(New page: Some definitions and explainations 1.Let A be the 3x3 matrix used to encrypt the message. <math>\,A=\left[ \begin{array}{ccc} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{3...) |
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Some definitions and explainations | Some definitions and explainations | ||
| − | + | == How can Bob Decrypt the Message? == | |
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| + | Let A be the 3x3 matrix used to encrypt the message. | ||
<math>\,A=\left[ \begin{array}{ccc} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{array} \right] \,</math> | <math>\,A=\left[ \begin{array}{ccc} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{array} \right] \,</math> | ||
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<math>\,B=\left[ \begin{array}{ccc} b_{11} & b_{12} & b_{13} \\ b_{21} & b_{22} & b_{23} \\ b_{31} & b_{32} & b_{33} \end{array} \right] \,</math> | <math>\,B=\left[ \begin{array}{ccc} b_{11} & b_{12} & b_{13} \\ b_{21} & b_{22} & b_{23} \\ b_{31} & b_{32} & b_{33} \end{array} \right] \,</math> | ||
| − | + | Correspondingly, C is the decrypted message | |
| − | + | From the poblem,we know: C=B*A | |
| − | + | Based on the multiplication of Matrix we can find that | |
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| + | <math>\,C_1 = B_1*A\,</math>, same for <math>C_2</math> and <math>C_3</math> | ||
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<math>\,e=mA\,</math> | <math>\,e=mA\,</math> | ||
Revision as of 10:47, 18 September 2008
Some definitions and explainations
Contents
How can Bob Decrypt the Message?
Let A be the 3x3 matrix used to encrypt the message.
$ \,A=\left[ \begin{array}{ccc} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{array} \right] \, $
Let B be the 3x3 matrix for the unencrypted message.
$ \,B=\left[ \begin{array}{ccc} b_{11} & b_{12} & b_{13} \\ b_{21} & b_{22} & b_{23} \\ b_{31} & b_{32} & b_{33} \end{array} \right] \, $
Correspondingly, C is the decrypted message
From the poblem,we know: C=B*A
Based on the multiplication of Matrix we can find that
$ \,C_1 = B_1*A\, $, same for $ C_2 $ and $ C_3 $
$ \,e=mA\, $
which is how the message is being encrypted. If we multiply both sides by the inverse of $ \,A\, $, we get
$ \,eA^{-1}=mAA^{-1}=mI=m\, $
Therefore, we can get the original message back if we multiply the encrypted message by $ \,A^{-1}\, $, given that the inverse of $ \,A\, $ exists.
Can Eve Decrypt the Message Without Finding the Inverse of A?
Yes, because of the fact $ \,e=mA\, $ is linear and Eve was given three linearly independent vector responses to the system and their corresponding inputs.
Proof of Linearity
Say we have two inputs $ \,m_1\, $ and $ \,m_2\, $ yielding outputs
$ \,e_1=m_1A\, $ and
$ \,e_2=m_2A\, $, respectively.
thus,
$ \,ae_1+be_2=am_1A+bm_2A $ for any $ \,a,b\in \mathbb{R}\, $
Now, apply $ \,am_1+bm_2\, $ to the system
$ \,(am_1+bm_2)A=am_1A+bm_2A\, $
Since the two results are equal
$ \,am_1A+bm_2A=am_1A+bm_2A\, $
the system is linear.
Main Proof
Since Eve was given that for the system
$ \,m_1=(1,0,4)\, $ yields $ \,e_1=(2,0,0)\, $
$ \,m_2=(0,1,0)\, $ yields $ \,e_2=(0,1,0)\, $
$ \,m_3=(1,0,1)\, $ yields $ \,e_3=(0,0,3)\, $
where $ \,e_1, e_2, e_3\, $ are clearly linearly independent vectors, Eve can take any encrypted message and write it as a linear combination of $ \,e_1, e_2, e_3\, $
$ \,\exists a,b,c\in \mathbb{R}\, $ such that $ \,e=ae_1+be_2+ce_3\, $, for any $ \,e\in \mathbb{R}^{3}\, $
Because the system is linear, we can write the input as
$ \,m=am_1+bm_2+cm_3\, $
thus, the message has been decrypted without knowing $ \,A^{-1}\, $.
What is the Decrypted Message?
The given encrypted message is
$ \,e=(2,23,3)\, $
This can be rewritten as a linear combination of the given system result vectors
$ \,e=ae_1+be_2+ce_3\, $
$ \,e=(2,23,3)=1\cdot (2,0,0)+23\cdot (0,1,0)+1\cdot (0,0,3)\, $
Because the system s linear, we can write the input as
$ \,m=am_1+bm_2+cm_3\, $
$ \,m=1\cdot (1,0,4)+23\cdot (0,1,0)+1\cdot (1,0,1)=(2,23,5)\, $
Therefore, the unencrypted message is "BWE".
