# 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

Survey

* Your assessment is very important for improving the work of artificial intelligence, which forms the content of this project

Document related concepts

Symmetric cone wikipedia, lookup

Rotation matrix wikipedia, lookup

Covariance and contravariance of vectors wikipedia, lookup

Capelli's identity wikipedia, lookup

Matrix (mathematics) wikipedia, lookup

Principal component analysis wikipedia, lookup

Determinant wikipedia, lookup

Non-negative matrix factorization wikipedia, lookup

System of linear equations wikipedia, lookup

Orthogonal matrix wikipedia, lookup

Singular-value decomposition wikipedia, lookup

Matrix calculus wikipedia, lookup

Four-vector wikipedia, lookup

Matrix multiplication wikipedia, lookup

Gaussian elimination wikipedia, lookup

Jordan normal form wikipedia, lookup

Cayley–Hamilton theorem wikipedia, lookup

Perron–Frobenius theorem wikipedia, lookup

Eigenvalues and eigenvectors wikipedia, lookup

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
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)
(•)
```
Related documents