Part 1

Bob can decrypt the message by multiplying the message he receives by the inverse of the encryption matrix

Part 2

Well eve could always do a left matrix divide of the (encryption matrix) \ (encrypted message) but that is the same as the inverse(encryption matrix) * (encrypted message). So Eve would most likely have to take the inverse of the matrix.

Part 3

The Decrypted matrix ends up being [2 23 5]


% input original message sent
message = [1 0 1;0 1 0;4 0 1]

% input encrypted matrix that corresponds to the original message
encryptMes = [2 0 0; 0 1 0; 0 0 3]

% input message that you want to decrypt
EncryptedMessage = [2;23;3]

% find what matrix is used to encrypt the message by taking the encrypted
% message and multiplying by the inverse of the message
EncryptionMatrix = encryptMes*inv(message)

% Take the inverse of the matrix used to encrypt the original message so
% that you can work backwards from and encrypted message to original
% message
InvEncryptionMatrix = inv(EncryptionMatrix)

% Decrypt the message by multiplying the inverse of the encryption matrix
% by the encrypted message.  Should leave you with the original decrypted
% message.
DecryptedMessage = InvEncryptionMatrix*EncryptedMessage

% Display of output for function...
% message =
% 
%      1     0     1
%      0     1     0
%      4     0     1
% 
% 
% encryptMes =
% 
%      2     0     0
%      0     1     0
%      0     0     3
% 
% 
% EncryptedMessage =
% 
%      2
%     23
%      3
% 
% 
% EncryptionMatrix =
% 
%    -0.6667         0    0.6667
%          0    1.0000         0
%     4.0000         0   -1.0000
% 
% 
% InvEncryptionMatrix =
% 
%     0.5000         0    0.3333
%          0    1.0000         0
%     2.0000         0    0.3333
% 
% 
% DecryptedMessage =
% 
%      2
%     23
%      5

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Basic linear algebra uncovers and clarifies very important geometry and algebra.

Dr. Paul Garrett