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Thanks Alan,

oh, I didn't see the oops bit before so it makes a little more sense now.

Where do the numbers 

0.7071 abd 3.5355 come from?  Are they input or output?

Does the following help?

Basically we are treating a vector as a complex number with the x-component as the real part and the y-component as the imaginary part.

Alan, I am not familiar with matlab but I am interested in this concept.

Can you try to explain it a little more please?  Maybe with a picture.  

Yes, welcome to the forum Sasini. I hope that you enjoy yourself here 

a brownie recipe asks for one and one third times as much sugar as chocolate chips. if one and two thirds cups of sugar is used, what quantity of chocolate chips would then be needed.

There is going to be more sugar than Choc chips (1 and 1/3 times as much)

Let C stand for the amount of choc chip and 

let S stand for the amount of sugar.

$$1\frac{1}{3}\times C =S$$

NOW if     $$S=1\frac{2}{3}$$     find C

You have to substitute   $$1\frac{2}{3}$$    into the the equation for S

i.e.

$$1\frac{1}{3}\times C =\;1\frac{2}{3}\\\\
\mbox{divide both sides by }1\frac{1}{3}\\\\
1\frac{1}{3}\div 1\frac{1}{3} \times C =\;1\frac{2}{3}\div 1\frac{1}{3}\\\\
C =\;1\frac{2}{3}\div 1\frac{1}{3}\qquad \\\\
\mbox{You could finish it on the calculator now but I'll keep going - it might help you}\\\\
C =\frac{5}{3}\div \frac{4}{3}\qquad \\\\
C =\frac{5}{\not3}\times \frac{\not3}{4}\qquad \\\\
C =\frac{5}{4} \\\\
C = 1\frac{1}{4}\quad cups$$

For 1 and 2/3 cups of sugar you will need 1 and 1/4 cups of choc chips.

Thanks. This is enough for this week. lol. 

Deleted deleted

Here's the result for a 3x3 matrix in case you ever want to use it.  (I didn't do this by hand!)

You cannot take anything to the right hand side because there is no right hand side.  This is NOT and equation,  it is just an expression. (I think that is the correct term)

You are also right Alan.

Since the original graph is not specified I am not sure that it is completely valid to talk about phase shift.

check the forum while you wait by using the search box at the top. I know there is lots on this (unless removed in the site overhaul.) 

I couldn't delete the message.

The negative signs do exist in my original post, but they can be difficult to see!  They are in line with the horizontal "divide" lines.  Look closely.

What is the maximum matrix dimension for the "method descibed" in this post. 

Technically, there isn't a maximum, but the method becomes increasingly cumbersome extremely rapidly as the size increases.  I think I've seen it used on a 3x3 in the past, but I wouldn't use it on anything other than a 2x2.

Is there a formula name please? 

I don't think it has a special name (though I could be wrong - I'm not too concerned about what things are called. "A rose by any other name is still a rose"!). 

Above you added a negative in the front of b and c in the last post ( I mistakingly said c and d) although in the originaly reply the negative do not exist. I see a difference but not why.

What is the maximum matrix dimension for the "method descibed" in this post. 

Is there a formula name please? I don't know that. 

Thank you.

I did define collinear however the diagram mentioning it had two vectors from the same point of origin. Hence my question. Is that defining them as collinear or it would have been something else? 

No, they're not collinear just because they start at the same point.  None of the forces in your diagrams that start at the same point are collinear.  They would all have to lie along the same straight line (not necessarily in the same direction) to be collinear. 

...I have been resolving the " hypotenues of the triangle for example to get the resulting force of Fxi and Fyj etc. But not adding them and just giving the answer. 

Resolving a force on to two perpendicular axes just gives you the components of that force.  If you do this separately  with two forces, you would need to add the corresponding components to find the coordinates of the resultant force.

Why wouldn't the |-11| be taken to RHS first? And the there is absolute value, then you cannot push the additional negative into the absoulte value right? So it drops away? Or else why are the absolute value signs there? I mean is this a trick question? If you take the -11 across first you'd have a different equation or isn't that possible? Why can't you put the additional negative into the absoulte value first? Thanks for clearing this up.

Yes, pi/2 shown on my graph also Melody!  You (and kitty) are right that there is a pi/2 shift between these curves. 

use a d. It's asked in the question. Alternatively express "Let x equal expenses". Good luck. 

Do the signs on c and d need to be changed when finding the determinant? 

No.  The determinant is the scalar value (i.e. a single number): ad-bc.

What was the 1/|A| I was explained to do?

This is just 1/(ad-bc).  Look at my expression for the inverse of A.  Every term has ad-bc on the bottom.  Because every term has a common factor this factor can be taken out of the matrix as a constant multiplier, so the inverse could be written as:

Look at the numerators of each term (including the signs).   

Is there another way to find I?

You don't really find I.  It's just defined as the unit matrix.  i.e. a square matrix with zeros everywhere except on the diagonal where there are 1s.

I was shown something multiply by first row, and add second row or something like that. Is that what I saw on Kahn Acadamy, reduced row echlon or something to solve for I? 

Yes there are several more complicated methods for finding the inverse of a matrix.  These are important, especially when large matrices are involved.  But for 2x2 matrices the first expression in the image above can be used (as long as ad-bc is not zero - if it is zero then A has no inverse).

I did define collinear however the diagram mentioning it had two vectors from the same point of origin. Hence my question. Is that defining them as collinear or it would have been something else? 

With regards to resultant force, I asked a lecturer today. When working a problem I have been resolving the " hypotenues of the triangle for example to get the resulting force of Fxi and Fyj etc. But not adding them and just giving the answer. Further example is 5sin45 = ... is the Fr (and times D is torque). Is this correct? The lecturer said the answer was correct, but it sounds like i'm only solving for force, not resulting force. Even today working with a parellelogram and multiple forces this was all I did. Or am I missing the definition of two different things here? 

That is right. 

Actually, pi is the phase.  A phase shift  means the difference in phase from another sine wave.  Since no other wave is specified here, there isn't really a shift involved.  

Thank you. Do the signs on c and d need to be changed when finding the determinant? What was the 1/|A| I was explained to do? Is there another way to find I? I was shown something multiply by first row, and add second row or something like that. Is that what I saw on Kahn Acadamy, reduced row echlon or something to solve for I? Thanks for the post.