Math 293, Homework 7
due in section on Thursday or Friday October 17 or 18, 1996.
(Postponed a day or so for fall break.)
Please hand in the numbered problems (like (1), (2), and (3)).
Do not " " " *'d " ( " a*), b*), " c*)).
MATLAB (e.g., see matrix sample on WWW page) may be useful to you, but
this whole assignment can be done by hand.
*) We now want to illustrate some facts about solution of the equation
Ax = b. Here in math 293 we just want to get the meaning of the ideas
by example. The reasoning behind most of these is a topic for math 294.
Consider the matrixes A = [0 3] and B = [1 4 ] .
[2 2] [3 12]
Consider also the colums vectors o = [0] , c = [4] and d = [2] .
[0] [7] [6]
The matrix A is 'non-singular' therefor the following things are
true. Compute and see for yourself.
a) det(A) not= 0 (calculate det(A)), and
b) Ax=o has the unique solution x=o (Solve Ax=o for x), and
c) The row reduced form of A has non-zero entry in
the lower right corner. (find the row reduced form of A), and
d) The equation Ax=b has a unique solution for all b
(solve Ax=c, and Ax=d for the c and d given above and note
that you don't have any choice in the solution),
e) None of the columns of A can be found by adding and subtracting
other columns (for a 2x2 this just means the columns aren't
multiples of each other).
(Show that column 2 is not a constant times column 1)
f) inv(A) exists (find inv(A) and check it by multiplying it by A).
The matrix B is 'singular' therefor the following things are true.
Compute and see for yourself.
aa) det(B) = 0 (calculate det(B)), and
bb) Bx=o has lots of solutions (find a family of solutions of Bx=o), and
cc) The row reduced form of B has a zero somewhere on the diagonal.
(find the row reduced form of B), and
dd) The equation Bx=b has no solutions for some b and multiple solutions
for the others (try to solve Bx=c and note the cause of
your inability to do so, find a family of solutions to
Bx=d)
ee) At least one of the columns of B can be found by adding and
subtracting multiples of the other columns
(for a 2x2 this just means one column is a multiple of the other)
(show that one column of B is a multiple of the other).
ff) inv(B) does not exist (try to find inv(B) and see that you
run into trouble).
1) Make up your own matrixes A and B, and your own column vectors
c and d and repeat the problem above. You need to pick matrices
and column vectors that make the parts work as they do above.
[A is non-singular, B is singular. c is not a multiple of one of the
columns of B, d is. ] Your matrices and column vectors should not
be the same as those from text, lecture, solutions, section, or of
any of your friends (they should be your own).
2) Optional.
Try to check part (dd) to the problem above with MATLAB (note any
warning messages you get and any solutions that look they are a
computers feeble attempt to tell you the answer is infinity just
cause the computer can't find the answer (Why can't the computer
find the answer?)). Make any other MATLAB checks that please you.
**) Consider the matrix A = [1 2]
[2 1]
Consider the colums vectors u = [1] , v = [ 1] and w = [2] .
[0] [-1] [2]
Multiply A by u, v and w to find out which of them is an
eigenvector and what the eigenvalue is for that eigenvector.
Do not set up or solve any quadratic equations.
3) Solve x' = Ax with A = [1 2] and xzero = [2].
[2 1] [3]