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

[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:

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

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 $= \begin{bmatrix} 3 & -7 \\ -2 & 5 \end{bmatrix} x \begin{bmatrix} -4 \\ 5 \end{bmatrix}$ ?

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

 

[Muphrid]

When you find the inverse, you do this:

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

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]

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

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:

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

The matrix A should be

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

with the negative sign on the 3.

 

[uperkurk]

The matrix A should be

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

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.