Exercise 3.C.9

Answers

Proof. Clearly, $Ac$ is an $m$ -by- $1$ matrix. Then

\begin{aligned} (Ac)_j, 1 & = \sum_{r = 1}^{n} A_{j, r} c_{r, 1} \\ = \sum_{r = 1}^{n} A_{j, r} c_{r} \\ = c_{1} A_{j, 1} + \dots + c_{n} A_{j, n} \\ = c_{1} (A_{., 1})_j, 1 + \dots + c_{n} (A_{., n})_j, 1 \\ = (c_{1} A_{., 1})_j, 1 + \dots + (c_{n} A_{., n})_j, 1 \\ = (c_{1} A_{., 1} + \dots + c_{n} A_{., n})_j, 1 . \end{aligned}

□

celiopassos

2017-10-06 00:00