Linear Algebra - Matrix Equations

In summary, the conversation discusses a linear algebra problem involving matrices and vectors. The goal is to prove that the matrix equation Ax=2v has a solution. The conversation explores different approaches to solving this problem, including using specific numbers and using variables. It is eventually determined that the solution is simply x=2u, given that A(u)=v and A(2u)=2v. Another question is then raised about determining what values of h would make v3 in the span of v1 and v2, and it is clarified that this involves solving for three variables using three linear equations.
  • #1
Jet1045
49
0

Homework Statement



Let A be a matrix of size 3x3. Let u and v be vectors in R3, such that Au = v. Prove that the matrix equation Ax=2v has a solution.

Homework Equations


The Attempt at a Solution



Alright, so this is for an intro linear algebra course, so the solution shouldn't be too coomplicated. But what I am confused about, is can i make A equal a matrix, and then solve this problem with actual numbers? Or to prove this, should i be using variables instead?
cause what i did was made:
A =
1 2 3
1 2 3
1 2 3

And then i mad u=
3
2
1

Then i multiplied A by u to get v, and got v=
10
10
10

Then i moved onto proving that Ax=2v has a solution.
So i wrote:
[1 2 3] [x1] = 20
[1 2 3] [x2] = 20
[1 2 3] [x3] = 20
Then i wrote out the matrix:
1 2 3 20
1 2 3 20
1 2 3 20

Passing to RREF i of course got
1 2 3 20
0 0 0 0
0 0 0 0

Which would therefore indicate that with the A and u that i picked, and the other info given that Ax = 2v has a solution because my RREF has 1 basic variable and 2 free variables, with no rows being 0 0 0 ≠ 0.

BUT i just don't know if i should be somehow proving this with variables instead of numbers, cause i see him saying, 'would this be true for a different A and u and v?'

Any help would be appreciated. :) THANKS!
 
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  • #2
You are way off down the wrong track. If A(u)=v, what's A(2u)? Just using the properties of linear transformations?
 
  • #3
Dick said:
You are way off down the wrong track. If A(u)=v, what's A(2u)? Just using the properties of linear transformations?

ahhh! my midterm is this friday and I am trying to do some practice now, and I am not getting anything :(

A(2u) = 2v right?
So when a Ax = 2v, does that mean x is 2u?
 
  • #4
Jet1045 said:
ahhh! my midterm is this friday and I am trying to do some practice now, and I am not getting anything :(

A(2u) = 2v right?
So when a Ax = 2v, does that mean x is 2u?

It means 2u is a solution to Ax=2v, sure. It might have more solutions depending on the matrix A, but you only need one.
 
  • #5
Dick said:
It means 2u is a solution to Ax=2v, sure. It might have more solutions depending on the matrix A, but you only need one.

Ohhh k, that makes sense. But i don't get how to prove it though?
i just say that since A(u) = v and A(2u) = 2v , Ax = 2v has a solution
 
  • #6
"for all y, there is a x=? such that f(x)=y" means f has a solution for every y.
"given y, there is an x=? such that f(x)=y" means f has a solution for that y.

you were only given one such y.
 
  • #7
xaos said:
"for all y, there is a x=? such that f(x)=y" means f has a solution for every y.
"given y, there is an x=? such that f(x)=y" means f has a solution for that y.

you were only given one such y.

what y was i given? the u? :/
 
  • #8
Jet1045 said:
what y was i given? the u? :/

If A(2u)=2v, which it does, given A(u)=v, then A(x)=2v has the solution x=2u. It's that simple. You are done with this one. As you suggested, it's not complicated. If you have a midterm on Friday, I'd suggest moving on to the next one.
 
  • #9
Dick said:
If A(2u)=2v, which it does, given A(u)=v, then A(x)=2v has the solution x=2u. It's that simple. You are done with this one. As you suggested, it's not complicated. If you have a midterm on Friday, I'd suggest moving on to the next one.

Oh wow, i totally over thought this one. Thanks so much. He gives these practice questions, then puts up the answer keys with ow many marks they'd be out of, and usually his qustions are worth like 10-20 marks, so i assumed there'd be more work to it.

I may as well ask another question here instead of making a new topic.

For what values of h is v3 in Span(v1, v2) where
v1 = [1 -5 3]
v2 = [-2 10 6]
v3 = [2 -9 h]
(of course those are supposed to be written vertically, i just can't do that haha)

K so i wrote out the matrix and began passing to REF.
and i came to here:
1 -2 2
0 0 1
0 12 -6+h

The second row has the form 0 0 ≠ 0, so doesn't that mean this system isn't going to have a solution? Where did i go wrong?
 
  • #10
Jet1045 said:
Oh wow, i totally over thought this one. Thanks so much. He gives these practice questions, then puts up the answer keys with ow many marks they'd be out of, and usually his qustions are worth like 10-20 marks, so i assumed there'd be more work to it.

I may as well ask another question here instead of making a new topic.

For what values of h is v3 in Span(v1, v2) where
v1 = [1 -5 3]
v2 = [-2 10 6]
v3 = [2 -9 h]
(of course those are supposed to be written vertically, i just can't do that haha)

K so i wrote out the matrix and began passing to REF.
and i came to here:
1 -2 2
0 0 1
0 12 -6+h

The second row has the form 0 0 ≠ 0, so doesn't that mean this system isn't going to have a solution? Where did i go wrong?

You are going wrong because you don't really understand the problem you are supposed to solve. If v3 is in the span of v1 and v2 then a*v1+b*v2=v3. If you split that into components then you have three linear equations for the three variables a, b and h. Solve them. Don't just blindly REF. You could also do this with a determinant, but it's better to understand the basics first.
 
  • #11
Dick said:
You are going wrong because you don't really understand the problem you are supposed to solve. If v3 is in the span of v1 and v2 then a*v1+b*v2=v3. If you split that into components then you have three linear equations for the three variables a, b and h. Solve them. Don't just blindly REF. You could also do this with a determinant, but it's better to understand the basics first.


Oh.. k yeah we haven't done the determinant yet so ill do the basic way.
So the three linear equations with the three variables would be the 3 rows in my augmented matrix correct? well like those would be the coefficients of the vairables. When you say solve them, do you mean solve each row as its own equation? I am a little confused, cause the only way i would think of solving equations is doing ref, that's why i always get confused with how to do questions cause i just assume i need to do ref :/
 
  • #12
Jet1045 said:
Oh.. k yeah we haven't done the determinant yet so ill do the basic way.
So the three linear equations with the three variables would be the 3 rows in my augmented matrix correct? well like those would be the coefficients of the vairables. When you say solve them, do you mean solve each row as its own equation? I am a little confused, cause the only way i would think of solving equations is doing ref, that's why i always get confused with how to do questions cause i just assume i need to do ref :/

Write down the equations you have to solve first.
 

Related to Linear Algebra - Matrix Equations

1. What is a matrix equation?

A matrix equation is a mathematical expression that uses matrices to represent a system of linear equations. It is written in the form Ax = b, where A is a coefficient matrix, x is a variable matrix, and b is a constant matrix.

2. How is linear algebra used to solve matrix equations?

Linear algebra techniques, such as Gaussian elimination and matrix operations, are used to manipulate matrices in order to solve for the variables in a matrix equation. These methods involve transforming the equation into an equivalent form until the values of the variables are determined.

3. What are the different types of solutions for a matrix equation?

There are three types of solutions for a matrix equation: unique, infinite, and no solution. A unique solution exists when there is a single set of values for the variables that satisfies the equation. An infinite solution occurs when there are infinitely many solutions that satisfy the equation. A no solution occurs when there are no values for the variables that satisfy the equation.

4. Can a matrix equation have more than one solution?

Yes, a matrix equation can have more than one solution. This occurs when the equation has infinitely many solutions, also known as an infinite solution. In this case, there are multiple sets of values for the variables that satisfy the equation.

5. How is linear algebra used in real-world applications of matrix equations?

Linear algebra and matrix equations are used in a variety of real-world applications, such as engineering, physics, economics, and computer science. They are used to model and solve complex systems and problems, such as calculating electrical currents in a circuit, predicting stock market trends, and creating computer graphics. They are also used in data analysis and machine learning algorithms.

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