How do I solve for matrix A in the equation y=mx+c?

[uperkurk]

So I have these two straight line equations:

y = -x + 1, y = 2x + 1 So now I have a single point of intersection of (1y,0x)

But the next part of the question is what confuses me but I've included a picture as it's just easier... would someone please be able to take the time to explain each step in the question?

I can do parts B and C but A is confusing me...

 

[Diffy]

This would probably get a better response in the homework question forums.

 

[berkeman]

This would probably get a better response in the homework question forums.

Thread moved.

 

[LCKurtz]

So I have these two straight line equations:

y = -x + 1, y = 2x + 1 So now I have a single point of intersection of (1y,0x)

But the next part of the question is what confuses me but I've included a picture as it's just easier... would someone please be able to take the time to explain each step in the question?

I can do parts B and C but A is confusing me...

Write your two equations like this:
$x+y=1$
$-2x+y=1$
Then put it in the form:$$
\left[ \begin{array}{cc}
? & ?\\ ? & ?
\end{array}\right]
\left[ \begin{array}{c}
x\\y
\end{array}\right] = \left[ \begin{array}{c}
?\\?
\end{array}\right]$$

 

[uperkurk]

Write your two equations like this:
$x+y=1$
$-2x+y=1$
Then put it in the form:$$
\left[ \begin{array}{cc}
? & ?\\ ? & ?
\end{array}\right]
\left[ \begin{array}{c}
x\\y
\end{array}\right] = \left[ \begin{array}{c}
?\\?
\end{array}\right]$$

\left[ \begin{array}{cc}<br /> 1 &amp; 1\\ -2 &amp; 1<br /> \end{array}\right]<br /> \left[ \begin{array}{c}<br /> x\\y<br /> \end{array}\right] = \left[ \begin{array}{c}<br /> 1\\1<br /> \end{array}\right]

That doesn't even look remotely correct :/

 

[HallsofIvy]

Why do you say that? We had kind of assumed that, since you talked about matrices and vectors, you knew what they are and knew how to multiply a matrix and a vector. is that not true?

Can you see that the matrix multiplication gives
\begin{bmatrix}1 &amp; 1 \\-2 &amp; 1\end{bmatrix}\begin{bmatrix}x \\ y\end{bmatrix}= \begin{bmatrix}x+ y\\ -2x+ y\end{bmatrix}= \begin{bmatrix}1 \\ 1\end{bmatrix}
and that two vectors are equal if and only if their corresponding components are equal.

That is, that matrix-vector equation is exactly the same as saying that x+ y= 1 and -2x+ y= 1.

 

[uperkurk]

Why do you say that? We had kind of assumed that, since you talked about matrices and vectors, you knew what they are and knew how to multiply a matrix and a vector. is that not true?

Can you see that the matrix multiplication gives
\begin{bmatrix}1 &amp; 1 \\-2 &amp; 1\end{bmatrix}\begin{bmatrix}x \\ y\end{bmatrix}= \begin{bmatrix}x+ y\\ -2x+ y\end{bmatrix}= \begin{bmatrix}1 \\ 1\end{bmatrix}
and that two vectors are equal if and only if their corresponding components are equal.

That is, that matrix-vector equation is exactly the same as saying that x+ y= 1 and -2x+ y= 1.

I've multiplied matrices together and I know how to do them but usually I get given them in the form

\begin{bmatrix}1 &amp; 1 \\-2 &amp; 1\end{bmatrix}\begin{bmatrix}2 &amp; 4 \\ 3 &amp; 1\end{bmatrix} For example and then you just multiply the 2 matrices together. But I understand how you explained it thanks.

 

[uperkurk]

OK I've got to the last stage, I've worked out the matrix inverse which is \begin{bmatrix}0.333 &amp; -0.333 \\0.666 &amp; 0.333\end{bmatrix} and I've verified it using the Identity matrix \begin{bmatrix}1 &amp; 0 \\0 &amp; 1\end{bmatrix} so now I just have to plot the 2 straight lines but I'm kinda lost in my own work now...

Am I plotting the original 2 straight lines using the equations or am I plotting them via the matrix and if so how do I do that?

EDIT: It says using the matrix inverse but I'm not sure how to do it

 

[rock.freak667]

You changed your equations into a matrix form of Ax=B

so if you pre-multiply everything by A^-1 you will have

A^-1Ax = A^-1B

And you worked out what A^-1A was. So just work out the right side of the equation.

 

[LCKurtz]

OK I've got to the last stage, I've worked out the matrix inverse which is \begin{bmatrix}0.333 &amp; -0.333 \\0.666 &amp; 0.333\end{bmatrix} and I've verified it using the Identity matrix \begin{bmatrix}1 &amp; 0 \\0 &amp; 1\end{bmatrix}

Get rid of the decimals. That is just an approximation to the inverse and it won't give you the identity when multiplying. There is absolutely no reason to use decimal approximations instead of exact fractional values in this problem.

 

[uperkurk]

I'm still stuck on this question, I kinda just forgot about it but I really need all the marks I can get, can someone please tell me how to plot the point of intersection of the 2 lines using the inverse matrix?

It is worth 5% so I need to know how to do it.

 

[LCKurtz]

I'm still stuck on this question, I kinda just forgot about it but I really need all the marks I can get, can someone please tell me how to plot the point of intersection of the 2 lines using the inverse matrix?

It is worth 5% so I need to know how to do it.

Did you read RockFreak's post #9?

 

[uperkurk]

Yes but I don't understand what he means... I have worked out what A^{-}1 is but I don't know how to plot the matrix elements as points on the graph

 

[rock.freak667]

Yes but I don't understand what he means... I have worked out what A^{-}1 is but I don't know how to plot the matrix elements as points on the graph

There is no need to plot a graph, you just need to calculate A^-1B.

 

[LCKurtz]

Yes but I don't understand what he means... I have worked out what A^{-}1 is but I don't know how to plot the matrix elements as points on the graph

Why don't you just plot the two straight lines the old fashioned way and look at how the intersection point compares with what you get when you solve the matrix equation by multiplying both sides by A^-1?

 

[uperkurk]

What is B though?

 

[LCKurtz]

Look at the matrices in your post #5. Use them.

 

[uperkurk]

2, -1?

 

[LCKurtz]

2, -1?

Do you really have to ask? Does that point work in both lines? If it does you are done, otherwise check your arithmetic.

 

[uperkurk]

I'm am losing my temper with this crap now, I don't know what is right, if you don't know what you're doing then you don't know if the answer is right or not... this is not homework, I am revising for a test so please can you just tell me?

 

[LCKurtz]

I'm sorry you are angry. You have been told more than enough to help you solve this problem. Perhaps if you can't solve it anyway you don't deserve the 5 marks you would get. I'm going to a movie now. Goodbye.

 

[uperkurk]

I am clearly confused, I did most of the entire question by myself and now you won't help me on the last bit? Maybe if you told me the method I could work out the answer... but you have said multiply the inverse of matrix A with B which I thought was

1
1

you're not giving me direction.

 

[rock.freak667]

I am clearly confused, I did most of the entire question by myself and now you won't help me on the last bit? Maybe if you told me the method I could work out the answer... but you have said multiply the inverse of matrix A with B which I thought was

1
1

you're not giving me direction.

\left[ \begin{array}{cc}<br /> 1 &amp; 1\\ -2 &amp; 1<br /> \end{array}\right]<br /> \left[ \begin{array}{c}<br /> x\\y<br /> \end{array}\right] = \left[ \begin{array}{c}<br /> 1\\1<br /> \end{array}\right]


That is the matrix equation which you formed in a previous post. This in the form Ax=B

(start from the left and match each letter to a matrix).

The matrix B is on the right side of the equation, from the equation you formed, what matrix is on the right side? Hence what is the matrix B (you already posted what it was).

You calculated A^-1.

I just told you how to get the matrix B, can you now calculate what A^-1B would be?

 

[uperkurk]

Yes... A^-1b would be a^-1 multiplied by B which gives 2, -1?

If I am calculating it wrong and 2, -1 is not the correct answer then would it be 0, 36?

I am literally multiplying the left hand side by

1
1

which is the right hand side (b)?

 

[Muphrid]

uperkurk, maybe it's time to take a step back and think about what's going on conceptually. You have a matrix equation Ax = b. What you want to find is the vector x, which describes the coordinates of the point where the two lines intersect.

The way you do this is by trying to "get rid of" that A on the left hand side. You do this by multiplying by A^{-1} on the left of both sides of the equation.

\begin{align*}<br /> Ax &amp;= b \\<br /> A^{-1} A x &amp;= A^{-1} b \\<br /> x &amp;= A^{-1} b \end{align*}

When you have calculated A^{-1} b, you have the point of intersection. You can plot both lines and check visually that this point is in the right neighborhood, and you should take both lines, plug in the coordinates, and check that the answers are the same.

Do you understand the basic process that's going on here?

 

[uperkurk]

This is ****ing ridiculous, look back at my first post, I posted the screenshot of the questions, it did not take me very long at all to figure out what needed to be done, but now I am going to be honest.

I have no idea of what you are talking about... whenever I have done y=mx+c equations I have always had 2 sets of coordinates. When you plot those coordinates the lines cross, the exact point at which they cross is the intersection.

The way you are explaining it is that

x = the intersection
A^-1 = the inverse of matrix A
b = the matrix \begin{bmatrix}1 \\ 1\end{bmatrix}

So I am reading that as "the point of intersection of x is found by multiplying the inverse of A with B.

Clearly this is incorrect because I even used a matrix calculator and the answer is most definitely \begin{bmatrix}2 \\ -1\end{bmatrix} or \begin{bmatrix}0 \\ 36\end{bmatrix}

So if you would be so kind as to tell me the answer, explain why the answer is what it is, then I can apply what you have showed me to solve some other problems, you can help some other people and I won't have a headache anymore... I am going to be honest I don't think any amount of typing is going to make me realize what I need to do so...

 

[Muphrid]

You don't seem to be multiplying out A^{-1} b correctly. Show the steps you're using to do this. Matrix multiplication of a matrix and a vector should give you

\begin{bmatrix}<br /> c &amp; d \\<br /> e &amp; f<br /> \end{bmatrix}<br /> \begin{bmatrix}<br /> g \\<br /> h<br /> \end{bmatrix}<br /> =<br /> \begin{bmatrix}<br /> cg + dh \\<br /> eg + fh<br /> \end{bmatrix}<br />

 

[uperkurk]

You don't seem to be multiplying out A^{-1} b correctly. Show the steps you're using to do this. Matrix multiplication of a matrix and a vector should give you

\begin{bmatrix}<br /> c &amp; d \\<br /> e &amp; f<br /> \end{bmatrix}<br /> \begin{bmatrix}<br /> g \\<br /> h<br /> \end{bmatrix}<br /> =<br /> \begin{bmatrix}<br /> cg + dh \\<br /> eg + fh<br /> \end{bmatrix}<br />

\begin{bmatrix}<br /> 1 &amp; 1 \\<br /> -2 &amp; 1<br /> \end{bmatrix}<br /> \begin{bmatrix}<br /> 1 \\<br /> 1<br /> \end{bmatrix}<br /> =<br /> \begin{bmatrix}<br /> 1*1 + 1*1 \\<br /> -2*1 + 1*1<br /> \end{bmatrix}<br />

2, -1 I assure you that is the correct answer and I can bet my life that 3 different websites + the windows calculator are not wrong.

 

[Muphrid]

You multiplied Ab, not A^{-1} b. You found A^{-1} already, remember? It was

A^{-1} = \frac{1}{3} \begin{bmatrix} 1 &amp; -1 \\ 2 &amp; 1 \end{bmatrix}

 

[uperkurk]

\begin{bmatrix} \frac{1}{3} & \frac{1}{3} \\ \frac {-2}{3} & \frac{1}{3} \end{bmatrix} multiplied by matrix B \begin{bmatrix} 1 \\ 1 \end{bmatrix} = \begin{bmatrix} 26 \\ -10 \end{bmatrix} you can check that on the calculator and online matrix calculation websites too...

Someone fix my tex tags please it's screwed up

 

[Muphrid]

You're still using the wrong matrix, man. You're not going to get any good answer when you're multiplying the wrong things. This is like trying to get the answer to (1/5)x4 when you're multiplying 5x4.

 

[uperkurk]

ok it's 2:35am, **** the 5% this really is not worth it, this is my retake which if I fail I will have to resit the year so every % matters but if nobody is going to give me the answer after I've attempted to understand 20+ posts then **** the 5%

Thank for your help anyway, if you wish to provide me with the answer, show me what I missed then I'd be extremely grateful, if not then thank you for your patience anyway

 

[rock.freak667]

ok it's 2:35am, **** the 5% this really is not worth it, this is my retake which if I fail I will have to resit the year so every % matters but if nobody is going to give me the answer after I've attempted to understand 20+ posts then **** the 5%

Thank for your help anyway, if you wish to provide me with the answer, show me what I missed then I'd be extremely grateful, if not then thank you for your patience anyway

Telling you the answer will not help you if in the exam they change the numbers. So far we've told you exactly what to multiply but you keep getting the wrong answer which means that you need to work on multiplying your matrices.

But if you keep getting frustrated, it might be time to step away from it for a while and then come back to it. The more you frustrate yourself, the more mistakes you'll make which will become an endless cycle.


You wrote the correct matrix in post #5 (HallsofIvy explained why it was correct)

You correctly found A^-1 in post #8 (just use the exact numbers and not the decimals)

uperkurk told you what to multiply to calculate A^-1B in post #30.


Just use those posts as your starting points when you come back to it.

 

[uperkurk]

OK I've decided to redo the entire question as I feel like last night I was just very tired. So step by step here is what I have worked out.

1. Given the two straight lines y=-x+1, y=2x+1 formulate the matrix vector equation representing the two lines in the form Ax=B

I rewrote the equations in the form x+y=1, -2+x=1 and put them into the form Ax=B
\left[ \begin{array}{cc}<br /> 1 &amp; 1\\ -2 &amp; 1<br /> \end{array}\right]<br /> \left[ \begin{array}{c}<br /> x\\y<br /> \end{array}\right] = \left[ \begin{array}{c}<br /> 1\\1<br /> \end{array}\right]

2. Find the inverse of the 2 by 2 matrix A obtained obove.

A^-1=\left[ \begin{array}{cc}<br /> 1 &amp; -1\\ 2 &amp; 1<br /> \end{array}\right] because I used the formula supplied to me. But this is not the actual inverse yet, I now must do the ad-bc part which means the real inverse of A is going to be A^-1=\left[ \begin{array}{cc}<br /> \frac{1}{3} &amp; \frac{-1}{3}\\ \frac{2}{3} &amp; \frac{1}{3}<br /> \end{array}\right]

3. Confirm this is the real inverse by showing that A * A^-1= I

So I multiplied A with the inverse of A which produced an identity matrix \left[ \begin{array}{cc}<br /> 1 &amp; 0\\ 0 &amp; 1<br /> \end{array}\right]

4. Using the matrix INVERSE find the POINT of intersection of the two lines.

After re-reading every post given to me, I am still convinced that to find the point of intersection I have to multiply A^-1\left[ \begin{array}{cc}<br /> \frac{1}{3} &amp; \frac{-1}{3}\\ \frac{2}{3} &amp; \frac{1}{3}<br /> \end{array}\right] which is the inverse, by B which is \left[ \begin{array}{cc}<br /> 1 \\ 1<br /> \end{array}\right]

After mutiplying these two matrices by the only way I have been taught, and checking and doubled checking my answer using an online matrix calculator yields the result 0, 36 (check the screenshots)

Now I don't know what more I can do to convince you all that I believe 100% that this is the correct answer, I'm not going to call you all liers I'm sure you know something that I do not.

Is there a special way to multiply matrices that have fractions?
Do I have the correct numbers?
Am I calculating the correct matrices?
Am I completely misunderstanding what a point of intersection is? Do I have to convert something back before doing the calculation?
Where exactly am I going wrong? Why am I going wrong?
Do I have to divide anything or perform some other calculation?

I know that you guys might think the best way for me to learn is to work the answer out by myself, but I think I have put in more than suffient effort myself, I'm attempted for 3 days to understand this question but my brain is not capable of understanding why my calculations are wrong.

I've had it before where I struggle to understand something and then someone has told me the answer, told my why the answer is what it is and where I went wrong and it helped me an insane amount.

But keep telling me "my calculations are wrong" is really really not going to help me.

 

[Muphrid]

You now have the matrices correct, but that calculator is completely wrong. While the x-coordinate is correct, the y coordinate isn't. What's (2/3)x1+(1/3)x1? Do you see whatever mistake there must've been in the multiplication now? This is what you should get for the answer.

I really wanted you to go through and post steps because nothing is remotely close in size to 36 in this problem-- how do you multiply two fractions by 1, add them, and get such a number? So I was skeptical the error would be apparent otherwise. Hopefully this shows you what you need to do.

 

[rock.freak667]

So far you have everything correct except for the final answer. Since you did it by hand and got (0 36), can you please show the exact steps you did?

Because if you notice the fractions you would multiply would not give a large number such as 36. (The online calculator I used does not give 0,36 as well).

 

[uperkurk]

0,1?

 

[LCKurtz]

Why don't you just plot the two straight lines the old fashioned way and look at how the intersection point compares with what you get when you solve the matrix equation by multiplying both sides by A^-1?

0,1?

Did you ever try my suggestion above and actually graph those two lines? Do they intersect at (0,1)?

 

[rock.freak667]

0,1?

A way to verify that is to substitute x= 0 into the equations and see if you get y=1.

 

[Muphrid]

Any answer you get should go into the two equations and yield some true statement like 0=0, 1=1, or such. It's important you know how to check like this on your own. Does your answer check?

Do you understand what you did wrong with the matrix multiplication?

 

[uperkurk]

So 0,1 is correct? Well I would be happy but about 20 posts back I posted the inverse matrix in decimals and one of you said there is no need to change the fractions into decimals just keep them in fractions, so when I worked out it by hand I got 0,1 but when I checked my working out using the online calculator I typed in 1/3, -1/3, 2/3 and 1/3 and the calculator gave me 0, 36 and I can only assume this because the calculator did not know that 1/3 was meant to be a fraction.

If I had typed in 0.333, -0.333, 0.666 and 0.333 into to calculator I would have got 0, 0.99999999 which I would have known to round up.

I would say thanks but if I'm totally honest none of you helped me... When someone is making such a tiny tiny error like I did telling them what error they're making really would help them out.

 

[Muphrid]

No good online calculator should've done that, and at no time did you mention that you had calculated that as an answer. Frankly, I find your over-reliance on a single online tool here crippling. It's okay to use tools, but use more than one and check them against each other. And frankly, none of us could've known that your tool couldn't understand fractions. That's why we ask for steps, not screenshotted computer output. You wanted help. We're not psychic, and people want answers given to them for grades without learning all the time.

You're welcome.

 

[uperkurk]

Without learning? I was merely asking for you to point out what EXACTLY I am doing wrong, but you all just kept beating around the bush...

Useless, but at least now I know to be careful checking my results when the matrix elements are in the form of fractions.

 

[rock.freak667]

Without learning? I was merely asking for you to point out what EXACTLY I am doing wrong, but you all just kept beating around the bush...

Useless, but at least now I know to be careful checking my results when the matrix elements are in the form of fractions.

The reason why we asked for the steps is because you posted this:

After mutiplying these two matrices by the only way I have been taught, and checking and doubled checking my answer using an online matrix calculator yields the result 0, 36 (check the screenshots)

Which also means most likely that you did it by hand and got the wrong answer. Muphrid pointed out why 36 was not the answer.

Ignoring all that, if you are given two different lines, do you think you can find the point of intersection using the matrix method?

 

[uperkurk]

If I got given a completely different question, I feel I could easily work out the point of intersection now. I only learned y=mx+c stuff a few weeks ago and I have dyscalculia so maths is pretty difficult for me to grasp.

But now I feel confident that if a similar question pops up in the exam I can easily do all the steps.

 

[Muphrid]

Without learning? I was merely asking for you to point out what EXACTLY I am doing wrong, but you all just kept beating around the bush...

Useless, but at least now I know to be careful checking my results when the matrix elements are in the form of fractions.

Sorry, but like I said, we're not psychic. We're more like remote tech support. We have no idea what's going wrong between the chair and the keyboard. We knew the answer you were getting was wrong. We did not know why other than that it was something in the matrix multiplication. Hence why we asked for steps.

And as yet, I remain unconvinced you actually can multiply matrices correctly and consistently. At no time in this thread did you mention (0,1) as a solution. At a couple times you mentioned two possible solutions, indicating you were uncertain. Since I basically gave you the answer, we will never know what you were really doing wrong, and I can only hope that you don't run into the same issue again later when you see a different problem involving matrix multiplication.

In short, I understand why you felt you should just be given the answer so you could see what you did wrong. I indulged the theory. I'm unconvinced it actually helped you.

Edit, if you want, though, here's another linear system:

\begin{align*} 5x+7y &amp;= -4 \\ 2x - 3y &amp;= 6\end{align*}

If you care to do this problem and get the right answer, then more power to you.

 

[uperkurk]

-54, 38

although I am not sure because when it comes to finding out the inverse I get

= \begin{bmatrix} \frac{1}{1} \end{bmatrix} which is just 1... so my matrix inverse becomes

A^-1 = \begin{bmatrix} 3 &amp; -7 \\ -2 &amp; 5 \end{bmatrix} x \begin{bmatrix} -4 \\ 5 \end{bmatrix} ?

(-4x3) + (-7x6) = -54
(-2) x (-4) + (5x6) = 38

 

[Muphrid]

When you find the inverse, you do this:

A = \begin{bmatrix} a &amp; b \\ c &amp; d \end{bmatrix} \to A^{-1} =\frac{1}{ad - bc} \begin{bmatrix} d &amp; -b \\ -c &amp; a \end{bmatrix}

So first, the sign on your 3 is wrong, and you didn't divide out by the determinant, which is (5x(-3) - 2x7).

Otherwise, you do seem to do the multiplication right for what you got from the step before. I suspect you may just need a little more practice and repetition then, and at that point, you'd be fine.

 

[uperkurk]

Our tutor provides us with all the formula's and I know to swap the elements for a and d and negate the elements b and c which is what I did.

So is -54, 38 correct? I don't dare check it on a calculator

Nevermind I know it's wrong... and I hate it when you have operators that conflict each other, for example 2 x -2 + - 4 I don't know what order to put stuff in.

 

[Muphrid]

I just said you changed the sign on 3 when you shouldn't have and you forgot to divide by the determinant, so no, the answer you got is not correct. You did multiply properly, though.

Standard order of operations is that multiplications come before additions unless parentheses explicitly tell you to do something else first. 2 \times (-2) + (-4) = -4 + (-4)= -8.