Alan

Represent the cost of an adult ticket by 'at' and the cost of a student ticket by'st'.

For the first family    2*at + 1*st  =  8.75

For the 2nd family    1*at + 4*st  = 10.50

From the 2nd family's equation, if we subtract 4*st from both sides we have 1*at = 10.50 - 4*st, which tells us the cost of an adult ticket in terms of the cost of student tickets.  The first family paid for 2 adult tickets which means they must have paid the equivalent of 2*(10.50 - 4*st) or 21 - 8*st for their 2 adult tickets.  Replacing the 2*at in their equation we get, for the first family, that:

 21 - 8*st + 1*st = 8.75

or 21 - 7*st = 8.75

Add 7*st to both sides and subtract 8.75 from both sides;

21 - 8.75 = 7*st

12.25 = 7*st

Divide both sides by 7 to find

st = 12.25/7 =  1.75

So the price of a student ticket is $1.75

Put this back in the equation for the 2nd family and we get

1*at + 4*1.75 = 10.50

at + 7 = 10.50

Subtract 7 from both sides

at = 10.50 - 7 = 3.50

so the price of an adult ticket is $3.50

Heron's formula can be used when you just know the length of each side. This says

Area = √[s*(s-a)*(s-b)*(s-c)]  where a, b and c are the lengths of the sides and s = (a+b+c)/2.

Here: s = (5+12+9)/2 = 13, so:

$${\mathtt{Area}} = {\sqrt{{\mathtt{13}}{\mathtt{\,\times\,}}\left({\mathtt{13}}{\mathtt{\,-\,}}{\mathtt{5}}\right){\mathtt{\,\times\,}}\left({\mathtt{13}}{\mathtt{\,-\,}}{\mathtt{12}}\right){\mathtt{\,\times\,}}\left({\mathtt{13}}{\mathtt{\,-\,}}{\mathtt{9}}\right)}} \Rightarrow {\mathtt{Area}} = {\mathtt{20.396\: \!078\: \!054\: \!371\: \!139\: \!3}}$$

Area ≈ 20.396

The rotation vector (rot) is e^i*pi/4

e^i*pi/4 = cos(pi/4)+i*sin(pi/4) = 1/√2 + i*1/√2 = 0.7071 + i*0.7071

Then vrot = v*rot = 

(2 + 3i)*(0.7071 + 0.7071i) = 2*0.7071-3*0.7071 +i*(2*0.7071 + 3*0.7071) = -0.7071 +i*5*0.7071 = -0.7071 + 3.5355i

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.

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

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"!). 

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.

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

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).

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.