solving unknown variables in matrices

So for un I've got to solve 3 variables from the equation y = k*x^2 + l*x + m, given are points P(-2, 4) Q(1,1) and R(2, -4). We are also instructed to use matrices.

I though a logical next step would be to make a matrix, matrix=[-2, 4;1, 1;2, -4]

From this you can make 3 linear equations, one for every row of the matrix. Is there a way to build and solve such an 'equation matrix' using Matlab? Or do I just have to do this by hand by treating it as a system of 3 coupled equations, which in this case isn't that difficult.

I've tried introducting the variables k, l and m into a matrix and making equations in there, but this hasnt worked.

Answer 1

[Lukas]

You need to solve an linear optimization problem A*x = b. Each line is one of your points P,Q,R.

• b is the solution vector and contains the y-variable of your three given points P,Q,R

b = [4 ; 1 ; -4]

• x is the vector of (unknown) parameters, this means that
• Each column of a corresponds to either x^0, x^1 or x^2 (for all points P,Q,R):

A = [1 -2 4; 
    1  1  1;
    1  2  4]

The solution (where x = [m;l;k] ) is:

x = A\b

[Loek Vossen]

why is the solution A\b rather than b\A, which would appear to be the answer when looking at the equation A*x=b.

sorry if its a stupid question but I am quite the beginner

[Torsten]

It's just a MATLAB operator; you should not think about the logic behind it:

https://de.mathworks.com/help/matlab/ref/mldivide.html

[Steven Lord]

Because that's how the \ and / operators are defined. See the first section of this documentation page for more information.

[John D'Errico]

Personally, if I did not know the answer already, I might have guessed x=b/A would have been the logical notational choice, as that is what you would write if A were a scalar variable.

But I do understand how you might have been confused. And I'm not absolutely positive that there is a better way to remember it, except to know the answer, which you will do now. ;-)

We can just blame it on Cleve of course, but I have found Cleve to know what he is doing. Anyway, what does the syntax B/A mean? If we read the help for (forward) slash, we find it solves this problem:

/   Slash or right division.
      B/A is the matrix division of A into B, which is roughly the
      same as B*INV(A) , except it is computed in a different way.

Thus x=B/A solves the problem

x*A = B

for a square matrix A. If A is non-square, then it solves the problem in a least squares sense.

Ok, so B/A is not available, since it already has a logical definition. So long ago, in a galaxy far, far away, Cleve defined backslash to work as it does. :)

Answer 2

[Torsten]

[4 -2 1; 1 1 1;4 2 1]*[k;l;m] = [4; 1; -4]