Linear Algebra and Its Applications, 4th Edition

Published by Brooks Cole
ISBN 10: 0030105676
ISBN 13: 978-0-03010-567-8

Chapter 1 - Section 1.4 - Matrix Notation and Matrix Multiplication - Problem Set - Page 26: 2

Answer

$(i) \begin{bmatrix} 7\\ 8\\ 9 \end{bmatrix}$ $(ii) \begin{bmatrix} 2\\ 5\\ 8 \end{bmatrix}$ $(iii) \begin{bmatrix} 3\\ 5\\ 7 \end{bmatrix}$

Work Step by Step

(i) $\begin{bmatrix} 4 & 1\\ 5 & 1\\ 6 & 1 \end{bmatrix} \begin{bmatrix} 1\\ 3 \end{bmatrix}$ $= 1 \begin{bmatrix} 4\\ 5\\ 6 \end{bmatrix} + 3 \begin{bmatrix} 1\\ 1\\ 1 \end{bmatrix}$ $= \begin{bmatrix} 1 \cdot 4 + 3 \cdot 1\\ 1 \cdot 5 + 3 \cdot 1\\ 1 \cdot 6 + 3 \cdot 1 \end{bmatrix} = \begin{bmatrix} 7\\ 8\\ 9 \end{bmatrix}$ (ii) $\begin{bmatrix} 1 & 2& 3\\ 4 & 5 & 6\\ 7 & 8 & 9 \end{bmatrix} \begin{bmatrix} 0\\ 1\\ 0 \end{bmatrix}$ $= 0 \begin{bmatrix} 1\\ 4\\ 7 \end{bmatrix} + 1 \begin{bmatrix} 2\\ 5\\ 8 \end{bmatrix} + 0 \begin{bmatrix} 3\\ 6\\ 9 \end{bmatrix}$ $= \begin{bmatrix} 2\\ 5\\ 8 \end{bmatrix}$ (iii) $\begin{bmatrix} 4 & 3\\ 6 & 6\\ 8 & 9 \end{bmatrix} \begin{bmatrix} 1/2\\ 1/3 \end{bmatrix}$ $= 1/2 \begin{bmatrix} 4\\ 6\\ 8 \end{bmatrix} + 1/3 \begin{bmatrix} 3\\ 6\\ 9 \end{bmatrix}$ $= \begin{bmatrix} 2+1\\ 3+2\\ 4+3 \end{bmatrix} = \begin{bmatrix} 3\\ 5\\ 7 \end{bmatrix}$
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