# Check if the matrix belongs to a specific set

#### diogomgf

##### Junior Member

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

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

Is that system consistent?

#### pka

##### Elite Member

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

##### Elite Member
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

##### Junior Member
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

##### Elite Member
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

##### Junior Member
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

##### Elite Member
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

##### Junior Member
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

#### Otis

##### Senior Member
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

##### Elite Member
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

##### Junior Member
@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.