false

Class Notes
(839,092)

Canada
(511,185)

University of Toronto St. George
(44,098)

Mathematics
(2,859)

MAT223H1
(295)

all
(5)

Lecture

Description

Department of Mathematics, University of Toronto
MAT223H1F - Linear Algebra I
Fall 2013
Tutorial Problems 1
1. Consider the system of linear equations
x1+ x 2 2
ax 1 bx =24
cx + dx = 8:
1 2
(a) Find non-zero values of a;b;c and d such that the system has no solutions.
(b) Find non-zero values of a;b;c and d such that the system has a unique solution.
(c) Find non-zero values of a;b;c and d such that the system has in▯nitely many solutions.
Give reasons in each case.
Solution #1:
We will give two solutions to this problem. In this ▯rst solution, we will work with the equations directly;
in the second, we will use matrices and some of the theory developed in the course so far.
(a) If we let a = b = 1 (and take any non-zero values for c and d), then the ▯rst two equations become
x1+ x 2 2
x + x = 4:
1 2
But then 2 = 4, which is nonsense. So the system has no solutions in this case.
(b) Notice that if we let c = d = 4 then the third equation becomes
4x1+ 4x =28
or equivalently, after dividing through by 4,
8
x 1 x =2 4 = 2;
which is exactly the ▯rst equation. So we may ignore the third equation in this case and consider the
simpler system
x1+ x 2 2
ax 1 bx =24:
Now it’s easy to ▯nd examples where this system has a unique solution. Keeping in mind that we want
to ▯nd non-zero values of a and b, let’s try a = ▯1 and b = 1. By solving the resulting simultaneous
equations, we get the unique solution x1= ▯1 and x = 2.
1 of 4 (c) Just as in the previous part, we may take c = d = 4 and work with the simpler system
x1+ x2= 2
ax1+ bx2= 4:
If we now take a = b = 2, then we see that the second equation becomes identical with the ▯rst, and
so our system reduces to just one equation:
x1+ x2= 2:
This equation clearly has in▯nitely many solutions, because for any2value 2f x , say x = t, we can
simply take1x = 2 ▯ t and get a solution to the equation.
Solution #2:
Now let’s approach this problem with matrices. The system of equations gives us the augmented matrix
2 3
1 1 2
4 5
a b 4 :
c d 8
(a) This system will have no solutions if, after reduction, some row is all zeros to the left of the bar and
nonzero to the right of the bar. For example, if we take a = b = 1, then we have
2 3 2 3
1 1 2 R2▯R1 1 1 2
4 1 1 45 ▯▯▯▯▯! 40 0 2 ;
4 4 8 c d 8
as desired.
(b) This system will have a unique solution if, after reduction, it has two leading 1s and a row of zeros. It’s
very easy to arrange that we get a row of zeros: simply take c = d = 4, and apply one row operation
to get 2 3 2 3
1 1 2 1 1 2
R 34R1
4a b 45 ▯▯▯▯▯! 4a b 45:
4 4 8 0 0 0
Now it’s easy to get a matrix with two leading 1s: let a = ▯1 and b = 1, and note that
2 3 2 3 2 3 2 3
1 1 2 1 1 2 R ▯ 1 1 1 2 1 0 ▯1
4▯1 1 45 ▯▯▯▯!1 40 2 65 ▯▯▯!2 40 1 35 ▯▯▯▯! 4 0 1 3 5 :
0 0 0 0 0 0 0 0 0 0 0 0
▯ ▯ ▯ ▯
x ▯1
This gives us the unique solution = .
x2 3
(c) This system will have in▯nitely many solutions if, after reduction, it has one leading 1 and two rows
of zeros. Mimicking what we did in the previous part, we can take a = b = 2 and c = d = 4, and then
row reduce: 2 3 2 3
1 1 2 R 22R1 1 1 2

More
Less
Unlock Document

Related notes for MAT223H1

Only page 1 are available for preview. Some parts have been intentionally blurred.

Unlock DocumentJoin OneClass

Access over 10 million pages of study

documents for 1.3 million courses.

Sign up

Join to view

Continue

Continue
OR

By registering, I agree to the
Terms
and
Privacy Policies

Already have an account?
Log in

Just a few more details

So we can recommend you notes for your school.

Reset Password

Please enter below the email address you registered with and we will send you a link to reset your password.