An equation like y = Xh can be solved by inversion of X (i.e., h = X-1y ). And, it can be solved manually or using Python libraries, when the X is square). But, if X is not square or invertible then there is a well known recipe to solve the system approximately called the pseudo inverse (i.e., Ordinary Least Squares , ) Using the below Python references, drag and drop a code for solving the ĥ, assuming X and y are numpy arrays/matrices, and all required modules are imported. numpy.linalg.inv(a)
    Compute the (multiplicative) inverse of a matrix.
numpy.linalg.pinv(a) //fyi, the second parameter can not be considered
    Compute the (Moore-Penrose) pseudo-inverse of a matrix.
numpy.dot(a, b, out=None)
    Dot product of two arrays.
numpy.transpose(a, axes=None)
    Permute the dimensions of an array.
numpy.ndarray.T
    Same as self.transpose(), except that self is returned if self.ndim

Question

An equation like  y = Xh  can be solved by inversion of X (i.e., h = X-1y ). And, it can be solved manually or using Python libraries, when the X is square). But, if X is not square or invertible then there is a well known recipe to solve the system approximately called the  pseudo inverse  (i.e.,  Ordinary Least Squares ,    ) Using the below Python references, drag and drop a code for solving  the  ĥ,  assuming X and y are  numpy  arrays/matrices, and all required modules are imported. numpy.linalg.inv(a) 
     Compute the (multiplicative) inverse of a matrix. 
 numpy.linalg.pinv(a)   //fyi, the second parameter can not be considered  
     Compute the (Moore-Penrose) pseudo-inverse of a matrix. 
 numpy.dot(a, b, out=None) 
     Dot product of two arrays. 
 numpy.transpose(a, axes=None) 
     Permute the dimensions of an array. 
 numpy.ndarray.T 
     Same as self.transpose(), except that self is returned if self.ndim < 2.

Answer

h_hat = np.dot (pinv(np.dot(X.T, X)), y)

Answer

h_hat = np.dot(pinv(np.dot(X.T, X)),X.T)

Answer

h_hat = np.dot (np.dot(pinv(np.dot(X.T, X)),X.T), y)

Answer

Not Answering.

Answer

h_hat = np.dot (np.dot(pinv(X.T)),X.T), y)

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An equation like y = Xh can be solved by inversion of X (i.e., h = X-1y ). And...

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