System of equations / matrices

[Mic :)]

Find all values of h and k such that the system has no solutions, a unique solution, infinitely many solutions

hx +6y =2
x (h+1)y =2k

I can't seem to augment the matrix. Am I allowed to multiply buy variables h / k?

I can find the determinant: h^2 +h -6

Then make it equal to 0 and solve; h = 3 and -2

I'm not entirely sure how that corresponds to the question.

Thanks a lot for any help; it's a little urgent.
:)

 

[BruceW]

how did you get that value of the determinant? I get something different. firstly, it is best to actually write down the matrix equation, since I'm guessing that is where you got the value of the determinant from.

 

[Mic :)]

how did you get that value of the determinant? I get something different. firstly, it is best to actually write down the matrix equation, since I'm guessing that is where you got the value of the determinant from.

h,6 II 2
1, (h+1) II 2k

I swapped the rows when I found mine, but if you don't you get: 6 -h^2-h
Unless I'm doing something terribly wrong.

 

[Mic :)]

Is it possible to apply gaussian elemination?

 

[ehild]

Is it possible to apply gaussian elemination?

Yes. Do it.

 

[BruceW]

h,6 II 2
1, (h+1) II 2k

I swapped the rows when I found mine, but if you don't you get: 6 -h^2-h
Unless I'm doing something terribly wrong.

Oh, right. I didn't see the y in the matrix when I looked at it before. Was it always there? I guess maybe I just didn't notice it. Anyway, yes I agree that you want h²+h-6 equal to zero. But are the solutions really 3 and -2 ?

I'm not entirely sure how that corresponds to the question.

Think about it case by case. What is possible when the determinant is nonzero, and what is possible when the determinant is zero?

 

[Mic :)]

Yes. Do it.

I don't seem to be able to. Is it possible to multiply a line by 1/h?

 

[ehild]

I don't seem to be able to. Is it possible to multiply a line by 1/h?

It is possible if h is not zero. But you can interchange row 1 with row 2.

 

[Mic :)]

It is possible if h is not zero. But you can interchange row 1 with row 2.

So I have swapped the rows and made the bottom left = 0

1, h+1 II 2k
0, 6/h -h -1 II 2/h -2k

Am I on the right tracks?

 

[ehild]

Do not divide by h. Subtract h times of the first line from the second one. You get 0, 6-h(h+1)||2-2kh

 

[Mic :)]

Do not divide by h. Subtract h times of the first line from the second one. You get 0, 6-h(h+1)||2-2kh

Should I attempt to solve 6-h(h+1) = 2-2kh?
(I arrived at this earlier)

 

[Mark44]

Should I attempt to solve 6-h(h+1) = 2-2kh?
(I arrived at this earlier)

The problem with working with matrices and augmented matrices is that it's too easy to lose sight of what the matrices represent. If your bottom row is
$\begin{bmatrix} 0 & h^2 + h - 6 & | & 2kh - 2 \end{bmatrix}$
what equation does this represent?

In light of your answer, would it be a good idea to set h² + h - 6 equal to 2kh - 2?

 

[Mic :)]

The problem with working with matrices and augmented matrices is that it's too easy to lose sight of what the matrices represent. If your bottom row is
$\begin{bmatrix} 0 & h^2 + h - 6 & | & 2kh - 2 \end{bmatrix}$
what equation does this represent?

In light of your answer, would it be a good idea to set h² + h - 6 equal to 2kh - 2?

When the x coeff. is 0, the y coeff. (h^2 + h - 6) is = to 2kh -2?

Probably not but I seem a little lost as to what needs to be done.

 

[Mark44]

When the x coeff. is 0, the y coeff. (h^2 + h - 6) is = to 2kh -2?

No. The coefficient of y is h² + h - 6. What equation does this represent?

 

[Mic :)]

No. The coefficient of y is h² + h - 6. What equation does this represent?

(h^2 + h - 6)y = 2kh -2?

 

[Mark44]

(h^2 + h - 6)y = 2kh -2?

Yes. Can you always solve this equation for y?

 

[Mic :)]

Yes. Can you always solve this equation for y?

Not if h = 3 or -2?

 

[Mark44]

OK, if h

Not if h = 3 or -2?

As was mentioned way back in post #6, the solutions are not h = 3, h = -2. What are the correct values?

Once you get those, how does k enter into things?

 

[Mic :)]

OK, if h

As was mentioned way back in post #6, the solutions are not h = 3, h = -2. What are the correct values?

Once you get those, how does k enter into things?

h=2, -3?

Sub the values into 1, h+1 II 2k?

Sorry about this, my head's not functioning.

 

[Mark44]

h=2, -3?

Sub the values into 1, h+1 II 2k?

How? That's not an equation.

Sorry about this, my head's not functioning.

You can't do this stuff if your head's not functioning. Maybe you should take a break and come back when it is functioning.

This is the equation.

(h^2 + h - 6)y = 2kh -2?

 

[Mic :)]

How? That's not an equation.You can't do this stuff if your head's not functioning. Maybe you should take a break and come back when it is functioning.

This is the equation.

Sub in 2 / -3 into the equation,

Find k.

k = 0.5 for 2
k= -1/3 for -3

So those are the values of h and k for which there are no solutions?

How would I go about this for a unique solution / infinitely many?

 

[Mark44]

Sub in 2 / -3 into the equation,

Find k.

k = 0.5 for 2
k= -1/3 for -3

So those are the values of h and k for which there are no solutions?

How would I go about this for a unique solution / infinitely many?

Write the equation for each case.
If h = 2 and k = 1/2, what is the equation?
If h = -3 and k = -1/3, what is the equation?
In each case, is there a) no solution, b) exactly one solution, c) infinite number of solutions?

After you do that, look at the situation when h ≠ 2 and h ≠ -3, still working with your equation. What value(s) of k give you a) no solution, b) exactly one solution, c) infinite number of solutions?

 

[Mic :)]

Do we use x + (h+1) = 2k to find infinitely many?
when h = -1 and k = any real no.?

 

[Mark44]

Do we use x + (h+1) = 2k to find infinitely many?
when h = -1 and k = any real no.?

No, this is "the equation."
(h^2 + h - 6)y = 2kh -2

Please answer the questions in post #22.

 

[Mic :)]

Write the equation for each case.
If h = 2 and k = 1/2, what is the equation?
If h = -3 and k = -1/3, what is the equation?
In each case, is there a) no solution, b) exactly one solution, c) infinite number of solutions?

After you do that, look at the situation when h ≠ 2 and h ≠ -3, still working with your equation. What value(s) of k give you a) no solution, b) exactly one solution, c) infinite number of solutions?

If h = 2 and k = 1/2,
If h = -3 and k = -1/3

There are no solutions for each case ( 0=0), so they are the values for a)?

If h or k = 0, there is one solution (h can be any value if k is 0?); b)?

Do we have c) when the left side = 0 or simply for h ≠ 2 and h ≠ -3?

 

[Mark44]

If h = 2 and k = 1/2,
If h = -3 and k = -1/3

There are no solutions for each case ( 0=0), so they are the values for a)?

For these two cases (h = 2, k = 1/2 and h = -3, k = -1/3) you have essentially one equation, the equation represented by the top row in your augmented matrix. This is either 2x + 6y = 2 or -3x + 6y = 2, either of which is a line.
The other equation, 0x + 0y = 0, is always true, no matter what x and y are.
Since you really have only one equation that represents two different lines, every point on either line is a solution to the system of equations.

Does that suggest that there are no solutions in this case?

If h or k = 0, there is one solution (h can be any value if k is 0?); b)?

All you need to start with is to say that h ≠ 2 and h ≠ -3. What is the equation in this case? I'm not looking for answers here - I'm looking for some reasoned evidence that results from working with an equation or two.

Do we have c) when the left side = 0 or simply for h ≠ 2 and h ≠ -3?

It looks like you're just guessing.

 

[Mic :)]

For these two cases (h = 2, k = 1/2 and h = -3, k = -1/3) you have essentially one equation, the equation represented by the top row in your augmented matrix. This is either 2x + 6y = 2 or -3x + 6y = 2, either of which is a line.
The other equation, 0x + 0y = 0, is always true, no matter what x and y are.
Since you really have only one equation that represents two different lines, every point on either line is a solution to the system of equations.

Does that suggest that there are no solutions in this case?

No, it suggests c)

All you need to start with is to say that h ≠ 2 and h ≠ -3. What is the equation in this case? I'm not looking for answers here - I'm looking for some reasoned evidence that results from working with an equation or two.

Is there a single definitive equation or is can we just sub any value of h that isn't 2 / -3? eg, 1 -----> x +6y =2
The equations do not result in solutions.

It looks like you're just guessing.

For h=0, the hx + 6y =2 gives y=1/3,
a unique solution?

 

[Mark44]

Let's back up a bit.
If h = -3 and k = -1/3, the second equation becomes 0x + 0y = 0, so the system represents a single line with an infinite number of solutions.
If h = 2 and k = 1/2, the second equation also becomes 0x + 0y = 0, so the system represents a single line with an infinite number of solutions.

What if h = -3 and k ≠ -1/3?
What does the equation look like then?

What if h = 2 and k ≠ 1/2.
What does the equation look like then?

After you answer these, we can look at the case when h ≠ -3 and h ≠ 2.

 

[Mic :)]

What if h = 2 and k ≠ 1/2. What if h = -3 and k ≠ -1/3?

0= a value
Which is not possible.

 

[Mark44]

What if h = 2 and k ≠ 1/2. What if h = -3 and k ≠ -1/3?

0= a value
Which is not possible.

OK. Can you summarize, in complete sentences, what you have found so far? Each sentence should show the specific values of h and k and should say whether there are a) no solutions, b) a unique solution, c) an infinite number of solutions.

Note that we still have some work to do, but I'm checking to see if you understand where we've gotten to at this point.