Linear algebra - matrix equations

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[SOLVED] linear algebra - matrix equations

Homework Statement


Consider the matrix A =
1 1 1
2 1 0
1 0 -1

a.)
Show that the equation Ax =
1
2
3
has no solutions in R^3, where R is the set of real numbers.

b.) Find two linearly independant vectors b in R^3 such that the equations Ax=b has a solution in R^3.

Homework Equations





The Attempt at a Solution



For part a.) i inserted a vector x of the form
a
b
c
into the equation and then multiplied this x by A. I then had 3 equations in a b and c:

(1) a + b + c = 1
(2) 2a + b =2
(3) a - c = 3

Rearranging (3) i get c = a - 3 and substituting this into (1), it becomes 2a + b = 4
But this contradicts (2) so i conclude that the equation has no solution in R^3.

I am stuck on part b.), if i change the original vector(1,2,3) to b equalling (x,y,z) i can simply replace the terms in the original set of euqations.

In doing this i find that 2a+b=x+z meaning that y = x+z

Is this the right way to do this, and how do i find the numbers in vector b? Could i use x = 1, y=3 and z=2 as one vecotr b as this satisfies my equations? The other being the same numbers but each is negative?

Any help is greatly appreciated, thanks
 

Answers and Replies

  • #2
Defennder
Homework Helper
2,591
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It's strange you're not using the techniques of linear algebra to do the problem for a). I suspect that you're probably unfamiliar with the theory behind linear algebra for Ax = b to have solutions.

I don't know how much you have learnt about linear algebra so far but have you learnt anything about column spaces yet? It'll help if you do.
 
  • #3
HallsofIvy
Science Advisor
Homework Helper
41,847
964
The "technique" Defennnder is talking about is, I think, "row reduction" and can be used to solve both of your problems in terms of matrix manipulations. However, what you have done is perfectly correct and is really the "basic" algebra version of those row reductions.

What you are saying, for (b), is that in order to have a solution, your <x, y, z> vector must satisfy y= x+ z. Choose values for two of those and solve for the third.
Could i use x = 1, y=3 and z=2 as one vecotr b as this satisfies my equations? The other being the same numbers but each is negative?
Yes, <1, 3, 2> works- in fact the equation then has an infinite number of solutions. It is also true that <-1, -3, -2> will also work but those two vectors are obviously not independent and you are asked for two linearly independent vectors that have that property.

I like to use "1" and "0" just because they are easy! Since, as you say, z= x+ y, if x= 1 and y= 0, z= 1. <1, 0, 1> is such a vector. If x= 0 and y= 1, then z= 1 so <0, 1, 1> is also such a vector. Can you see that they are obviously independent?
 
  • #4
213
0
The "technique" Defennnder is talking about is, I think, "row reduction" and can be used to solve both of your problems in terms of matrix manipulations. However, what you have done is perfectly correct and is really the "basic" algebra version of those row reductions.

What you are saying, for (b), is that in order to have a solution, your <x, y, z> vector must satisfy y= x+ z. Choose values for two of those and solve for the third.

Yes, <1, 3, 2> works- in fact the equation then has an infinite number of solutions. It is also true that <-1, -3, -2> will also work but those two vectors are obviously not independent and you are asked for two linearly independent vectors that have that property.

I like to use "1" and "0" just because they are easy! Since, as you say, z= x+ y, if x= 1 and y= 0, z= 1. <1, 0, 1> is such a vector. If x= 0 and y= 1, then z= 1 so <0, 1, 1> is also such a vector. Can you see that they are obviously independent?

Thanks for your help guys, ive been looking over my course sylabus and i don't see row reduction in it but i think it is similar to Guassian elimination which is covered in my course. So i have used that to get it to row echelon form to show that in part a.) the matrix is inconsistent.

The row echelon form i get is:
1 0.5 0¦ b2/2
0 1 2 ¦ 2b1-b2
0 0 0 ¦ b1+b3-b2\right][/tex]

By rearranging and substitution i get:
(x1+x2+x3) = b1
(2x1+x2) = b2
(x1-x3) = b3

So using b1+b3 = b2 i see that there are solutions of b as (1, 0, -1), (0,1,1), (1,1,0) and the null space solution (0,0,0).

From this i get solutions of Ax=b for these particular b values as:
x = (-1,2,0), (0,1,-1), (0,1,0) and the null space solution (1,-2,1)

So i am asked for 2 linearly independant solutions. Which ones do i choose, and with each one do i include the nullspace solution as a part of each? Many thanks

p.s i just reread the question and i am only asked to find 2 linearly independant vectors, b, that produce a solution in R^3. So 1,0,-1 and 0,1,1 are independent. I might just go for those. Thanks again
 
Last edited:

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