Let's break down each part step by step. We are given: \[ [s_{ij}] = \begin{bmatrix} 1 & & 2 \\ & 1 & 2 \\ 3 & & 3 \end{bmatrix}, \quad [a_i] = \begin{bmatrix} 1 \\ 2 \\ 3 \end{bmatrix} \] Let’s evaluate each item: --- ## (a) \( s_{ii} \) This means the **diagonal elements** of \( s_{ij} \): - \( s_{11} = 1 \) - \( s_{22} = 1 \) - \( s_{33} = 3 \) ### **Answer:** \( s_{ii} = \{1,\, 1,\, 3\} \) --- ## (b) \( s_{ij} s_{ij} \) This means **sum over all \(i\) and \(j\)** of the squares of each entry: \[ s_{ij} s_{ij} = \sum_{i=1}^3 \sum_{j=1}^3 (s_{ij})^2 \] Calculate each entry squared and sum: \[ \begin{align*} \text{Row 1: } & 1^2,\, ^2,\, 2^2 \rightarrow 1 + + 4 = 5 \\ \text{Row 2: } & ^2,\, 1^2,\, 2^2 \rightarrow + 1 + 4 = 5 \\ \text{Row 3: } & 3^2,\, ^2,\, 3^2 \rightarrow 9 + + 9 = 18 \\ \end{align*} \] Sum: \( 5 + 5 + 18 = 28 \) ### **Answer:** \( s_{ij} s_{ij} = 28 \) --- ## (c) \( s_{jk} s_{kj} \) This is \(\sum_{j=1}^3 \sum_{k=1}^3 s_{jk} s_{kj}\), or the sum of each entry times its transpose entry: \[ \sum_{j,k} s_{jk} s_{kj} \] Let's list all pairs: - For (1,1): \(s_{11} s_{11}\) - For (1,2): \(s_{12} s_{21}\) - For (1,3): \(s_{13} s_{31}\) - For (2,1): \(s_{21} s_{12}\) - For (2,2): \(s_{22} s_{22}\) - For (2,3): \(s_{23} s_{32}\) - For (3,1): \(s_{31} s_{13}\) - For (3,2): \(s_{32} s_{23}\) - For (3,3): \(s_{33} s_{33}\) Now substitute the values: \[ \begin{align*} s_{11}s_{11} &= 1 \times 1 = 1 \\ s_{12}s_{21} &= \times = \\ s_{13}s_{31} &= 2 \times 3 = 6 \\ s_{21}s_{12} &= \times = \\ s_{22}s_{22} &= 1 \times 1 = 1 \\ s_{23}s_{32} &= 2 \times = \\ s_{31}s_{13} &= 3 \times 2 = 6 \\ s_{32}s_{23} &= \times 2 = \\ s_{33}s_{33} &= 3 \times 3 = 9 \\ \end{align*} \] Sum: \( 1 + + 6 + + 1 + + 6 + + 9 = 23 \) ### **Answer:** \( s_{jk} s_{kj} = 23 \) --- ## (d) \( a_m a_m \) This is the sum of the squares of all elements in \( a_i \): \[ a_m a_m = \sum_{m=1}^3 (a_m)^2 = 1^2 + 2^2 + 3^2 = 1 + 4 + 9 = 14 \] ### **Answer:** \( a_m a_m = 14 \) --- ## (e) \( s_{mn} a_m a_n \) This is a **quadratic form**: \( a^T S a \) \[ s_{mn} a_m a_n = \sum_{m=1}^3 \sum_{n=1}^3 s_{mn} a_m a_n \] Let's compute this step by step: First, write values: - \( a = \begin{bmatrix}1\\2\\3\end{bmatrix} \) - \( S = \begin{bmatrix}1 & & 2 \\ & 1 & 2 \\ 3 & & 3 \end{bmatrix} \) Compute \( S a \): \[ S a = \begin{bmatrix} 1 \times 1 + \times 2 + 2 \times 3 \\ \times 1 + 1 \times 2 + 2 \times 3 \\ 3 \times 1 + \times 2 + 3 \times 3 \end{bmatrix} = \begin{bmatrix} 1 + + 6 \\ + 2 + 6 \\ 3 + + 9 \end{bmatrix} = \begin{bmatrix} 7 \\ 8 \\ 12 \end{bmatrix} \] Now, \( a^T (Sa) = 1 \times 7 + 2 \times 8 + 3 \times 12 = 7 + 16 + 36 = 59 \) ### **Answer:** \( s_{mn} a_m a_n = 59 \) --- ## **Summary Table** | Part | Expression | Value | |------|-----------------------|--------| | (a) | \( s_{ii} \) | \( \{1, 1, 3\} \) | | (b) | \( s_{ij} s_{ij} \) | \( 28 \) | | (c) | \( s_{jk} s_{kj} \) | \( 23 \) | | (d) | \( a_m a_m \) | \( 14 \) | | (e) | \( s_{mn} a_m a_n \) | \( 59 \) | If you have questions about any step, let me know!