Exercise For ` u` = (1, 2),

Answer

x_{1}[ |
1 | ] + x_{2}[ |
3 | ] + x_{3}[ |
5 | ] = [ | x_{1} |
] + [ | 3x_{2} |
] + [ | 5x_{3} |
] = [ | x_{1} + 3x_{2} + 5x_{3} |
] |

2 | 4 | 6 | 2x_{1} |
4x_{2} |
6x_{3} |
2x_{1} + 4x_{2} + 6x_{3} |

x_{1}[ |
1 | ] + x_{2}[ |
4 | ] + x_{3}[ |
7 | ] = [ | x_{1} |
] +[ | 4x_{2} |
] + [ | 7x_{3} |
] = [ | x_{1} + 4x_{2} + 7x_{3} |
] |

2 | 5 | 8 | 2x_{1} |
5x_{2} |
8x_{3} |
2x_{1} + 5x_{2} + 8x_{3} |
||||||||

3 | 6 | 9 | 3x_{1} |
6x_{2} |
9x_{3} |
3x_{1} + 6x_{2} + 9x_{3} |

The results are of the form ** Ax**, with the matrix

[ | 1 | 3 | 5 | ] = []u v w |

2 | 4 | 6 |

[ | 1 | 4 | 7 | ] = [a_{1}
a_{2} a_{3}] |

2 | 5 | 8 | ||

3 | 6 | 9 |

In general, we have

`x`_{1}**a**_{1} + `x`_{2}**a**_{2}
+ ... + ` x_{n}a_{n}` =

where ` A` = [