Check if the matrix belongs to a specific set

diogomgf

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I don't know where to start with this exercise:

Check if \(\displaystyle u\) belongs to the set \(\displaystyle C\):
a. \(\displaystyle u = \begin{bmatrix} -1\\ 2 \\ 1 \end{bmatrix}\) and \(\displaystyle C = \left \{ a\begin{bmatrix} -2\\ 1 \\ 0 \end{bmatrix} + b\begin{bmatrix} 3\\ 0 \\ 1 \end{bmatrix} : a,b \in R \right \}\)

b. \(\displaystyle u = \left ( 1,0,1 \right )\) and \(\displaystyle C = \left \{ a_{1}(2,0,1) + a_{2}(-3,-1,1) : a_{1}, a_{2} \in R \right \}\)
 

Otis

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Hello diogomgf. For part (a), I would think about the system of three equations that is C = u.

Is that system consistent?

😎
 

pka

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I don't know where to start with this exercise:

Check if \(\displaystyle u\) belongs to the set \(\displaystyle C\):
a. \(\displaystyle u = \begin{bmatrix} -1\\ 2 \\ 1 \end{bmatrix}\) and \(\displaystyle C = \left \{ a\begin{bmatrix} -2\\ 1 \\ 0 \end{bmatrix} + b\begin{bmatrix} 3\\ 0 \\ 1 \end{bmatrix} : a,b \in R \right \}\)

b. \(\displaystyle u = \left ( 1,0,1 \right )\) and \(\displaystyle C = \left \{ a_{1}(2,0,1) + a_{2}(-3,-1,1) : a_{1}, a_{2} \in R \right \}\)
a) Note that the \(y\)-coordinate of \(\vec{\bf{u}}\) tells us that \({\bf a}=2\)
and that the \(z\)-coordinate of \(\vec{\bf{u}}\) tells us that \({\bf b}=1\).
Then use those two values to verify the \(x\)-coordinate of of \(\vec{\bf{u}}\).
Now you do part b) and post the result.
 

Jomo

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You need to understand what C means. In the fist part, C contains all the linear combinations of (-2, 1, 0) and (3, 0, 1).

Now u = (-1, 2, 1) is in C if u can be expressed as a linear combination of (-2, 1, 0) and (3, 0, 1).

Since the 3rd component of (-2, 1, 0) we know that we need to multiply (3, 0, 1) by 1 to get the 3rd entry of u = (-1, 2, 1) which is 1. This is exactly what pka said.

Now can you find the value a to multiply (-2, 1, 0) by so that u = (-1, 2, 1) = a(-2, 1, 0) + 1(3, 0, 1).

If yes, then u is in C. If not, then u is not in C.
 

diogomgf

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You need to understand what C means. In the fist part, C contains all the linear combinations of (-2, 1, 0) and (3, 0, 1).

Now u = (-1, 2, 1) is in C if u can be expressed as a linear combination of (-2, 1, 0) and (3, 0, 1).

Since the 3rd component of (-2, 1, 0) we know that we need to multiply (3, 0, 1) by 1 to get the 3rd entry of u = (-1, 2, 1) which is 1. This is exactly what pka said.

Now can you find the value a to multiply (-2, 1, 0) by so that u = (-1, 2, 1) = a(-2, 1, 0) + 1(3, 0, 1).

If yes, then u is in C. If not, then u is not in C.
a) Note that the \(y\)-coordinate of \(\vec{\bf{u}}\) tells us that \({\bf a}=2\)
and that the \(z\)-coordinate of \(\vec{\bf{u}}\) tells us that \({\bf b}=1\).
Then use those two values to verify the \(x\)-coordinate of of \(\vec{\bf{u}}\).
Now you do part b) and post the result.
I have not studied linear combinations yet, though I have seen some videos on the subject so I understand what you are saying.


Hello diogomgf. For part (a), I would think about the system of three equations that is C = u.

Is that system consistent?

😎
I followed your suggestion along with gauss-jordan elimination and solved the problem.
What would happen though If the system (a) was possible but undefined? Would \(\displaystyle u\) still belong to \(\displaystyle C\)?
 

Jomo

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I have not studied linear combinations yet, though I have seen some videos on the subject so I understand what you are saying.




I followed your suggestion along with gauss-jordan elimination and solved the problem.
What would happen though If the system (a) was possible but undefined? Would \(\displaystyle u\) still belong to \(\displaystyle C\)?
What does the system (a) mean. What does it mean if a system is undefined but possible??

I assume that you mean that if there are no a and b that satisfies the equation, that is the system is undefined, then it is possible.

What is it? Do you mean that u will still be in C? I thought that C is defined to only contain linear combinations of (-2, 1, 0) and (3, 0, 1).

Why did you use gauss-jordan elimination? Usually that is the way to go but not in this case. The possible answer is staring you in the face.

Ignoring the 1st components for a moment we want to solve ( 2, 1) = a(1, 0) + b( 0, 1). Notice that no matter what a is, a(1,0) will always have a 0 in the last position. So the only way to get the 1 in (2, 1) will be if b=1.

A similar thing is true for the 1st component. b(0,1) will not change that 0. So a will have to be 2 as 2(1,0) = (2,0).

This means that the only possible solution is a=2 and b=1. It just remains to verify if these two values also satisfies the 1st components. That is does (-1, 2, 1) = 2(-2, 1, 0) + 1(3, 0, 1)? All that needs to be checked is if (-1) = 2(-2) + 1 (3) and it does!
 

diogomgf

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Why did you use gauss-jordan elimination? Usually that is the way to go but not in this case. The possible answer is staring you in the face.

Ignoring the 1st components for a moment we want to solve ( 2, 1) = a(1, 0) + b( 0, 1). Notice that no matter what a is, a(1,0) will always have a 0 in the last position. So the only way to get the 1 in (2, 1) will be if b=1.

A similar thing is true for the 1st component. b(0,1) will not change that 0. So a will have to be 2 as 2(1,0) = (2,0).

This means that the only possible solution is a=2 and b=1. It just remains to verify if these two values also satisfies the 1st components. That is does (-1, 2, 1) = 2(-2, 1, 0) + 1(3, 0, 1)? All that needs to be checked is if (-1) = 2(-2) + 1 (3) and it does!
What you are doing there is simply solving a system of linear equations... wich can be done with gauss-jordan elimination and thats what I did...

What does the system (a) mean. What does it mean if a system is undefined but possible??

I assume that you mean that if there are no a and b that satisfies the equation, that is the system is undefined, then it is possible.

What is it? Do you mean that u will still be in C? I thought that C is defined to only contain linear combinations of (-2, 1, 0) and (3, 0, 1).
An undefined but possible system is one where there are more than 1 set of solutions... e.g If \(\displaystyle a = x_{1} + b\) and \(\displaystyle b = x_{2}\)

EDIT: typo.
 

Jomo

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Yes, maybe what I am doing is solving a system of equations which can be done using gauss-jordan elimination.
The point is that you can just read of the answer using my method.

You should not call a system undefined but possible.
As my method showed there is just one solution. Using gauss-jordan should have produced the same results.
 

diogomgf

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You should not call a system undefined but possible.
I am using the literal translation from the expression used here in Portugal... Maybe it is wrong?

As my method showed there is just one solution. Using gauss-jordan should have produced the same results.
It did :LOL:
 

Otis

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I followed your suggestion [to look at the system of equations] along with gauss-jordan elimination …
That's fine, diogomgf. Unless exercise instructions say otherwise, students are free to solve problems using any skills they know.

I actually like the mechanics of row reduction, yet I would have used the substitution method (in the first equation) because inspection shows the value of each parameter as "given".

Code:
-2a + 3b = -1
  a      =  2
       b =  1
😎
 

Jomo

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I agree with Otis that using any method is fine. And believe it or not I do support allowing a student to use any method that they like. I also like to get the student to think beyond what their teacher teaches them. I only pushed the OP to try an alternative method to get the OP to think a bit.
 

diogomgf

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@Jomo @Otis
Usually there are many ways to solve the same problem... I just try to use the subject I'm studying to solve the problems posted in that area of the book.
At the moment I'm learning gauss-jordan elimination so I try to stick to that.
 
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