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Linear Algebra Homework # 4
YOUR NAME
Instructions: You can either print out the homework and write your answer on it(recommended),
or other ways that clearly shows your work and computations. Write your UNI in the right top
corner of each page. and your name in the left top corner. For computational excercise, circle the
final answer. This homework is due at 4:30pm June, 10th.
Problem 1. Find out the solution set of the following system of linear equation. Represent the set
of solution by the way you like(Try multible way is recommended)
(1)

 3x + 7y + 5z = 13
5x + 2y
= 5

x+z
= 3
(2)
3x + 3y + 5z = 11
x + 2y
= 3
Date: Printed on:June 7, 2015;
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Linear Algebra Homework # 4
Your UNI:
(3)
x+y+z =6
(4)

= 5
 3x + 2y + z
4x + y + 6z
= 7

10x + 5y + 8z = 17
2
YOUR NAME
Your UNI:
(5)

= 5
 3x + 2y + z
4x + y + 6z
= 8

10x + 5y + 8z = 17
Problem 2. Find all possible value of λ for each of the following problem,
(1) such that this linear Equation have unique solution

 (3 + λ)x + 2y + z = 5
4x + y + 6z
= 8

10x + 5y + 8z
= 17
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Linear Algebra Homework # 4
(2) such that this lienar Equation have non-zero solution
(1 + λ)x − 2y = 0
2x + (5 + λ)y = 0
Your UNI:
Problem 3. Suppose you have two bottles of alcohol solution. The first bottle the concentration of
alcohol is 30%, in the second bottle the concertration of alcohol is 50%, in which ratio will you mix
them to get an alcohol solution have concentration 45%?
4
YOUR NAME
Your UNI:
Problem 4. Now m, n is two fixed number, in a discussion titled ”which m × n matrix do you
like”. The two people says”
Ricky:
I like all the matrix Am×n , such that the system of lienar equation Ax = b always has solution
no matter what b is! Let’s call this kind of matrix as Ricky’s Matrix
Michel:
I like all the matrix Am×n , such that AT x = 0 only have 0 solution! Let’s call this kind of matrix
as Michel’s Matrix
Host:
Well, I claim that every Ricky’s matrix is Michel’s matrix. and every Michel’s matrix is also
Ricky’s matrix.
Is that true? if it is, proof.
(Hint: suppose y is a solution, consider 0 = 0x = (AT y)T x = y T Ax = y T b, bacause we can find
x which could enable b to be any vector, that forces y to be 0)
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