Choose h and k so that the matrix has a unique solution

[Arnoldjavs3]

Homework Statement

$$
A = \begin{bmatrix}
1 & 2\\
2 & h\\ = k
\end{bmatrix}
$$
Mod note:
Corrected augmented matrix:
$\begin{bmatrix} 1 & 2 & | & 2 \\ 2 & h & | & k \end{bmatrix}$

Homework Equations

The Attempt at a Solution

Ok, so apparently it's a bad idea to bring this into row reduced echelon form. How can I formulate this into having a unique solution then? I nkow that if you want infinite solutions, just make the two rows dependent on each other but I'm not sure how to approach it for a unique solution. (And if we want no solutions, make it so that the system is inconsistent by getting an answer like 0!=1)

What conditions do i need to fulfill here for it to have a unique solution?

 

[member 587159]

Your matrix has a mistake in it. Can you fix it? Can you give the full problem statement? What is the system you are working with? Why is it a bad idea to put it into row echelon form?

 

[Arnoldjavs3]

hmm.. I'm unsure of how to make an augmented matrix in latex.

it's meant to be :
row one: 1x + 2y = 2
row two: 2x + h = k

 

[member 587159]

Is it 2x + h = k or 2x + hy = k?

 

[Arnoldjavs3]

http://prntscr.com/e4kqrc

i'm assuming 2x + hy = k, sorry

 

[member 587159]

Okay. Now, we can get started. Put that matrix in row reduced echelon form. This system has a unique solution if it has rank 2 (why?). Try to find the values for h and k such that the system has rank 2. If you saw determinants, there is an easier method.

 

[Arnoldjavs3]

the determinant is h-4 correct? how does this help me?

 

[Arnoldjavs3]

for row reduced echelon form i got:

x = (2h-2-k)/h
y = (2+k)/h

I believe rank indicates the linearly independent columns in a matrix, in this case rank 2? Do I just input values for h and k so that x = 1 and y = 1?

 

[member 587159]

Yes, you calculated the determinant correctly. There is a theorem that says that this matrix has rank 2 if and only if the determinant is non zero. What can you conclude? Did you see such theorem?

 

[Arnoldjavs3]

I have not seen such a theorem before. But I'm having troubles drawing the connections here - I believe that the two rows must be independent of each other, but they also need to be consistent. How does rank play into this? If it is of rank 2, then they will have two rows/columns that are independent right? But how does this information help me into finding values for h, and k?

 

[member 587159]

If it has rank 2, then the system has a unique solution. Just what you are looking for! Yes, this is equivalent with saying that the rows/colums must be linear independent. By using that, you can find the values of h and k for which the system has a unique solution.

Also, the row reduced form is a matrix. Not another system (yet it corresponds to an equivalent system). I got something though, so you might want to look at that again.

 

[Arnoldjavs3]

hmm... how about x = (2h-2k)/h
and y = k/h. Do i just choose random values?>

 

[Mark44]

hmm.. I'm unsure of how to make an augmented matrix in latex.

I edited your post to add the corrected augmented matrix. You can right click on what I added to see how it's done.

 

[member 587159]

hmm... how about x = (2h-2k)/h
and y = k/h. Do i just choose random values?>

This isn't a guess game! Show your work so we can see where things go wrong.

 

[Arnoldjavs3]

$\begin{bmatrix} 1 & 2 & | & 2 \\ 2 & h & | & k \end{bmatrix}$

$\begin{bmatrix} 1 & 2 & | & 2 \\ 0 & h-4 & | & k-4 \end{bmatrix}$

$\begin{bmatrix} 1 & 2 & | & 2 \\ 0 & 1 & | & (k-4)/(h-4) \end{bmatrix}$

$\begin{bmatrix} 1 & 0 & | & 2 - ((2k-8)(h-4)) \\ 0 & 1 & | & (k-4)/(h-4) \end{bmatrix}$

and as such i got h-4 = 2h-2k

and the other equation gets h-4 = k-4
so h=k and then substituting that in equation one, i get h=4.

I think this calls for infinite solutoins and not a unique one though

 

[member 587159]

$\begin{bmatrix} 1 & 2 & | & 2 \\ 2 & h & | & k \end{bmatrix}$

$\begin{bmatrix} 1 & 2 & | & 2 \\ 0 & h-4 & | & k-4 \end{bmatrix}$

$\begin{bmatrix} 1 & 2 & | & 2 \\ 0 & 1 & | & (k-4)/(h-4) \end{bmatrix}$

$\begin{bmatrix} 1 & 0 & | & 2 - ((2k-8)(h-4)) \\ 0 & 1 & | & (k-4)/(h-4) \end{bmatrix}$

and as such i got h-4 = 2h-2k

and the other equation gets h-4 = k-4
so h=k and then substituting that in equation one, i get h=4.

I think this calls for infinite solutoins and not a unique one though

Well done! That seems correct. Now look at the second step you did:

$\begin{bmatrix} 1 & 2 & | & 2 \\ 0 & h-4 & | & k-4 \end{bmatrix}$

From here, you have enough information to solve the problem. What happens when $h = 4, k = 4, h \neq 4, k \neq 4$? It is clear that if $h = 4$, we either have an infinite amount of solutions ($k = 4$) or the system is not consistent ($k \neq 4$). So we deduce that $h \neq 4$. Does the value of k matter?

 

[Arnoldjavs3]

Well done! That seems correct. Now look at the second step you did:

$\begin{bmatrix} 1 & 2 & | & 2 \\ 0 & h-4 & | & k-4 \end{bmatrix}$

From here, you have enough information to solve the problem. What happens when $h = 4, k = 4, h \neq 4, k \neq 4$? It is clear that if $h = 4$, we either have an infinite amount of solutions ($k = 4$) or the system is not consistent ($k \neq 4$). So we deduce that $h \neq 4$. Does the value of k matter?

Just a question - does k still have to correspond to what h is(it should right - because it would be inconsistent otherwise)? Its bothering me that I can't algebraically find specific, unique values. What if h was 3?
-y = k - 4 - > would k have to be 3?

 

[member 587159]

Just a question - does k still have to correspond to what h is(it should right - because it would be inconsistent otherwise)? Its bothering me that I can't algebraically find specific, unique values. What if h was 3?
-y = k - 4 - > would k have to be 3?

You are finding the values of h and k such that the system has a unique solution. The only restriction on h is that it can't be equal to 4. Thus $h \in \mathbb{R} - \{4\}$. We can solve the system then, and we deduce that the value of k won't affect whether the system has a unique solution. So $k \in \mathbb{R}$.

Conclusion: The system has a unique solution if:

$h \in \mathbb{R} - \{4\}$
$k \in \mathbb{R}$.

As for your question what happens when h = 3? Try it out! You will find a unique solution for the system. The same goes when h is any real number not equal to 4.

 

[Ray Vickson]

I have not seen such a theorem before. But I'm having troubles drawing the connections here - I believe that the two rows must be independent of each other, but they also need to be consistent. How does rank play into this? If it is of rank 2, then they will have two rows/columns that are independent right? But how does this information help me into finding values for h, and k?

Whether the rows are independent refers only to the left-hand sides of the equations, so depends only on $h.$ If the two rows on the left are linearly independent, it does not matter what you have on the right, so instead of $2$ and $k$ you could have any $a$ and $b$; the solution would still be unique, but of course, would vary as for different $a,b$.

If the two rows on the left are linearly dependent, the equations are consistent for some right-hand-sides $a,b$ (so, for some value of $k$ in your case), but the solution would NOT be unique. For other right-hand sides the equations would be inconsistent, having no solutions at all.

 

[Arnoldjavs3]

Whether the rows are independent refers only to the left-hand sides of the equations, so depends only on $h.$ If the two rows on the left are linearly independent, it does not matter what you have on the right, so instead of $2$ and $k$ you could have any $a$ and $b$; the solution would still be unique, but of course, would vary as for different $a,b$.

If the two rows on the left are linearly dependent, the equations are consistent for some right-hand-sides $a,b$ (so, for some value of $k$ in your case), but the solution would NOT be unique. For other right-hand sides the equations would be inconsistent, having no solutions at all.

Yeah this makes a lot more sense now. I didn't understand what made rows linear independent, I wish my textbook had gone over this. I was wondering why it had to be h that couldn't be 4, and not k but this clarifies that.

 

[Arnoldjavs3]

You are finding the values of h and k such that the system has a unique solution. The only restriction on h is that it can't be equal to 4. Thus $h \in \mathbb{R} - \{4\}$. We can solve the system then, and we deduce that the value of k won't affect whether the system has a unique solution. So $k \in \mathbb{R}$.

Conclusion: The system has a unique solution if:

$h \in \mathbb{R} - \{4\}$
$k \in \mathbb{R}$.

As for your question what happens when h = 3? Try it out! You will find a unique solution for the system. The same goes when h is any real number not equal to 4.

Hi guys, I'm coming back to this problem and I wanted to know why h can't be 4 in this scenario. Is it because the second row will be 0,0? Why does that make this not a unique solution?

 

[member 587159]

Hi guys, I'm coming back to this problem and I wanted to know why h can't be 4 in this scenario. Is it because the second row will be 0,0? Why does that make this not a unique solution?

If $h = 4$, and we have $k \neq 4$

We have that $0x + 0y = c$ for a certain real number $c$ that is non zero. Hence, the system has no solution.

If $h = 4$, and we have $k = 4$

Then, we find that $0x + 0y = 0$. This means, we can leave this equation out of the system as any $(x,y)$ is suited. Thus, we end up with a system in 2 unknowns and 1 equation, thus, we have an infinite amount of solutions (so not a unique one, that's for sure)