 4 2
2 −4 −2
1.
(•)
(b)
(c)
(d)
(e)
2.
(•)
(b)
(c)
(d)
(e)
3.
(a)
(•)
(c)
(d)
(e)
4.
(a)
(b)
(c)
(•)
(e)
5.
(a)
(b)
(c)
(d)
(•)
6.
(a)
(b)
(c)
(d)
(•)
7.
(a)
(b)
(•)
(d)
(e)
8.
(a)
(•)
(c)
(d)
(e)
9.
(a)
(b)
(•)
(d)
(e)
10.
(a)
(b)
(c)
(•)
(e)
11.
(a)
(b)
(•)
(d)
(e)
12.
(a)
(b)
(c)
(d)
(•)