Remark The following exercises show that, for the matrix product, ** AB** =

Exercise
For each of the following matrix ` A`,
find all the two column matrices

1)

= [A |
1 | 2 | ] |

Answer The matrix ` X` has to be 2 by 2. Let

= [X |
x |
y |
] |

z |
w |

Then

= [AX |
x + 2z |
y + 2w |
] |

so that ** AX** =

`x` + 2`z` = 0,

`y` + 2`w` = 0.

This further means `x` = -2`z`, `y` = -2`w`, with `z` and `w` arbitrary.
We conclude that

= [X |
- 2z |
- 2w |
] = z[ |
-2 | 0 | ] + w[ |
0 | -2 | ] |

z |
w |
1 | 0 | 0 | 1 |

2)

= [A |
1 | ] |

2 |

Answer The matrix ` X` has to be 1 by 2. Let

= [AX |
x |
y |
] |

2x |
2y |

so that ** AX** =

`x` = 0, 2`x` = 0,

`y` = 0, 2`y` = 0.

We conclude that ` X` =

3)

= [A |
1 | 1 | ] |

1 | -1 |

Answer The matrix ` X` has to be 2 by 2. Let

= [X |
x |
y |
] |

z |
w |

Then

= [AX |
x + z |
y + w |
] |

x - z |
y - w |

so that ** AX** =

`x` + `z` = 0, `x` - `z` = 0,

`y` + `w` = 0, `y` - `w` = 0.

The first two equations tell us `x` = `z` = 0. The last two equations tell us `y` = `w` = 0.
We conclude that ** AX** =

4)

= [A |
2 | 4 | ] |

-1 | -2 |

Answer The matrix ` X` has to be 2 by 2. Let

= [AX |
2x + 4z |
2y + 4w |
] |

- x - 2z |
- y - 2w |

so that ** AX** =

2`x` + 4`z` = 0, - `x` - 2`z` = 0,

2`y` + 4`w` = 0, - `y` - 2`w` = 0.

The first two equations tell us `x` = - 2`z`, `z` arbitrary.
The last two equations tell us `y` = - 2`w`, `w` arbitrary.
We conclude that

= [X |
- 2z |
- 2w |
] = z[ |
-2 | 0 | ] + w[ |
0 | -2 | ] |

z |
w |
1 | 0 | 0 | 1 |

5)

= [A |
0 | 0 | ] |

0 | 0 |

Answer For any 2 by 2 matrix ` X`, we always have

6)

= [A |
1 | -2 | 3 | ] |

-2 | 3 | -4 |

Answer The matrix ` X` has to be 3 by 2. Let

= [X |
x |
y |
] |

z |
w |
||

u |
v |

Then ** AX** =

`x` - 2`z` + 3`u` = 0, - 2`x` + 3`z` - 4`u` = 0,

`y` - 2`w` + 3`v` = 0, - 2`y` + 3`w` - 4`v` = 0.

The first two equations tell us `x` = `u`, `z` = 2`u`, `u` arbitrary.
The last two equations tell us `y` = `v`, `w` = 2`v`, `v` arbitrary.
We conclude that

= [X |
u |
v |
] = u[ |
1 | 0 | ] + v[ |
0 | 1 | ] |

2u |
2v |
2 | 0 | 0 | 2 | ||||

u |
v |
1 | 0 | 0 | 1 |

7)

= [A |
1 | -2 | ] |

-2 | 3 | ||

3 | -4 |

Answer The matrix ` X` has to be 2 by 2. Let

= [X |
x |
y |
] |

z |
w |

Then ** AX** =

`x` - 2`z` = 0, - 2`x` + 3`z` = 0, 3`x` - 4`z` = 0,

`y` - 2`w` = 0, - 2`y` + 3`w` = 0, 3`y` - 4`w` = 0.

From the equations, we conclude that ` X` has to be the zero matrix.

Remark Here are some further problems for you to think about.

- Let
= [**X**]. Then`x y`= [`AX`] by this formula. Thus`Ax Ay`=`AX`means solving**O**=`Ax`**0**and=`Ay`**0**. More precisely, we solve the homogeneous system=`Ax`**0**and put two independent solutions as the two columns of. Check out the discussion with the concrete examples we worked out above.**X** - When can we find nonzero
so that**X**=`AX`**0**? Answer the question in terms of properties of the systems=`Ax`.**b** - What if
has 3 columns? four columns?**X** - For a given matrix
, find**A**such that**X**=`XA`. First try the problem with the matrices given in the exercise.**O**