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

[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.

 

[uperkurk]

A = \begin{bmatrix} 5 &amp; 7 \\ 2 &amp; 3 \end{bmatrix} \to A^{-1} =\frac{1}{ad - bc} \begin{bmatrix} 3 &amp; -7 \\ -2 &amp; 5 \end{bmatrix}

Which is what I wrote? To be honest whoever invented this crap could not have made it more complicated if they tried.

A^{-1} =\frac{1}{ad - bc} is where I go wrong because I get confused on my operators, yes I'm at uni, but I have dyscalculia and have never ever been good at remembering numbers or anything to do with maths. I failed GCSE math and the only reason I'm doing math now is because it's part of my computer science course so I have no option but to learn it.

 

[Muphrid]

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

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

The matrix A should be

A = \begin{bmatrix} 5 &amp; 7 \\ 2 &amp; -3 \end{bmatrix}

with the negative sign on the 3.

 

[uperkurk]

The matrix A should be

A = \begin{bmatrix} 5 &amp; 7 \\ 2 &amp; -3 \end{bmatrix}

with the negative sign on the 3.

Ohhhh my bad I didn't realize that. anyway this thread has gone on way too long i'll let you go help some other people.

It's a shame it took so long to figure out what I was actually doing wrong I could have spend those 2 days on something else but not to worry still have 2 more days to revise.