how would I go about solving this matrix equation with a non-square

>>8519805
Write matrix X with some arbitrary coefficients, then multiply the matrices on the left side and solve the system of equations

First X must be a 2x3 matrix

3a+2d = 3
4a + 3d = 1
1a +0d = 7 => a= 7
...

keep solving from there.

>>8519805
you might try

A x = B
A^t A x = A^t B
x = (A^t A)^(-1) A^t B
If there is an x that works, this will give it to you. You have to check that it works though.

>>8519836
this is great, thanks

by X working you just mean that there is a solution, right?

>>8519836
That's the fun way:D

>>8519805
You're mistaken, it does have a left inverse.

[math]\begin{pmatrix} a^\top \\ b^\top \end{pmatrix} \begin{pmatrix}u & v \end{pmatrix} = \begin{pmatrix} a^\top u & a^\top v \\ b^\top u & b^\top v\end{pmatrix} = I[/math]

You just need to find 2 other 3D vectors that are mutually orthonormal to these 2.

>>8519838
I mean you have to take the x you get and check that Ax = B, i.e. plug it back in and check. This is because the implication in the math above doesn't go both ways

>>8519805
The inverse of A exists, it's:
[math]\begin{pmatrix}3 & -2& 0 \\ -4 & 3 & 0\end{pmatrix}[/math]

Try it out. Left multiply and solve.

>>8520779
there are infinitely many inverses tho. do all inverses give the same result?

>>8520794
oh then parametrize the inverse then to get all the answers then maybe

>>8520808
could try but I'm lazy.
doesn't matrix multiplication associativity assert that no matter what matrix you use, it will give the same result?

>>8519805
this guy has the answer youre looking for>>8519823

Need me to just solve it for you?

>>8519836

A=[3 2; 4 3; 1 0];
B=[3 9 7; 1 11 7; 7 5 7];
x=inv(A'*A)*A'*B
error=max(max(abs(A*x-B)))

gives

x =

7.0000 5.0000 7.0000
-9.0000 -3.0000 -7.0000

error = 4.4409e-14


It works.

>>8519805
The system is over-determined (9 equations in 6 unknowns). Reduce it to 6 equations by discarding any one row from both matrices. Then you have A'.X=B' where A' is 2x2 and B' is 2x3. A' can be inverted, and you can just use X=A'^-1.B.

Multiplying the discarded row of A by X should then give you the discarded row of B. If it doesn't, the system is inconsistent (i.e. there is no matrix X satisfying the equation).