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How MATLAB makes the distinction between P-Cores and E-Cores?

  It is known that modern CPUs have both Performance cores (P-cores) and efficiency cores (E-cores), different types of CPU cores that have different purposes and are designed for different tasks. P-cores typically have higher clock speeds and designed for high-performance tasks, while E-cores operate at lower clock speeds and focus on energy-efficient processing. In MATLAB, maxNumCompThreads returns the current maximum number of computational threads. Currently, the maximum number of computational threads is equal to the number of physical cores on your machine. How MATLAB makes the distinction between P-Cores and E-Cores ? NOTE:- Matlabsolutions.com  provide latest  MatLab Homework Help, MatLab Assignment Help  ,  Finance Assignment Help  for students, engineers and researchers in Multiple Branches like ECE, EEE, CSE, Mechanical, Civil with 100% output.Matlab Code for B.E, B.Tech,M.E,M.Tech, Ph.D. Scholars with 100% privacy guaranteed. Get MATLAB projects...
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Why does a matrix have to be square to get an inverse matrix?

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Recall that a matrix A is said to be invertible if there exists a matrix B such that A⋅B=I and B⋅A=I, where a priori these identity matrices may have different sizes.  In order for this to make sense, if A is an n×m matrix, then B must be an m×n matrix, so we could write this as

A⋅B=In
B⋅A=Im
A priori it might be possible for this to happen when m≠n, in which case a non-square matrix would have an inverse.  However, this is not actually possible.  Since it is not possible, the condition for invertibility is sometimes shortened to AB=BA=I, as however, this is not the real reason why only square matrices have inverses.
As is often the case with linear algebra concepts, it is better to think of matrix inversion in terms of linear transformations instead of matrices.  In the language of linear transformations, the fact that only square matrices are invertible is a consequence of the fact that "dimension of vector spaces is invariant under invertible linear transformations." 

Suppose we have two finite-dimensional vector spaces V and W and that f:V→W is a linear transformation.  Then the following three things are equivalent:
  1. f is one-to-one and onto:  for every vector w∈W there exists a unique v∈V with f(v)=w. 
  2. There exists an inverse linear transformation g:W→V to f, i.e. g∘f=idV and f∘g=idW.
  3. After choosing bases for V and W, the matrix representing f is invertible.

These equivalences are easy to see.  Condition 1 implies condition 2 since we can define g to be the inverse function f−1, and then it turns out that g is actually linear.  In the opposite direction, Condition 2 implies condition 1 since g∘f=idV implies f is one-to-one and f∘g=idW implies f is onto.  Condition 2 implies condition 3 since the matrix corresponding to f has the matrix corresponding to g as its inverse; conversely, if the matrix corresponding to f has an inverse matrix, then the linear transformation corresponding to that matrix will be an inverse linear transformation to f.

A linear transformation satisfying any of the above 3 equivalent conditions is often called an isomorphism.

Coming back down to earth a bit, suppose we have an n×m matrix A that admits an inverse.  The corresponding linear transformation f:Rm→Rn is one-to-one and onto.

Claim: If f:V→W is a one-to-one and onto linear transformation, then dim⁡V=dim⁡W.

In other words, dimension is an isomorphism invariant. 

Since the dimension of Rn is n, this will imply that n=m whenever A is an invertible n×m matrix.

Proof of claim: If {e1,…,em} is a basis for V then {f(e1),…,f(em)} is a basis for W. 

Indeed, the set {f(e1),…,f(em)} is linearly independent since f is one-to-one--any relation

a1f(e1)+a2f(e2)+⋯+amf(em)=0

can be rearranged to

f(a1e1+a2e2+⋯+amem)=0=f(0)

by linearity, which then forces

a1e1+⋯+amem=0,

since f is one-to-one, giving ai=0 for all i since e1,…,em are a basis for V. 

On the other hand {f(e1),…,f(em)} spans W.  Since f is onto, for any w∈W we can find some v∈V with f(v)=w. Then we can write

v=a1e1+⋯+amem

and hence

w=f(v)=a1f(e1)+⋯+amf(em)

by linearity.

Thus both V and W are m-dimensional.   

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