1. The problem presents three matrices A, B, and C with numeric values arranged in 10 rows and 5 columns each. 2. Since no specific question is asked, we can analyze or summarize the data, such as calculating means or identifying patterns. 3. For example, to find the mean of each matrix, sum all elements and divide by the total number of elements (10 rows \times 5 columns = 50 elements). 4. The formula for the mean of matrix $M$ is $$\text{mean}(M) = \frac{\sum_{i=1}^{n}\sum_{j=1}^{m} M_{ij}}{n \times m}$$ where $n=10$, $m=5$. 5. Calculate sums: - Sum of A: $\sum A = 5.68 + 4.72 + 1.71 + 7.52 + 2.03 + 0.47 + 9.48 + 6.92 + 6.83 + 0.58 + 0.36 - 3.86 + 0.27 - 3.20 + 8.41 + 4.20 + 0.45 + 0.30 + 1.58 + 0.97 + 3.94 + 0.67 + 4.85 - 9.96 + 0.38 + 6.12 - 6.30 + 0.04 + 0.74 + 7.15 + 0.76 + 2.59 + 2.56 + 0.07 + 0.97 + 0.21 + 0.71 + 0.75 - 0.35 + 0.34 + 8.30 - 0.54 + 0.38 + 2.72 + 1.23 + 1.47 + 3.12 + 0.42 + 34 + 0.65 = 102.88$ - Sum of B: $\sum B = 4.43 + 7.42 + 4.89 + 2.97 + 2.08 + 2.95 - 3.13 + 7.51 - 1.68 + 9.35 + 0.60 + 6.80 + 0.08 + 5.02 + 0.82 + 7.14 + 1.42 + 1.56 - 4.26 + 0.37 + 0.79 + 0.72 + 4.43 + 0.79 + 3.42 + 1.82 - 0.83 + 0.72 + 1.84 + 2.17 + 6.08 + 0.03 + 5.64 + 0.30 + 5.42 + 5.47 + 9.11 + 0.38 + 2.75 + 0.26 + 0.31 + 0.76 + 0.71 + 0.47 + 0.31 + 0.97 + 0.39 + 0.62 + 1.16 + 1.74 = 100.88$ - Sum of C: $\sum C = 2.41 + 9.36 + 9.80 + 4.56 + 9.16 + 9.85 - 7.80 + 4.95 - 1.23 + 0.98 + 0.69 + 0.05 + 7.76 + 0.75 + 1.50 + 0.53 + 3.19 + 0.75 + 6.81 + 4.24 + 1.27 - 1.47 + 1.39 + 0.19 + 0.72 + 3.16 + 0.32 + 3.04 + 0.90 + 0.69 + 0.90 + 0.45 + 0.16 + 7.75 + 3.54 + 0.41 + 2.81 + 0.39 + 0.16 + 0.36 + 2.17 + 0.72 + 1.47 - 0.37 + 0.65 + 0.36 + 3.13 + 0.56 + 5.54 + 6.39 = 100.88$ 6. Calculate means: - Mean of A: $\frac{102.88}{50} = 2.0576$ - Mean of B: $\frac{100.88}{50} = 2.0176$ - Mean of C: $\frac{100.88}{50} = 2.0176$ 7. These means give a general idea of the average values in each matrix. Final answer: - Mean of matrix A is approximately $2.06$. - Mean of matrix B is approximately $2.02$. - Mean of matrix C is approximately $2.02$.