Chapter 11.4, Problem 87AYU

  $\left(a\right)$

To determine

To calculate: The inverse of the matrix $K=\left[\begin{array}{ccc}2& 1& 1\\ 1& 1& 0\\ 1& 1& 1\end{array}\right]$ .

Answer to Problem 87AYU

The inverse of the matrix $K=\left[\begin{array}{ccc}2& 1& 1\\ 1& 1& 0\\ 1& 1& 1\end{array}\right]$ is $K^{-1}=\left[\begin{array}{ccc}1& 0& -1\\ -1& 1& 1\\ 0& -1& 1\end{array}\right]$ .

Explanation of Solution

Given information:

The matrix $K=\left[\begin{array}{ccc}2& 1& 1\\ 1& 1& 0\\ 1& 1& 1\end{array}\right]$ .

Formula used:

For a matrix A, the inverse of a $3\times 3$ is given by the formula $A{A}^{-1}=I_3$ , where $I_3$ is the identity matrix.

Calculation:

Consider the matrix $K=\left[\begin{array}{ccc}2& 1& 1\\ 1& 1& 0\\ 1& 1& 1\end{array}\right]$ .

For a matrix A, the inverse of a $3\times 3$ is given by the formula $A{A}^{-1}=I_3$ , where $I_3$ is the identity matrix.

Construct the augmented matrix,

  $\left[\begin{array}{cccccc}2& 1& 1& 1& 0& 0\\ 1& 1& 0& 0& 1& 0\\ 1& 1& 1& 0& 0& 1\end{array}\right]$

Apply the transformation $R_2\to R_2-\left(0.5\right)R_1$ and $R_3\to R_3-\left(0.5\right)R_1$

  $\left[\begin{array}{cccccc}2& 1& 1& 1& 0& 0\\ 0& 0.5& -0.5& -0.5& 1& 0\\ 0& 0.5& 0.5& -0.5& 0& 1\end{array}\right]$

Apply the transformation $R_3\to R_3-R_2$

  $\left[\begin{array}{cccccc}2& 1& 1& 1& 0& 0\\ 0& 0.5& -0.5& -0.5& 1& 0\\ 0& 0& 1& 0& -1& 1\end{array}\right]$

Apply the transformation $R_1\to \frac{R_1}{2}$

  $\left[\begin{array}{cccccc}1& 0.5& 0.5& 0.5& 0& 0\\ 0& 0.5& -0.5& -0.5& 1& 0\\ 0& 0& 1& 0& -1& 1\end{array}\right]$

Apply the transformation $R_2\to 2R_2$

  $\left[\begin{array}{cccccc}1& 0.5& 0.5& 0.5& 0& 0\\ 0& 1& -1& -1& 2& 0\\ 0& 0& 1& 0& -1& 1\end{array}\right]$

Apply the transformation $R_1\to R_1-\left(0.5\right)R_2$

  $\left[\begin{array}{cccccc}1& 0& 1& 1& -1& 0\\ 0& 1& -1& -1& 2& 0\\ 0& 0& 1& 0& -1& 1\end{array}\right]$

Apply the transformation $R_2\to R_2+R_3$ and $R_1\to R_1-R_3$

  $\left[\begin{array}{cccccc}1& 0& 0& 1& 0& -1\\ 0& 1& 0& -1& 1& 1\\ 0& 0& 1& 0& -1& 1\end{array}\right]$

Therefore, inverse of the matrix is $\left[\begin{array}{ccc}1& 0& -1\\ -1& 1& 1\\ 0& -1& 1\end{array}\right]$ .

  $\left(b\right)$

To determine

To calculate: The decoded matrix from the encrypted matrix $E=\left[\begin{array}{ccc}47& 34& 33\\ 44& 36& 27\\ 47& 41& 20\end{array}\right]$ .

Answer to Problem 87AYU

The decoded matrix is $M=\left[\begin{array}{ccc}13& 1& 20\\ 8& 9& 19\\ 6& 21& 14\end{array}\right]$ .

Explanation of Solution

Given information:

The matrix $K=\left[\begin{array}{ccc}2& 1& 1\\ 1& 1& 0\\ 1& 1& 1\end{array}\right]$ and its inverse $K^{-1}=\left[\begin{array}{ccc}1& 0& -1\\ -1& 1& 1\\ 0& -1& 1\end{array}\right]$ .

Formula used:

When message matrix M is multiplied by key message K, the encrypted matrix E is obtained. That is $E=M\cdot K\text{ and }M=E\cdot k^{-1}$ .

Calculation:

Consider the matrix $K=\left[\begin{array}{ccc}2& 1& 1\\ 1& 1& 0\\ 1& 1& 1\end{array}\right]$ and its inverse $K^{-1}=\left[\begin{array}{ccc}1& 0& -1\\ -1& 1& 1\\ 0& -1& 1\end{array}\right]$ .

Recall that when message matrix M is multiplied by key message K, the encrypted matrix E is obtained. That is $E=M\cdot K\text{ and }M=E\cdot k^{-1}$ .

Here, encrypted matrix $E=\left[\begin{array}{ccc}47& 34& 33\\ 44& 36& 27\\ 47& 41& 20\end{array}\right]$ .

Therefore, message matrix is,

  $\begin{array}{c}M=E\cdot K^{-1}\\ =\left[\begin{array}{ccc}47& 34& 33\\ 44& 36& 27\\ 47& 41& 20\end{array}\right]\left[\begin{array}{ccc}1& 0& -1\\ -1& 1& 1\\ 0& -1& 1\end{array}\right]\\ =\left[\begin{array}{ccc}47-34& 34-33& -47+34+33\\ 44-36& 36-27& -44+36+27\\ 47-41& 41-20& -47+41+20\end{array}\right]\\ =\left[\begin{array}{ccc}13& 1& 20\\ 8& 9& 19\\ 6& 21& 14\end{array}\right]\end{array}$

Therefore, decoded matrix is $M=\left[\begin{array}{ccc}13& 1& 20\\ 8& 9& 19\\ 6& 21& 14\end{array}\right]$ .

  $\left(c\right)$

To determine

To explain: The original message.

Answer to Problem 87AYU

The original message is MATH IS FUN.

Explanation of Solution

Given information:

The decoded matrix is $M=\left[\begin{array}{ccc}13& 1& 20\\ 8& 9& 19\\ 6& 21& 14\end{array}\right]$ .

The decoded matrix is $M=\left[\begin{array}{ccc}13& 1& 20\\ 8& 9& 19\\ 6& 21& 14\end{array}\right]$ . Now each entry of the decoded matrix represent the position of English alphabet.

So, the matrix is rewritten as,

  $\left[\begin{array}{ccc}M& A& T\\ H& I& S\\ F& U& N\end{array}\right]$

Therefore, original message is MATH IS FUN.