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$\textbf{Exercise 13 Commentary:}$ This exercise provides an explicit formula for the $(j, k)$-entry of the $n$-th power $A^n$ of an $n \times n$ matrix $A$, in terms of the entries of $A$. Specifically, it shows that $(A^n)_{j,k} = \sum_{i_1, i_2, \ldots, i_n = 1}^n A_{j, i_1}A_{i_1, i_2} \cdots A_{i_{n-1}, i_n}A_{i_n, k}$.
The formula is derived by considering the definition of matrix multiplication and expanding the entries of the product $A^n = A \cdots A$ using the entries of $A$. It highlights how the $(j, k)$-entry of $A^n$ depends on all possible length-$n$ products of entries from $A$, tracing paths from row $j$ to column $k$ through intermediate rows and columns.
$\textbf{Exercise 13 Examples:}$
1) Let $A = \begin{pmatrix}1&2\\3&4\end{pmatrix}$. Then $A^2 = \begin{pmatrix}7&10\\15&22\end{pmatrix}$, and the formula gives $(A^2)_{1,1} = 1 \cdot 1 + 2 \cdot 3 = 7$, $(A^2)_{1,2} = 1 \cdot 2 + 2 \cdot 4 = 10$, $(A^2)_{2,1} = 3 \cdot 1 + 4 \cdot 3 = 15$, $(A^2)_{2,2} = 3 \cdot 2 + 4 \cdot 4 = 22$.
2) Let $A = \begin{pmatrix}0&1&0\\0&0&1\\1&0&0\end{pmatrix}$. Then $A^3 = \begin{pmatrix}1&0&0\\0&1&0\\0&0&1\end{pmatrix} = I_3$, and the formula correctly gives $(A^3)_{j,k} = 1$ if $j = k$, and 0 otherwise.
3) Let $A$ be the $4 \times 4$ matrix with $1