# Math help

#### dougbenham

##### Member
I need a little matrix math help. I'm attempting to do some matrix algebra including division. I have found several online resources that say the division of matrices is done like so:
AX=B
X = A^(-1) B

And in order to take the inverse of a matrix it must be a square...

So lets say I have the following:
AB = CD
Where A has dimensions (4, 4), B has the dimensions (4, 1), C has dimensions (4, 4), and D has dimensions (4, 1).

I am solving for A. So I would guess the first step is to "divide" by B. How do I do this since B doesn't have square dimensions? Any suggestions would be appreciated..

#### Rubikgenius

##### Member
Ummm... I am good at math, but matrixes are kind of hard. I kinda forgot how to do them

#### qqwref

##### Member
Here's the thing - both AB and CD are 4x1 matrices, right? They only have 4 'data points'. So you are trying to solve for 16 variables but you only have 4 equations. That's why you can't just use an inverse - if you could, it would imply there was a unique solution, which there isn't.

The best way to get something resembling a solution would be to write out CD (in terms of numbers) and AB (in terms of variables and whatnot); then you would get four equations. For instance the first one would be something like
B_1 A_11 + B_2 A_12 + B_3 A_13 + B_4 A_14 = (CD)_1.
Each of the four equations contains information about every variable in a given row of A. So, unless you know almost everything about A, there are an infinite number of solutions.

#### rjohnson_8ball

##### Member
Wouldn't the inverse of a 4x1 just be a 1x4 where each of the elements are reciprocals? That would cause

B B^(-1) = I

which is what you want. You might test this out on some samples to make sure it works. Let me know what the answer turns out to be, if I am mistaken.

@qqwref, you might be right. Maybe the question is to provide any A that solves it? If so, A could be a diagonal matrix?

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#### JBCM627

##### Member
Wouldn't the inverse of a 4x1 just be a 1x4 where each of the elements are reciprocals?
Multiplying those won't give you a 4x4 identity matrix.

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#### rjohnson_8ball

##### Member
Wouldn't the inverse of a 4x1 just be a 1x4 where each of the elements are reciprocals?
Multiplying those won't give you a 4x4 identity matrix.
It might give a 1x1 matrix with a value of 1?

#### JBCM627

##### Member
It might give a 1x1 matrix with a value of 1?
4x1.1x4 matrix gives you a 4x4 matrix, which is what you would need. You can't multiply a 4x1.1x4 and end up with an identity matrix, although you can by taking a 1x4.4x1 matrix.

#### Kian

##### Member
Ummm... I am good at math, but matrixes are kind of hard. I kinda forgot how to do them
But you're a genius. How can this be?

#### dougbenham

##### Member
Here's the thing - both AB and CD are 4x1 matrices, right? They only have 4 'data points'. So you are trying to solve for 16 variables but you only have 4 equations. That's why you can't just use an inverse - if you could, it would imply there was a unique solution, which there isn't.

The best way to get something resembling a solution would be to write out CD (in terms of numbers) and AB (in terms of variables and whatnot); then you would get four equations. For instance the first one would be something like
B_1 A_11 + B_2 A_12 + B_3 A_13 + B_4 A_14 = (CD)_1.
Each of the four equations contains information about every variable in a given row of A. So, unless you know almost everything about A, there are an infinite number of solutions.
Okay I'll tell you the practical side to the problem. I'm doing 3D transformational geometry for the CubeTeacher program. I'm making a function that finds the appropriate rotation matrix to obtain the new position without actually changing the current position variable. Hmm this is hard to explain. Lets say a point is at [0, 1, 0] in the format [x, y ,z].

Lets define the term 'applied point' as when we take the rotation matrix and multiply it by the point [0, 1, 0], therefore applying the rotation to the point.

Now lets define the starting rotation matrix as the identity matrix:
1 0 0
0 1 0
0 0 1

When we calculate the 'applied point', we should get the same matrix, [0, 1, 0].

Now lets say I want to know what rotation matrix to use to get the 'applied point' to be [1, 0, 0] (rotated clockwise 90 degrees along the y-axis plane). So lets identify what we know and what we are looking for. We know the point ([0, 1, 0]). And we know the new point location ([1, 0, 0]). We want to know the new rotation matrix.

Therefore we can come up with an equation:
AB = C

Where A is the new rotation matrix, B is the original point location ([0, 1, 0]), and C is the new point location ([1, 0, 0]).

So how would I solve for the new rotation matrix? You can't divide both sides by B because B is not a square matrix.

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#### JBCM627

##### Member
So how would I solve for the new rotation matrix? You can't divide both sides by B because B is not a square matrix.
Oh hey, we are going over SO3 a bit in my GR class right now. What qq suggested is probably similar to what you are looking for. You should be able to reduce A (your rotation matrix) down to a matrix containing 2 variables (since in rotating a point you only have 2 degrees of freedom), which gives you 3 systems of equations to solve with 2 variables after evaluating AC.

I'll be a bit more specific... you can accomplish any 3 space rotation by using just the Rx and Rz matrices, say. So Rz.Rx.B = C... and (Rz.Rx) will be your rotation matrix with 2 variables... the ammount you rotate around the x and z axes. (These rotations will be through angles we can call theta and phi, respectively). You should be able to at least approximately solve for theta and phi then.

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#### dougbenham

##### Member
you can accomplish any 3 space rotation by using just the Rx and Rz matrices, say. So Rz.Rx.B = C... and (Rz.Rx) will be your rotation matrix with 2 variables... the ammount you rotate around the x and z axes. (These rotations will be through angles we can call theta and phi, respectively). You should be able to at least approximately solve for theta and phi then.
Ah.. Thank you for the explanation of this. I had not realized that any 3D space rotation was possible with only Rx and Rz.

Except my rotation matrix is just a single matrix. There is no Rx and Rz.. How would I get those from the total rotation matrix?

And also, I'm not quite sure how I would be able to approach this problem. I understand the logic behind it, however, I don't understand how I can "solve for theta and phi". Could you give me some guidance or perhaps an example with numbers?

#### ThatGuy

##### Member
Multiply C and D. Then create the matrix A of all variables. Then multiply it with B and set each equal. Of course, that gives you a bunch of system of equations. These types of problems don't really appear anywhere much.

#### JBCM627

##### Member
Except my rotation matrix is just a single matrix. There is no Rx and Rz.. How would I get those from the total rotation matrix?
If you rotate the point B about Rx, you will have:
Rx.B
Then rotating that about Rz, you have:
Rz.(Rx.B)
And by associativity, you have:
(Rz.Rx).B
So your rotation matrix is R = (Rz.Rx)

Explicitly, this is... much easier to type out in mathematica on here. See the bottom of attached file for what R is.

And also, I'm not quite sure how I would be able to approach this problem. I understand the logic behind it, however, I don't understand how I can "solve for theta and phi". Could you give me some guidance or perhaps an example with numbers?
Unless the case is simple (like [1,0,0] -> [0,1,0]), I'm not sure you'll be able to solve for theta and phi analytically.

For simplicity's sake, I'm only going to do this explicitly for the aforementioned matrices. Also, if I type out a 1x3 matrix/vector, assume it should be transposed.

For a vector [x,y,z], it is obvious that in R:[1,0,0] -> [0,1,0], R is a 90-degree rotation about the z-axis. Assuming we don't know that though, we have
R.[1,0,0] = [Cos[phi], Sin[phi], 0] = [0,1,0].
So Sin[phi] = 1, phi = pi/2.

plugging in phi = pi/2 gives us the rotation matrix:
Code:
0  -Cos[theta] Sin[theta]
1       0         0
0  Sin[theta]  Cos[theta]

And it doesn't matter what theta is, so if you like, it can be something like this:

0  0  1
1  0  0
0  1  0

Or even in this special case,

0  0  0
1  0  0
0  0  0
Again, I'm guessing most cases can't be solved analytically, so you would need some way to approximate theta and phi. This also only works for normalized vectors; as in, B and C will have to have the same magnitude.

I'll state the following explicitly too:
if you have B = [x,y,z] and C = [x',y',z']
Then you have the following system of equations to solve for theta and phi:
x = x'*Cos[phi] + y'*Sin[phi]
y = z'*Sin[theta] + Cos[theta]*(y'*Cos[phi] - x'*Sin[phi])
z = z'*Cos[theta] + Sin[theta]*(x'*Sin[phi] - y'*Cos[phi])
...and an approximation would just entail minimizing these functions.

.

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#### dougbenham

##### Member
Except my rotation matrix is just a single matrix. There is no Rx and Rz.. How would I get those from the total rotation matrix?
If you rotate the point B about Rx, you will have:
Rx.B
Then rotating that about Rz, you have:
Rz.(Rx.B)
And by associativity, you have:
(Rz.Rx).B
So your rotation matrix is R = (Rz.Rx)

Explicitly, this is... much easier to type out in mathematica on here. See the bottom of attached file for what R is.

And also, I'm not quite sure how I would be able to approach this problem. I understand the logic behind it, however, I don't understand how I can "solve for theta and phi". Could you give me some guidance or perhaps an example with numbers?
Unless the case is simple (like [1,0,0] -> [0,1,0]), I'm not sure you'll be able to solve for theta and phi analytically.

For simplicity's sake, I'm only going to do this explicitly for the aforementioned matrices. Also, if I type out a 1x3 matrix/vector, assume it should be transposed.

For a vector [x,y,z], it is obvious that in R:[1,0,0] -> [0,1,0], R is a 90-degree rotation about the z-axis. Assuming we don't know that though, we have
R.[1,0,0] = [Cos[phi], Sin[phi], 0] = [0,1,0].
So Sin[phi] = 1, phi = pi/2.

plugging in phi = pi/2 gives us the rotation matrix:
Code:
0  -Cos[theta] Sin[theta]
1       0         0
0  Sin[theta]  Cos[theta]

And it doesn't matter what theta is, so if you like, it can be something like this:

0  0  1
1  0  0
0  1  0

Or even in this special case,

0  0  0
1  0  0
0  0  0
Again, I'm guessing most cases can't be solved analytically, so you would need some way to approximate theta and phi. This also only works for normalized vectors; as in, B and C will have to have the same magnitude.

I'll state the following explicitly too:
if you have B = [x,y,z] and C = [x',y',z']
Then you have the following system of equations to solve for theta and phi:
x = x'*Cos[phi] + y'*Sin[phi]
y = z'*Sin[theta] + Cos[theta]*(y'*Cos[phi] - x'*Sin[phi])
z = z'*Cos[theta] + Sin[theta]*(x'*Sin[phi] - y'*Cos[phi])
...and an approximation would just entail minimizing these functions.

.
Wow this is difficult stuff. I've reread your post like 10 times to completely understand it How did you get that system of equations at the end?

#### JBCM627

##### Member
Wow this is difficult stuff. I've reread your post like 10 times to completely understand it How did you get that system of equations at the end?
Glad it made sense eventually I probably could've been a bit clearer. What is actually a really cool extension of this though is to do hyperbolic rotations (replacing sin with sinh and cos with cosh) in one of the matrices... this gives rise to a lot of the math behind special relativity.

For those equations, I had mathematica solve for x, y, and z... I probably should've done it the other way around and just had it evaluate R.B, to give x',y', and z'. That would've made more sense. I'll do that real fast...
x' = x*Cos[Phi] - y*Cos[Theta]*Sin[Phi] + z*Sin[Theta]*Sin[Phi]
y' = y*Cos[Theta]*Cos[Phi] - z*Cos[Phi]*Sin[Theta] + x*Sin[Phi]
z' = z*Cos[Theta] + y*Sin[Theta]

Also, I played around with it a bit more, and you can explicitly solve for Sin[theta] and Sin[phi]. You can just substitute Sqrt[1-Sin[phi]^2] in for Cos[phi], and similarly for Cos[theta]. The exact answers for theta and phi (phi in particular) are horrendously long though, so I'm not going to paste it in here... if you are interested I can email it to you.

Another important thing to note is that this only gives you one particular solution. There are infinitely many ways you could move one point to another on a circle... rotating about the x and z axes (as opposed to other arbitrary ones) just happens to look a bit nicer.

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#### Lord Voldemort

##### He-Who-Must-Not-Be-Named
Ummm... I am good at math, but matrixes are kind of hard. I kinda forgot how to do them
I feel like I'm adding to the problem, but what was the point of the post?

#### dougbenham

##### Member
For those equations, I had mathematica solve for x, y, and z... I probably should've done it the other way around and just had it evaluate R.B, to give x',y', and z'. That would've made more sense. I'll do that real fast...
x' = x*Cos[Phi] - y*Cos[Theta]*Sin[Phi] + z*Sin[Theta]*Sin[Phi]
y' = y*Cos[Theta]*Cos[Phi] - z*Cos[Phi]*Sin[Theta] + x*Sin[Phi]
z' = z*Cos[Theta] + y*Sin[Theta]
Thats exactly what I derived from the rotation matrix you supplied. And I figured I'd be trying to solve for theta in the bottom equation first since it only has theta (not phi). So for the past day I've been sitting and looking at that equation trying to think of how to solve for theta I couldn't think of any useful trig identities that would help.

You can just substitute Sqrt[1-Sin[phi]^2] in for Cos[phi], and similarly for Cos[theta].
Thats what I was looking for haha. And so from that you can derive the value of theta (attached image).

Now I just got mathematica today and I'm a bit curious of how to use it.. I typed in 'Solve[a == zcos[t] + ysin[t], t]' and evaluated it (I used a instead of z'). It gave the response '{{}}'. What did I do wrong? Shouldn't it have given me the long expression I included in the attachment?

#### JBCM627

##### Member
Now I just got mathematica today and I'm a bit curious of how to use it.. I typed in 'Solve[a == zcos[t] + ysin[t], t]' and evaluated it (I used a instead of z'). It gave the response '{{}}'. What did I do wrong? Shouldn't it have given me the long expression I included in the attachment?
Yeah, careful on the syntax... capital Sin and Cos. You'll need to explicitly type out a space or "*" between the variables and functions too, as in, "z*Cos[t]".

It also doesn't give me exactly what is in your attachment, but perhaps your expression somehow reduces to what it gave me. But haha, just wait til you see how long the expression for phi is... I had to scroll.

#### dougbenham

##### Member
Now I just got mathematica today and I'm a bit curious of how to use it.. I typed in 'Solve[a == zcos[t] + ysin[t], t]' and evaluated it (I used a instead of z'). It gave the response '{{}}'. What did I do wrong? Shouldn't it have given me the long expression I included in the attachment?
Yeah, careful on the syntax... capital Sin and Cos. You'll need to explicitly type out a space or "*" between the variables and functions too, as in, "z*Cos[t]".

It also doesn't give me exactly what is in your attachment, but perhaps your expression somehow reduces to what it gave me. But haha, just wait til you see how long the expression for phi is... I had to scroll.
Ah thanks. I didn't realize the capital would matter I attached what it gave me for theta. Could you tell me how it got that? Like a step-by-step perhaps?

#### JBCM627

##### Member
I attached what it gave me for theta. Could you tell me how it got that? Like a step-by-step perhaps?
Just algebra...
Let c = Cos[t].
a = s*c+y*Sqrt[1-c^2]
(a-zc)^2 = y^2-(yc)^2
(z^2-y^2)*c^2 - (2az)*c + (a^2+y^2) = 0
Then just use the quadratic formula, and it comes out.