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(MATH152)[2000](f)final.pdf
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Math 152 Final Exam
December 15, 2000 Instructor: Guowu Meng
Your Name
Student Number
1.
This is an open-book exam. For more space, write on the opposite side.

2.
Do not use calculator. If your answer is too complicated, you must have made a mistake.

3.
Show all your work. Cross o. (instead of erase) the undesirable part.

4.
Provide all the details. Your reason counts most of the points.


Number Score
1
2
3
4
5
6
Total

(1) (20 points) Consider the matrix
..
110 A = 110 .
..
001
1) Find linearly independent vectors that span NulA, i.e., the solution set of equation Ax = 0. 2) Find linearly independent vectors that span ColA, i.e., the column space of A.
..
1 3) Find the distance from v =5 . to NulA.
.
2 4) Find an orthogonal matrix P a diagonal matrix D such that A = PDP t .
110 . . . .1 .
Solution.1) A . 001 ..So NulA is the span of . 1 ..
000 0
. . . .
1 0
2) ColA is the span of . 1 . and . 0 ..
0 1
. . . . . .
.1 .2 3

3) ProjNulAv =
4
2
.
.. ..
1=2 ..So Dist(v, NulA)= ||v . ProjNulAv|| = || 3 || = 23. 00 2
4) det(A . I)= .( . 1)( . 2)=0, so =0, 1, 2.
..
..
. .1 .

When =0, E = NulA = span 1.
.. ..
0
.. 010 . 0 . .. ..
.. ..
When =1, E = Nul 100 = span 0.
000 . 1 .
. . . . . . . .
.1 1 0 1 .10 . 1 .
When =2, E = Nul . 1 .1 0 . = Nul . 0 0 1 . = span . 1 . .
0 0 .1 0 0 0 . 0 .

00 It is clear that
.
. ..
.
0
.
. and 0 . are eigenvectors of A with eigenvalue 2, 0 and 1 respectively, and they
1
., .
2 1
0
21
0 0
=
.
.
.
2
1 21
form an orthonormal basis for R3, soifwelet P 0 ., D = . 000 ., then P is an orthogonal
1
matrix, D is a diagonal matrix, and A = PDP t .
(2) (20 points) 1) Find the inverse of the matrix
..
100 A = 110
..
101
2) Find the inverse of A2 .
3) Solve the matrix equation

..
11 AX = 10
..
11 (Here X is an unknown 3 2 matrix.)
Solution. Easy, no solution provided.
(3) (15 points) 1) Find an example of two non-invertible matrices A and B such that AB is invertible. 2) Find an example of two diagonalizable 2 2 matrices A and B such that AB is not diagonalizable. 3) Show that rank AB rank B. (Hint: The rank of a matrix is de.ned to be the maximum number of linearly
independent columns of the matrix. )
1
Solution.1) Say A =[1 0] and B =.
0
10 11
2) Say A =, B =.
0 12 02
3) Suppose that the set of pivotal columns of B is {v1, ,vr}and the set of non-pivotal columns of B is {u1, ..., us}. Then the set of columns of AB is {Av1, ,Avr,Au1, ..., Aus}. Since each ui is in the span of {v1, ,vr}, so each Aui is in the span of {Av1, ,Avr}. Therefore, ColAB = span{Av1, ,Avr}. Note that the rank of AB is the number of vectors in a minimal spanning set, so rankAB r = rankB.
(4) (20 points) Consider di.er