MATLAB: Hello everyone, I am facing one problem,

equationStatistics and Machine Learning Toolbox

I have 3 different non linear data. Let's say x,y and z.for different values of Z I have different data set of x and y. If I fix the value of Z I am able to find the relation between x and y, but if I change the value of Z that relation won't work, I need to find relation between x,y and z independently. Can anyone please help! Thank you in advance.

for z=500 I have x = 1145 to 1240 with the gap of 1,y = (17.75,16.2,15.1,14.25,13.45…. respectively

for z=400, x values remains the same 1145-1240 y values differs.

I need to find a relation between x,y and z independently. in which y = fun(x,z). thank you.

Best Answer

Try the attached code where I fit an exponential growth curve to each curve, which belongs to each x value. It uses fitnlm() of the Statistics and Machine Learning Toolbox.

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MATLAB: Problem in calculating a determinant for a matrix

Nobody bothered to answer this homework question yet? Return to linear algebra 101.

1. If the determinant of the nxn matrix A is D, then the determinant of the matrix k*A is k^n*D.

2. DON'T USE THE DETERMINANT!!!!!!!!!!!!!! EVER!!!!!!!!!!!!!!!!!!!!!!!!!!!!

Because of 1, it is trivial to create a matrix that will result in any garbage answer that you desire, at least in terms of floating point arithmetic. This is exactly what you have discovered, i.e., a matrix that should be singular, yet returns numerical trash from the determinant.

3. The determinant of a matrix can be written as the product of the eigenvalues. (See 1!) It also can be computed as the product of the diagonals of the U factor from an LU factorization. This is what det does.

Look at the diagonals from the LU factor of the matrix that you have created.

[L,U] = lu(A);
diag(U)
ans =
        100
       -100
       -200
       -300
       -300
       -500
       -600
       -700
       -700
        100
        100
       -200
       -300
        100
       -400
 1.4211e-14
prod(diag(U))
ans =
  -9.0245e+21

See that the matrix is clearly singular. Yet, the one tiny diagonal element, which is numerically zero compared to the rest in terms, is overwhelmed when you form the product.

det(A/100)
ans =
   7.0504e-11

Is the matrix singular? OF COURSE IT IS!

rank(A)
ans =
    15

4. Use rank to determine the rank of a matrix. If you know a little more about linear algebra, then you can use other tools like cond or svd.

cond(A)
ans =
   6.0814e+17

In the end, unless you thoroughly understand det, and know why and when not to use it, you should never use det. And IF you do understand all of this about det, then you should also know why you almost never need to use det!

MATLAB: How to find y value from the 3d graph with a known x and z values

Seems fairly simply, just build a scattered interpolant and use that to interpolate your query points.

x=[2.6,5.2,14,23.3,28.3,2.6,5.2,14,23.3,28.3,2.6,5.2,14,23.3,28.3];	
y=[3.5,6.9,17.7,29.1,35.2,3.5,6.8,17.5,28.7,34.8,3.4,6.6,16.9,27.6,33.5];
z=[450,450,450,450,450,350,350,350,350,350,250,250,250,250,250];
YfromXZ = scatteredInterpolant(x(:), z(:), y(:));  %interpolant that lets you calculate y knowing x and z
queryx = 20;    %can be a vector/matrix
queryz = 367;   %can be a vector/matrix as long as it's the same size as queryx
matchingy = YfromXZ(queryx, queryz);
%for visualisation
figure; scatter3(x, y, z); hold('on'); plot3(queryx, matchingy, queryz, 'r*');

Best Answer

Related Solutions

MATLAB: Problem in calculating a determinant for a matrix

MATLAB: How to find y value from the 3d graph with a known x and z values

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