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Why is there no Matrix3x3 in the System.Numerics namespace? c#

I am trying to make a little GUI library in c#, but after doing some research on matrix transformations I found out that I need a Matrix3x3 to store rotation, scale, and translation of a Vector2. But in the C# System.Numerics there is only a Matrix3x2 or Matrix4x4? Could I use one of those instead? If so how would I go about it? And why isnt there a Matrix3x3 in the standard library?
I am very new to Matrix and Vector programming, So sorry if this is a stupid question.
Thanks in advance.

You can use Matrix4x4. Start with an identity matrix and fill the top left 3×3 elements with your matrix.
For example to solve a 3×3 system of equations:
// Find the barycentric coordinates
//
// Solve for (wA,wB,wC):
// | px | | ax bx cx | | wA |
// | py | = | ay by cy | | wB |
// | pz | | az bz cz | | wC |
var m = new Matrix4x4(
A.X, B.X, C.X, 0,
A.Y, B.Y, C.Y, 0,
A.Z, B.Z, C.Z, 0,
0, 0, 0, 1);
if (Matrix4x4.Invert(m, out Matrix4x4 u))
{
var w = new Vector3(
u.M11*P.X + u.M12*P.Y + u.M13*P.Z,
u.M21*P.X + u.M22*P.Y + u.M23*P.Z,
u.M31*P.X + u.M32*P.Y + u.M33*P.Z);
// ...
}
As for the reasons, the intent of System.Numerics has to be related to computer graphics since it utilizes Homogeneous Coordinates in which 3D vectors contain 4 elements. Three regular coordinates and a scalar weight factor. The math with homogeneous coordinates for computer graphics is vastly simplified. The only reason there is a Vector3 is because a Vector4 should be treated as a vector of 3 elements plus a scalar, and thus Vector3 should be used in composition and decomposition of the homogeneous coordinates. It means that not all 4 elements can be treated equally, and sometimes you need to do things with the vector (first three elements) separately from the scalar value (fourth element).
Also System.Numerics uses single precision float elements, which are almost never used in scientific computing, but universally applied to computer graphics for their speed. Hopefully one day when the CLR supports AVX512 there will be double precision numeric intrinsics that scientists can actually use.

Decompose 3x3 Matrix

I have a 3x3 matrix I'm using to track movement in 2D. I need to extract from that the translation, rotation and scale matrices. Can anyone suggest how I would do this? I've had no luck searching online so far (possibly I'm using the wrong terms).

This is just off the top of my head so there may be an error in here, but:
Assuming your matrix is row-major (just transpose everything if you're using column major):
| cos(t) -sin(t) 0 |
| sin(t) cos(t) 0 |
| tx ty 1 |
The translation vector will be the last row in the matrix [tx ty 1]. Extracting scale and rotation in a composed matrix is a bit trickier.
Looking at a 3x3 rotation matrix,
| cos(t) -sin(t) 0 |
| sin(t) cos(t) 0 |
| 0 0 1 |
And a scale matrix
| vx 0 0 |
| 0 vy 0 |
| 0 0 1 |
The combined rotation & scale matrix might look like (ct = cos(t), st = sin(t))
| vx*ct -vx*st 0 |
| vy*st vy*ct 0 |
| 0 0 1 |
For uniform scaling, vx=vy.
| v*ct -v*st 0 |
| v*st v*ct 0 |
| 0 0 1 |
Remembering the trig identity
ct^2 + st^2 = 1
We can see that the
(v*ct)^2 + (v*st)^2 = v^2
or
v^2*ct^2 + v^2*st^2 = v^2
... all the terms of which are in the composite (scale,rotation,translation or SRT for short) matrix,
So,
v = sqrt((v*ct)^2 + (v*st)^2)
or
v = sqrt(M[0,0]^2 + M[0,1]^2);
Theta, then is just
t = acos(vct/v)
or
t = acos(M[0,0]/v)
or, this might work much easier but I haven't tried it:
theta = atan2(vst,vct),
scale = sqrt(vst^2+vct^2)
where ^ is an exponent, not an XOR.
... and you can work out the rest.
It would be wiser to keep your scale, rotation and translation values around and use those values both to build the matrix and for whatever other tasks you require. Relying on the matrix as the only container for that information will eventually lead to compound floating-point errors and other drama.
Some notes:
Theta is an angle. It's a fun, easy-to-draw Greek symbol, and all us engineers love greek characters. It's mostly Stockholm syndrome since some of them look like 5's, and some of them are impossible to draw unless you're Greek.
This works great for 2D, where there is only one possible rotation axis. If you're working in three (or higher! It's a thing now!) dimensions, the concept is the same but the implementation becomes much, much messier.
st, ct, vst, vct, etc. are all contained in the composite matrix you have available. (Composite == concatenated == combined. The terminology in use depends on who you're reading.) The non-intuitive part is extracting the elements that you need from the matrix. This requires understanding what is in the matrix, where, and why.
This is a solvable problem. Normally, I'm a sit-down-and-start-coding kind of guy, and in 25 years of professional development, I'd have to say that it's worked out pretty well so far. But, there are times where a whiteboard or a notebook and a mechanical pencil are better friends than your keyboard.

For a column major matrix, given:
X vector of < Xx, Xy >;
Y vector of < Yx, Yy > and
Translation/position of < Tx, Ty >,
The corresponding column major transformation matrix is:
The translation is:
For the Scaling, the scale factors for the X and Y vectors are:
The scaling transformation is then:
For Rotation, the normalized vectors are:
X Vector: < Xx/Sx, Xy/Sx >
Y Vector: < Yx/Sy, Yy/Sy >
And the rotation transformation "R" is:
RECOMPOSITION:
Given translation "T", rotation "R", and scaling "S" transformation in column major order, the original matrix/transform can be recreated with:
T * R * S
Assumption, the original transformation has X and Y vectors that are:
Not parallel;
Not a length of zero;
Not mirrored; and
Not skewed/sheared.
The rotation can be further decomposed into an angle as described by 3Dave (which is row major, and needs to be transposed to be column major).

Project 3D coordinates to a 2D plane

I have a plane defined by a normal vector and another normalalised direction vector that is moving along that plane, both in 3D space.
I'm trying to figure out how to project that normal direction 3D vector onto the plane such that it ends up being a 2D vector with x/y coordinates.

It sounds like you need to find the angle between the direction vector and the plane. The size of the projection is going to scale with the cosine of that angle. Since the normal vector of the plane is perpendicular, I think you can find the sine between the normal vector and your direction vector.
The angle between the two vectors is given by the dot product of the vectors over the magnitudes multiplied together. That gives us our theta. Take the sin of theta, and we have the scaling factor (I'll call it s)
Next, you need to define unit size vectors on the plane to project onto. It's probably easiest to do this by setting one of the unit vectors in the direction of the projection to move forward...
If you set the unit vector in the direction of the projection, then you know the length of the projection in that unit space by using the scaling factor and multiplying by the length of the vector.
After that, with the unit vector, multiply in the length and find your vector relative to your normally defined xyz axis.
I hope this helps.

Try something like this. I wrote a paper on this exact method a while ago and can provide you with a copy if you would like.
PointF Transform32(Point3 P)
{
float pX = (float)(((V.J * sxy) - V.I * cxy) * zoom);
float pY = (float)(((V.K * cz) - (V.I * sxy * sz) - (V.J * sz * cxy)));
return new PointF(Origin.X + pX, Origin.Y - pY);
}
cxy is the cosine of the x-y camera angle, measured in radians from the positive x-axis on the xy plane.
sxy is the sine of the x-y camera angle.
cz is the cosine of the z camera angle, measured in radians from the x-y plane (so the angle is zero if the camera rests on that plane).
sz is the sine of the z camera angle.
Alternatively:
Vector3 V = new Vector3(P.X, P.Y, P.Z);
Vector3 R = Operator.Project(V, View);
Vector3 Q = V - R;
Vector3 A = Operator.Cross(View, zA);
Vector3 B = Operator.Cross(A, View);
int pY = (int)(Operator.Dot(Q, B) / B.GetMagnitude());
int pX = (int)(Operator.Dot(Q, A) / A.GetMagnitude());
pY and pX should be your coordinates. Here, vector V is the position vector of the point in question, R is the projection of that vector onto your viewing vector, Q is the component of V orthogonal to the viewing Vector, A is an artificial X-axis formed by the cross-product of the viewing vector with the vector (0,0,1), and B is an artificial Y-axis formed by the cross product of A and (0,0,1).
It sounds like what you're looking for is something like a simple rendering engine, similar to this, which used the above formulae:
Hope this helps.

Relationship between projected and unprojected Z-Values in Direct3D

I've been trying to figure this relationship out but I can't, maybe I'm just not searching for the right thing. If I project a world-space coordinate to clip space using Vector3.Project, the X and Y coordinates make sense but I can't figure out how it's computing the Z (0..1) coordinate. For instance, if my nearplane is 1 and farplane is 1000, I project a Vector3 of (0,0,500) (camera center, 50% of distance to far plane) to screen space I get (1050, 500, .9994785)
The resulting X and Y coordinates make perfect sense but I don't understand where it's getting the resulting Z-value.
I need this because I'm actually trying to UNPROJECT screen-space coordinates and I need to be able to pick a Z-value to tell it the distance from the camera I want the world-space coordinate to be, but I don't understand the relationship between clip space Z (0-1) and world-space Z (nearplane-farplane).
In case this helps, my transformation matrices are:
World = Matrix.Identity;
//basically centered at 0,0,0 looking into the screen
View = Matrix.LookAtLH(
new Vector3(0,0,0), //camera position
new Vector3(0,0,1), //look target
new Vector3(0,1,0)); //up vector
Projection = Matrix.PerspectiveFovLH(
(float)(Math.PI / 4), //FieldOfViewY
1.6f, // AspectRatio
1, //NearPlane
1000); //FarPlane

Standard perspective projection creates a reciprocal relationship between the scene depth and the depth buffer value, not a linear one. This causes a higher percentage of buffer precision to be applied to objects closer to the near plane than those closer to the far plane, which is typically desired. As for the actual math, here's the breakdown:
The bottom-right 2x2 elements (corresponding to z and w) of the projection matrix are:
[far / (far - near) ] [1]
[-far * near / (far - near)] [0]
This means that after multiplying, z' = z * far / (far - near) - far * near / (far - near) and w' = z. After this step, there is the perspective divide, z'' = z' / w'.
In your specific case, the math works out to the value you got:
z = 500
z' = z * 1000 / (1000 - 999) - 1000 / (1000 - 999) = 499.499499499...
w' = z = 500
z'' = z' / w' = 0.998998998...
To recover the original depth, simply reverse the operations:
z = (far / (far - near)) / ((far / (far - near)) - z'')

How to get coords inside a transformed sprite?

I am trying to the get the x and y coordinates inside a transformed sprite. I have a simple 200x200 sprite which rotates in the middle of the screen - with an origin of (0,0) to keep things simple.
I have written a piece of code that can transform the mouse coordinates but only with a specified x OR y value.
int ox = (int)(MousePos.X - Position.X);
int oy = (int)(MousePos.Y - Position.Y);
Relative.X = (float)((ox - (Math.Sin(Rotation) * Y /* problem here */)) / Math.Cos(Rotation));
Relative.Y = (float)((oy + (Math.Sin(Rotation) * X /* problem here */)) / Math.Cos(Rotation));
How can I achieve this? Or how can I fix my equation?

The most general way is to express the transformation as a matrix. This way, you can add any other transformation later, if you find you need it.
For the given transformation, the matrix is:
var mat = Matrix.CreateRotationZ(Rotation) * Matrix.CreateTranslation(Position);
This matrix can be interpreted as the system transformation from sprite space to world space. You want the inverse transformation - the system transformation from world space to sprite space.
var inv = Matrix.Invert(mat);
You can transform the mouse coordinates with this matrix:
var mouseInSpriteSpace = Vector2.Transform(MousePos, inv);
And you get the mouse position in the sprite's local system.
You can check if you have the correct matrix mat by using the overload of Spritebatch.Begin() that takes a matrix. If you pass the matrix, draw the sprite at (0, 0) with no rotation.