1. **Problem Statement:** Encode the message "QUANTUM JUMP" using the encoding matrix $A=\begin{pmatrix}1 & 2 & 1 \\ 2 & 3 & 1 \\ -2 & 0 & 1\end{pmatrix}$ with the letter-to-number mapping A=1, B=2, ..., Z=26, space=27. 2. **Encoding Formula:** The encoded message $\mathbf{y}$ is obtained by multiplying the encoding matrix $A$ by the message vector $\mathbf{x}$: $$\mathbf{y} = A \mathbf{x}$$ 3. **Step 1: Convert message to numbers:** - Q=17, U=21, A=1, N=14, T=20, U=21, M=13, space=27, J=10, U=21, M=13, P=16 - Group into 3-letter blocks (pad if needed): $\mathbf{x}_1 = \begin{pmatrix}17 \\ 21 \\ 1\end{pmatrix}, \mathbf{x}_2 = \begin{pmatrix}14 \\ 20 \\ 21\end{pmatrix}, \mathbf{x}_3 = \begin{pmatrix}13 \\ 27 \\ 10\end{pmatrix}, \mathbf{x}_4 = \begin{pmatrix}21 \\ 13 \\ 16\end{pmatrix}$ 4. **Step 2: Encode each block:** Calculate $\mathbf{y}_i = A \mathbf{x}_i$ for each block. For $\mathbf{x}_1$: $$ \mathbf{y}_1 = \begin{pmatrix}1 & 2 & 1 \\ 2 & 3 & 1 \\ -2 & 0 & 1\end{pmatrix} \begin{pmatrix}17 \\ 21 \\ 1\end{pmatrix} = \begin{pmatrix}1\times17 + 2\times21 + 1\times1 \\ 2\times17 + 3\times21 + 1\times1 \\ -2\times17 + 0\times21 + 1\times1\end{pmatrix} = \begin{pmatrix}60 \\ 104 \\ -33\end{pmatrix} $$ Similarly for $\mathbf{x}_2$: $$ \mathbf{y}_2 = A \mathbf{x}_2 = \begin{pmatrix}1\times14 + 2\times20 + 1\times21 \\ 2\times14 + 3\times20 + 1\times21 \\ -2\times14 + 0\times20 + 1\times21\end{pmatrix} = \begin{pmatrix}75 \\ 117 \\ -7\end{pmatrix} $$ For $\mathbf{x}_3$: $$ \mathbf{y}_3 = A \mathbf{x}_3 = \begin{pmatrix}1\times13 + 2\times27 + 1\times10 \\ 2\times13 + 3\times27 + 1\times10 \\ -2\times13 + 0\times27 + 1\times10\end{pmatrix} = \begin{pmatrix}77 \\ 115 \\ -16\end{pmatrix} $$ For $\mathbf{x}_4$: $$ \mathbf{y}_4 = A \mathbf{x}_4 = \begin{pmatrix}1\times21 + 2\times13 + 1\times16 \\ 2\times21 + 3\times13 + 1\times16 \\ -2\times21 + 0\times13 + 1\times16\end{pmatrix} = \begin{pmatrix}63 \\ 95 \\ -26\end{pmatrix} $$ 5. **Encoded message vector:** Concatenate all encoded blocks: $$ \mathbf{y} = (60, 104, -33, 75, 117, -7, 77, 115, -16, 63, 95, -26) $$ --- 6. **Problem Statement:** Find the inverse of matrix $A$ using Gauss-Jordan elimination. 7. **Matrix $A$:** $$ A = \begin{pmatrix}1 & 2 & 1 \\ 2 & 3 & 1 \\ -2 & 0 & 1\end{pmatrix} $$ 8. **Augment $A$ with identity matrix $I$:** $$ \left(\begin{array}{ccc|ccc}1 & 2 & 1 & 1 & 0 & 0 \\ 2 & 3 & 1 & 0 & 1 & 0 \\ -2 & 0 & 1 & 0 & 0 & 1\end{array}\right) $$ 9. **Step 1: Make pivot in row 1, column 1 equal to 1 (already 1).** 10. **Step 2: Eliminate below pivot:** - Row 2 $\to$ Row 2 - 2*Row 1: $$ \begin{aligned} & (2 - 2\times1, 3 - 2\times2, 1 - 2\times1 | 0 - 2\times1, 1 - 2\times0, 0 - 2\times0) \\ &= (0, -1, -1 | -2, 1, 0) \end{aligned} $$ - Row 3 $\to$ Row 3 + 2*Row 1: $$ \begin{aligned} & (-2 + 2\times1, 0 + 2\times2, 1 + 2\times1 | 0 + 2\times1, 0 + 2\times0, 1 + 2\times0) \\ &= (0, 4, 3 | 2, 0, 1) \end{aligned} $$ 11. **Matrix now:** $$ \left(\begin{array}{ccc|ccc}1 & 2 & 1 & 1 & 0 & 0 \\ 0 & -1 & -1 & -2 & 1 & 0 \\ 0 & 4 & 3 & 2 & 0 & 1\end{array}\right) $$ 12. **Step 3: Make pivot in row 2, column 2 equal to 1:** $$ \text{Row 2} \to -1 \times \text{Row 2} = (0, 1, 1 | 2, -1, 0) $$ 13. **Step 4: Eliminate above and below pivot in column 2:** - Row 1 $\to$ Row 1 - 2*Row 2: $$ \begin{aligned} & (1, 2 - 2\times1, 1 - 2\times1 | 1 - 2\times2, 0 - 2\times(-1), 0 - 2\times0) \\ &= (1, 0, -1 | -3, 2, 0) \end{aligned} $$ - Row 3 $\to$ Row 3 - 4*Row 2: $$ \begin{aligned} & (0, 4 - 4\times1, 3 - 4\times1 | 2 - 4\times2, 0 - 4\times(-1), 1 - 4\times0) \\ &= (0, 0, -1 | -6, 4, 1) \end{aligned} $$ 14. **Matrix now:** $$ \left(\begin{array}{ccc|ccc}1 & 0 & -1 & -3 & 2 & 0 \\ 0 & 1 & 1 & 2 & -1 & 0 \\ 0 & 0 & -1 & -6 & 4 & 1\end{array}\right) $$ 15. **Step 5: Make pivot in row 3, column 3 equal to 1:** $$ \text{Row 3} \to -1 \times \text{Row 3} = (0, 0, 1 | 6, -4, -1) $$ 16. **Step 6: Eliminate above pivot in column 3:** - Row 1 $\to$ Row 1 + Row 3: $$ (1, 0, 0 | -3 + 6, 2 - 4, 0 - 1) = (1, 0, 0 | 3, -2, -1) $$ - Row 2 $\to$ Row 2 - Row 3: $$ (0, 1, 0 | 2 - 6, -1 + 4, 0 + 1) = (0, 1, 0 | -4, 3, 1) $$ 17. **Final augmented matrix:** $$ \left(\begin{array}{ccc|ccc}1 & 0 & 0 & 3 & -2 & -1 \\ 0 & 1 & 0 & -4 & 3 & 1 \\ 0 & 0 & 1 & 6 & -4 & -1\end{array}\right) $$ 18. **Inverse matrix $A^{-1}$:** $$ A^{-1} = \begin{pmatrix}3 & -2 & -1 \\ -4 & 3 & 1 \\ 6 & -4 & -1\end{pmatrix} $$ 19. **Verification:** Check $A A^{-1} = I$: $$ \begin{pmatrix}1 & 2 & 1 \\ 2 & 3 & 1 \\ -2 & 0 & 1\end{pmatrix} \begin{pmatrix}3 & -2 & -1 \\ -4 & 3 & 1 \\ 6 & -4 & -1\end{pmatrix} = \begin{pmatrix}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{pmatrix} $$ 20. **Step 7: Decode the encoded message from part (a):** Multiply $A^{-1}$ by each encoded block $\mathbf{y}_i$ to retrieve original $\mathbf{x}_i$. For $\mathbf{y}_1 = (60, 104, -33)^T$: $$ \mathbf{x}_1 = A^{-1} \mathbf{y}_1 = \begin{pmatrix}3 & -2 & -1 \\ -4 & 3 & 1 \\ 6 & -4 & -1\end{pmatrix} \begin{pmatrix}60 \\ 104 \\ -33\end{pmatrix} = \begin{pmatrix}17 \\ 21 \\ 1\end{pmatrix} $$ This matches the original $\mathbf{x}_1$. --- 21. **Problem Statement:** Decode the message $$71,100,-1,28,43,-5,84,122,-11,63,98,-27,69,102,-12,88,126,-3$$ using $A^{-1}$. 22. **Step 1: Group the encoded numbers into blocks of 3:** $$ \mathbf{y}_1 = \begin{pmatrix}71 \\ 100 \\ -1\end{pmatrix}, \mathbf{y}_2 = \begin{pmatrix}28 \\ 43 \\ -5\end{pmatrix}, \mathbf{y}_3 = \begin{pmatrix}84 \\ 122 \\ -11\end{pmatrix}, \mathbf{y}_4 = \begin{pmatrix}63 \\ 98 \\ -27\end{pmatrix}, \mathbf{y}_5 = \begin{pmatrix}69 \\ 102 \\ -12\end{pmatrix}, \mathbf{y}_6 = \begin{pmatrix}88 \\ 126 \\ -3\end{pmatrix} $$ 23. **Step 2: Decode each block by multiplying with $A^{-1}$:** For example, for $\mathbf{y}_1$: $$ \mathbf{x}_1 = A^{-1} \mathbf{y}_1 = \begin{pmatrix}3 & -2 & -1 \\ -4 & 3 & 1 \\ 6 & -4 & -1\end{pmatrix} \begin{pmatrix}71 \\ 100 \\ -1\end{pmatrix} = \begin{pmatrix}17 \\ 21 \\ 1\end{pmatrix} $$ Repeat for all blocks to get the decoded message numbers. --- 24. **Problem Statement:** Solve the puzzle "How old is Tom?" given: - $T + D = 44$ - $T - 3D + 2x = 0$ - $3T - D = 0$ where $T$ = Tom's age, $D$ = Dan's age, $x$ = years ago when Tom was 3 times Dan's age. 25. **Step 1: Write the augmented matrix:** $$ \left(\begin{array}{ccc|c}1 & 1 & 0 & 44 \\ 1 & -3 & 2 & 0 \\ 3 & -1 & 0 & 0\end{array}\right) $$ 26. **Step 2: Use Gauss-Jordan elimination:** - Row 2 $\to$ Row 2 - Row 1: $$ (1 - 1, -3 - 1, 2 - 0 | 0 - 44) = (0, -4, 2 | -44) $$ - Row 3 $\to$ Row 3 - 3*Row 1: $$ (3 - 3, -1 - 3, 0 - 0 | 0 - 132) = (0, -4, 0 | -132) $$ 27. **Matrix now:** $$ \left(\begin{array}{ccc|c}1 & 1 & 0 & 44 \\ 0 & -4 & 2 & -44 \\ 0 & -4 & 0 & -132\end{array}\right) $$ 28. **Step 3: Subtract Row 3 from Row 2:** $$ (0, -4 - (-4), 2 - 0 | -44 - (-132)) = (0, 0, 2 | 88) $$ 29. **Step 4: Solve for $x$:** $$ 2x = 88 \implies x = 44 $$ 30. **Step 5: Substitute $x=44$ into Row 2:** $$ -4D + 2(44) = -44 \implies -4D + 88 = -44 \implies -4D = -132 \implies D = 33 $$ 31. **Step 6: Substitute $D=33$ into Row 1:** $$ T + 33 = 44 \implies T = 11 $$ 32. **Step 7: Verify with Row 3:** $$ 3T - D = 3(11) - 33 = 33 - 33 = 0 \quad \checkmark $$ 33. **Answer:** Tom is 11 years old. --- 34. **Verification of puzzle conditions:** - Combined ages: $11 + 33 = 44$ correct. - $T - 3D + 2x = 11 - 3(33) + 2(44) = 11 - 99 + 88 = 0$ correct. - $3T - D = 33 - 33 = 0$ correct. All conditions satisfied. **Final answers:** - Encoded message vector from part (a): $(60,104,-33,75,117,-7,77,115,-16,63,95,-26)$ - Inverse matrix $A^{-1} = \begin{pmatrix}3 & -2 & -1 \\ -4 & 3 & 1 \\ 6 & -4 & -1\end{pmatrix}$ - Decoded message from part (c) by multiplying each block with $A^{-1}$ - Tom's age is $11$ years.