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pogonyaev
4 years ago
12

One of the simplest and best known polyalphabetic ciphers is _________ cipher. In this scheme, the set of related monoalphabetic

substitution rules consists of the 26 Caesar ciphers with shifts of 0 through 25. Each cipher is denoted by a key letter which is the ciphertext letter that substitutes for the plaintext letter a.
Computers and Technology
1 answer:
OleMash [197]4 years ago
3 0

Answer:

Vigenere

Explanation:

Vigenere Cipher is an encryption method of alphabetic symbols that utilizes an easy form of polyalphabetic exchange which is carried out through Vigenere table.

Hence, in this situation, the correct answer is VIGENERE Cipher because it is the form of cipher whereby the set of related monoalphabetic substitution rules comprises of the 26 Caesar ciphers with shifts of 0 through 25. Each cipher is indicated by a key letter which is the ciphertext letter that substitutes for the plaintext letter.

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What would you have to know about the pivot columns in an augmented matrix in order to know that the linear system is consistent
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Answer:

The Rouché-Capelli Theorem. This theorem establishes a connection between how a linear system behaves and the ranks of its coefficient matrix (A) and its counterpart the augmented matrix.

rank(A)=rank\left ( \left [ A|B \right ] \right )\:and\:n=rank(A)

Then satisfying this theorem the system is consistent and has one single solution.

Explanation:

1) To answer that, you should have to know The Rouché-Capelli Theorem. This theorem establishes a connection between how a linear system behaves and the ranks of its coefficient matrix (A) and its counterpart the augmented matrix.

rank(A)=rank\left ( \left [ A|B \right ] \right )\:and\:n=rank(A)

rank(A)

Then the system is consistent and has a unique solution.

<em>E.g.</em>

\left\{\begin{matrix}x-3y-2z=6 \\ 2x-4y-3z=8 \\ -3x+6y+8z=-5  \end{matrix}\right.

2) Writing it as Linear system

A=\begin{pmatrix}1 & -3 &-2 \\  2& -4 &-3 \\ -3 &6  &8 \end{pmatrix} B=\begin{pmatrix}6\\ 8\\ 5\end{pmatrix}

rank(A) =\left(\begin{matrix}7 & 0 & 0 \\0 & 7 & 0 \\0 & 0 & 7\end{matrix}\right)=3

3) The Rank (A) is 3 found through Gauss elimination

(A|B)=\begin{pmatrix}1 & -3 &-2  &6 \\  2& -4 &-3  &8 \\  -3&6  &8  &-5 \end{pmatrix}

rank(A|B)=\left(\begin{matrix}1 & -3 & -2 \\0 & 2 & 1 \\0 & 0 & \frac{7}{2}\end{matrix}\right)=3

4) The rank of (A|B) is also equal to 3, found through Gauss elimination:

So this linear system is consistent and has a unique solution.

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