(→‎Part C: Application of Linearity)
(→‎Part C: Application of Linearity)
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2.  No.  <math>[Secret Message]*[Secret Matrix]=[Encoded Message]\!</math>.  Thus the only way to solve for the secret message if the encoded message were known would be to multiply both sides by the inverse of the 3-by-3 secret matrix.
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2.  Yes.  <math>[Secret Message]*[Secret Matrix]=[Encoded Message]\!</math>.  <br>
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Revision as of 11:34, 18 September 2008

Part C: Application of Linearity

1. Bob can decrypt the message by multiplying it (in groups of 3 numbers) by the inverse of the 3-by-3 secret matrix.

2. Yes. $ [Secret Message]*[Secret Matrix]=[Encoded Message]\! $.

3.

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

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