I am using the Identity matrix to find an inverse of a given matrix, A. I would like to pull out just the inverse matrix after the rref function, and assign it to its own variable, but I don't know Matlab well enough to figure it out on my own. Here's my code:
A = [1,2,-2; 1,1,1; 0,0,1;]
I = [1,0,0; 0,1,0; 0,0,1;]
Ainv = [A,I]
rref(Ainv)
Any help would be greatly appreciated!
You could use getfield(rref(Ainv), { [1:3] [4:6] }) but inv(A), or using a temporary variable is better as mentioned in the comments.
Related
I am trying to replicate this formula:
I have gathered all variables in my workspace. However estimating vec(Theta') does not seem to work and so I am a little bit stuck.
Theta = A*B-C;
vTheta = vec(Theta');
A, B and C are defined. The problem is that MATLAB does not seem to know the function vec to do what I would like to do with Theta as in the formula.
How to fix this?
I don't know where you got that equation from, but vec is a function in R, maybe it's related to that? If you want to convert a matrix Theta into a vector, do
Theta(:)
Edit: If you need to transpose the matrix first, MATLAB might not let you do Theta'(:). Instead do it in two steps:
tmp = Theta'; tmp(:)
As written above the Colon Operator is the way vectorize defined variable.
Yet, sometime we want to vectorize a sub set of a variable.
Let's say we have a matrix - mA and we'd like to vectorize a sub section of it - mA(2:3, 4:7).
One way is to define a new variable and vectorize it:
vA = mA(2:3, 4:7);
vA = vA(:);
Yet, what if we only wanted to use this inside another expression and only once?
Could we escape the need to generate explicit variable?
Well, unfortunately MATLAB doesn't have the view() functionality like in Julia.
Yet if you want to avoid explicitly naming new variable (I'm not sure if MATLAB's JIT Engine can also void the memory allocation as Julia) you can do:
reshape(mA(2:3, 4:7), [], 1)
This will always yield a column vector.
You can also use:
reshape(mA(2:3, 4:7), 1, [])
To generate row vector.
For instance you can do:
reshape(mA(2:3, 4:7), 1, []) * reshape(mA(2:3, 4:7), [], 1, )
This will be the sum of squared values of those elements.
I am trying to solve a large system of linear equations in matlab (about 3600 equations!) Since the number of variables is too many, I have to define them in a matrix. So I use this code to define the variables:
A = sym('A',[60,60]);
It has to be a 60x60 matrix because the variables correspond to a finite difference problem (i.e. a mesh grid).
After writing the corresponding equations in a for loop, I use the solve function in this way:
mysol = solve(eq,A);
Where eq is the equations matrix.
My problem is that, when I try to solve this system as shown in MATLAB help, i.e. writing something like this:
C(1,1) = mysol.A(1,1)
I get an error that says: "Reference to non-existent field 'A'". But if I write something like:
C(1,1) = mysol.A1_1
then it works.
Does anyone know how I can fix this? I don't want to do this for every variable!
You could try accessing the fields in the structure dynamically like so:
for ii = 1:60
for jj = 1:60
field = sprintf('A%d_%d', ii, jj);
C(ii, jj) = mysol.(field);
end
end
I have image, and I want to do up sampling. First of all I need to plug zeros between pixels such that [1,2,3] transforms to [1,0,2,0,3]. Can anyone tell me how to do it without using paddarray and without using for loops?
Thank you in advance!
Something like this?:
B=zeros(size(img)*2);
B(1:2:end,1:2:end)=img;
However there are ways of up-sampling in matlab without having the need of doing it by hand, for example interp2
You could also make use of MATLAB's way of dynamically allocating variables if you don't specify a number for an index into the array. By omitting indexing into certain locations in your array, MATLAB will fill in these values with zeroes by default. As such:
B(1:2:5) = 1:3
B =
1 0 2 0 3
V = [1,2,3];
padded(numel(V)*2) = 0;
padded(1:2:end) = V
And then just deal with the trailing zero if numel(V) was odd
There is a function upsample that does exactly this from Octave-Forge -- see docs on upsample.
Or you can look at the source of upsample to see what implements it. Are you opposed to using a package or a function?
I am trying to compute the following expected value for Z being lognormally distributed
E[Z^eta w(F_Z (Z))^-eta]
where eta is a real number, F_Z the distribution function of Z and w:[0,1]->[0,1] an increasing function.
First of all, I am pretty new to Matlab so I don't know which way of integrating is the better one, numerically or symbolically. I tried symbolically.
My idea was to subsequently define functions:
syms x;
g_1(x) = x^eta;
g_2(x) = logncdf(x);
g_2(x) = w(x)^-eta;
g_4(x) = g_1(x) * g_3(g_2(x));
And then
exp = int(g_4(x),x,0,inf)
Unfortunately this doesn't work and MATLAB just posts the whole expression of g_4...
Is it better to use the numerical integration quadqk? What am I doing wrong here? I already read something about MATLAB not being the best program for integration but I have to use it so switching to a different program does not help.
Thanks a lot!
In Matlab I'm trying to find points in a 3d matrix whose coordinates are smaller than some function.
If these coordinates are equal to some functions than I can write:
A(some_function1,some_function2,some_function3)=2;
But what if I want to do something like:
A(<some_function1,<some_function2,<some_function3)=2;
This isn't working - so what is the other way of finding such points without using "for" loop? Unfortunately with "for" loop my code takes a lot of time to compute. Thank you for your help!
How about something along the lines of
A( ceil(min(some_function1,size(A,1))),...
ceil(min(some_function2,size(A,2))),...
ceil(min(some_function3,size(A,3))) );
This will cap the indicies to the end of each array dimension
You can just use regular indexing to achieve this:
A(1:floor(some_function1),1:floor(some_function2),1:floor(some_function3)) = 2;
assuming you check / ensure that floor(some_function*) is smaller than the dimensions of A
Try:
A(1:size(A,1)<some_function1, 1:size(A,2)<some_function2, 1:size(A,3)<some_function3) = 2
I hope I got your question correctly.