1. **Problem Statement:** Decode the message transmitted by the matrix $$M' = \begin{pmatrix} 51 & 40 & 36 & 51 & 37 & 42 \\ 85 & 81 & 71 & 114 & 64 & 84 \\ 126 & 130 & 113 & 196 & 100 & 135 \end{pmatrix}$$ encoded by the matrix $$A = \begin{pmatrix} 1 & 1 & 1 \\ 1 & 2 & 3 \\ 1 & 3 & 6 \end{pmatrix}$$ with the given correspondence table between letters and numbers. 2. **Formula and Approach:** The encoding is done by multiplying the encoding matrix $$A$$ by the message matrix $$M$$: $$M' = A \times M$$ To decode, we need to find $$M$$: $$M = A^{-1} \times M'$$ 3. **Calculate the inverse of matrix $$A$$:** First, compute the determinant of $$A$$: $$\det(A) = 1(2 \times 6 - 3 \times 3) - 1(1 \times 6 - 3 \times 1) + 1(1 \times 3 - 2 \times 1) = 1(12 - 9) - 1(6 - 3) + 1(3 - 2) = 3 - 3 + 1 = 1$$ Since $$\det(A) = 1$$, the inverse exists and is the adjugate matrix. Calculate the adjugate matrix of $$A$$: $$\text{adj}(A) = \begin{pmatrix} (2 \times 6 - 3 \times 3) & -(1 \times 6 - 3 \times 1) & (1 \times 3 - 2 \times 1) \\ -(1 \times 6 - 3 \times 1) & (1 \times 6 - 1 \times 1) & -(1 \times 3 - 1 \times 1) \\ (1 \times 3 - 2 \times 1) & -(1 \times 3 - 1 \times 1) & (1 \times 2 - 1 \times 1) \end{pmatrix} = \begin{pmatrix} 3 & -3 & 1 \\ -3 & 5 & -2 \\ 1 & -2 & 1 \end{pmatrix}$$ Since $$\det(A) = 1$$, $$A^{-1} = \text{adj}(A)$$: $$A^{-1} = \begin{pmatrix} 3 & -3 & 1 \\ -3 & 5 & -2 \\ 1 & -2 & 1 \end{pmatrix}$$ 4. **Calculate $$M = A^{-1} \times M'$$:** Multiply $$A^{-1}$$ (3x3) by $$M'$$ (3x6): For each column $$j$$ of $$M'$$, compute: $$M_{:,j} = A^{-1} \times M'_{:,j}$$ Calculate each element of $$M$$: - Column 1: $$\begin{pmatrix}3 & -3 & 1 \\ -3 & 5 & -2 \\ 1 & -2 & 1 \end{pmatrix} \times \begin{pmatrix}51 \\ 85 \\ 126 \end{pmatrix} = \begin{pmatrix}3 \times 51 - 3 \times 85 + 1 \times 126 \\ -3 \times 51 + 5 \times 85 - 2 \times 126 \\ 1 \times 51 - 2 \times 85 + 1 \times 126 \end{pmatrix} = \begin{pmatrix}153 - 255 + 126 \\ -153 + 425 - 252 \\ 51 - 170 + 126 \end{pmatrix} = \begin{pmatrix}24 \\ 20 \\ 7 \end{pmatrix}$$ - Column 2: $$\begin{pmatrix}3 & -3 & 1 \\ -3 & 5 & -2 \\ 1 & -2 & 1 \end{pmatrix} \times \begin{pmatrix}40 \\ 81 \\ 130 \end{pmatrix} = \begin{pmatrix}120 - 243 + 130 \\ -120 + 405 - 260 \\ 40 - 162 + 130 \end{pmatrix} = \begin{pmatrix}7 \\ 25 \\ 8 \end{pmatrix}$$ - Column 3: $$\begin{pmatrix}3 & -3 & 1 \\ -3 & 5 & -2 \\ 1 & -2 & 1 \end{pmatrix} \times \begin{pmatrix}36 \\ 71 \\ 113 \end{pmatrix} = \begin{pmatrix}108 - 213 + 113 \\ -108 + 355 - 226 \\ 36 - 142 + 113 \end{pmatrix} = \begin{pmatrix}8 \\ 21 \\ 7 \end{pmatrix}$$ - Column 4: $$\begin{pmatrix}3 & -3 & 1 \\ -3 & 5 & -2 \\ 1 & -2 & 1 \end{pmatrix} \times \begin{pmatrix}51 \\ 114 \\ 196 \end{pmatrix} = \begin{pmatrix}153 - 342 + 196 \\ -153 + 570 - 392 \\ 51 - 228 + 196 \end{pmatrix} = \begin{pmatrix}7 \\ 25 \\ 19 \end{pmatrix}$$ - Column 5: $$\begin{pmatrix}3 & -3 & 1 \\ -3 & 5 & -2 \\ 1 & -2 & 1 \end{pmatrix} \times \begin{pmatrix}37 \\ 64 \\ 100 \end{pmatrix} = \begin{pmatrix}111 - 192 + 100 \\ -111 + 320 - 200 \\ 37 - 128 + 100 \end{pmatrix} = \begin{pmatrix}19 \\ 9 \\ 9 \end{pmatrix}$$ - Column 6: $$\begin{pmatrix}3 & -3 & 1 \\ -3 & 5 & -2 \\ 1 & -2 & 1 \end{pmatrix} \times \begin{pmatrix}42 \\ 84 \\ 135 \end{pmatrix} = \begin{pmatrix}126 - 252 + 135 \\ -126 + 420 - 270 \\ 42 - 168 + 135 \end{pmatrix} = \begin{pmatrix}9 \\ 24 \\ 9 \end{pmatrix}$$ 5. **Decoded matrix $$M$$:** $$M = \begin{pmatrix} 24 & 7 & 8 & 7 & 19 & 9 \\ 20 & 25 & 21 & 25 & 9 & 24 \\ 7 & 8 & 7 & 19 & 9 & 9 \end{pmatrix}$$ 6. **Decode numbers to letters using the correspondence table:** Correspondences: C=17, Z=3, B=26, S=27, T=0, A=25, Y=16, D=6, P=23, I=19, J=8, L=13, R=12, W=1, K=20, Q=14, H=24, V=15, N=9, F=4, O=10, U=2, E=7, M=11, X=21, G=18 Decode each number in $$M$$: - Row 1: 24(H), 7(E), 8(J), 7(E), 19(I), 9(N) - Row 2: 20(K), 25(A), 21(X), 25(A), 9(N), 24(H) - Row 3: 7(E), 8(J), 7(E), 19(I), 9(N), 9(N) 7. **Final decoded message:** Row-wise reading: **H E J E I N K A X A N H E J E I N N** This is the decoded message from the matrix multiplication and correspondence table.