1. **Problem Statement:** Given the matrix $$S = \begin{bmatrix}1 & 0 & 2 \\ 0 & 1 & 2 \\ 3 & 0 & 3\end{bmatrix}$$ and the vector $$a = \begin{bmatrix}1 \\ 2 \\ 3\end{bmatrix},$$ evaluate the following expressions: (a) $$S_{ii}$$ (sum of diagonal elements of $$S$$), (b) $$S_{ij} S_{ij}$$ (sum of squares of all elements of $$S$$), (c) $$S_{ji} S_{ji}$$ (same as (b) because summation indices are dummy), (d) $$S_{jk} S_{kj}$$ (sum over indices $$j,k$$), (e) $$a_m a_m$$ (sum of squares of elements of $$a$$), (f) $$S_{mn} a_m a_n$$ (double sum with $$S$$ and $$a$$), (g) $$S_{nm} a_m a_n$$ (similar to (f) but indices swapped). 2. **Important Rules and Notation:** - Repeated indices imply summation (Einstein summation convention). - $$S_{ii}$$ means sum of diagonal elements: $$\sum_i S_{ii}$$. - For sums like $$S_{ij} S_{ij}$$, sum over all $$i,j$$. - Vector $$a$$ components are $$a_1=1, a_2=2, a_3=3$$. 3. **Step-by-step Evaluation:** (a) $$S_{ii} = S_{11} + S_{22} + S_{33} = 1 + 1 + 3 = 5$$. (b) Calculate $$S_{ij} S_{ij} = \sum_{i=1}^3 \sum_{j=1}^3 (S_{ij})^2$$: - Row 1: $$1^2 + 0^2 + 2^2 = 1 + 0 + 4 = 5$$ - Row 2: $$0^2 + 1^2 + 2^2 = 0 + 1 + 4 = 5$$ - Row 3: $$3^2 + 0^2 + 3^2 = 9 + 0 + 9 = 18$$ Sum total: $$5 + 5 + 18 = 28$$. (c) $$S_{ji} S_{ji}$$ is the same as (b) because indices are dummy, so answer is $$28$$. (d) Evaluate $$S_{jk} S_{kj} = \sum_{j=1}^3 \sum_{k=1}^3 S_{jk} S_{kj}$$. Calculate the matrix product $$S \times S^T$$ and sum all elements: - Compute $$S S^T$$: $$S = \begin{bmatrix}1 & 0 & 2 \\ 0 & 1 & 2 \\ 3 & 0 & 3\end{bmatrix}, S^T = \begin{bmatrix}1 & 0 & 3 \\ 0 & 1 & 0 \\ 2 & 2 & 3\end{bmatrix}$$ - Multiply: $$S S^T = \begin{bmatrix}1*1+0*0+2*2 & 1*0+0*1+2*2 & 1*3+0*0+2*3 \\ 0*1+1*0+2*2 & 0*0+1*1+2*0 & 0*3+1*0+2*3 \\ 3*1+0*0+3*2 & 3*0+0*1+3*2 & 3*3+0*0+3*3\end{bmatrix} = \begin{bmatrix}5 & 4 & 9 \\ 4 & 1 & 6 \\ 9 & 6 & 18\end{bmatrix}$$ - Sum all elements: $$5 + 4 + 9 + 4 + 1 + 6 + 9 + 6 + 18 = 62$$ (e) $$a_m a_m = a_1^2 + a_2^2 + a_3^2 = 1^2 + 2^2 + 3^2 = 1 + 4 + 9 = 14$$. (f) Evaluate $$S_{mn} a_m a_n = \sum_{m=1}^3 \sum_{n=1}^3 S_{mn} a_m a_n$$. Calculate each term: - For $$m=1$$: $$a_1=1$$ - $$n=1$$: $$S_{11} a_1 a_1 = 1*1*1=1$$ - $$n=2$$: $$S_{12} a_1 a_2 = 0*1*2=0$$ - $$n=3$$: $$S_{13} a_1 a_3 = 2*1*3=6$$ - For $$m=2$$: $$a_2=2$$ - $$n=1$$: $$S_{21} a_2 a_1 = 0*2*1=0$$ - $$n=2$$: $$S_{22} a_2 a_2 = 1*2*2=4$$ - $$n=3$$: $$S_{23} a_2 a_3 = 2*2*3=12$$ - For $$m=3$$: $$a_3=3$$ - $$n=1$$: $$S_{31} a_3 a_1 = 3*3*1=9$$ - $$n=2$$: $$S_{32} a_3 a_2 = 0*3*2=0$$ - $$n=3$$: $$S_{33} a_3 a_3 = 3*3*3=27$$ Sum all: $$1 + 0 + 6 + 0 + 4 + 12 + 9 + 0 + 27 = 59$$. (g) Evaluate $$S_{nm} a_m a_n$$. Since $$m,n$$ are dummy indices, this is the same as (f), so answer is $$59$$. **Final answers:** - (a) 5 - (b) 28 - (c) 28 - (d) 62 - (e) 14 - (f) 59 - (g) 59