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

1)

= [A |
1 | 2 | ] |

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

= [X |
x |
] |

z |

Then ** AX** = [

= [X |
1 - 2z |
] = [ | 1 | ] + z[ |
-2 | ] |

z |
0 | 1 |

where `z` is arbitrary.

2)

= [A |
1 | ] |

2 |

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

= [AX |
x |
y |
] |

2x |
2y |

so that ** AX** =

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` = 1, `x` - `z` = 0,

`y` + `w` = 0, `y` - `w` = 1.

The first two equations tell us `x` = `z` = 1/2.
The last two equations tell us `y` = 1/2, `w` = -1/2.
We conclude that ** AX** =

= [X |
1/2 | 1/2 | ] |

1/2 | -1/2 |

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` = 1, - `x` - 2`z` = 0,

2`y` + 4`w` = 0, - `y` - 2`w` = 1.

The first two equations are contradictory.
The last two equations are also contradictory.
We conclude that there is no ` X` such that

5)

= [A |
0 | 0 | ] |

0 | 0 |

Answer We have ** AX** =

= [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` = 1, - 2`x` + 3`z` - 4`u` = 0,

`y` - 2`w` + 3`v` = 0, - 2`y` + 3`w` - 4`v` = 1.

The first two equations tell us `x` = 1 + `u`, `z` = 2`u`, `u` arbitrary.
The last two equations tell us `y` = - 2 + `v`, `w` = -1 + 2`v`, `v` arbitrary.
We conclude that

= [X |
1 + u |
-2 +v |
] = [ | 1 | -2 | ] + u[ |
1 | 0 | ] + v[ |
0 | 1 | ] |

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

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

where `u` and `v` are arbitrary.

7)

= [A |
1 | -2 | ] |

-2 | 3 | ||

3 | -4 |

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

= [X |
x |
y |
z |
] |

u |
v |
w |

Then ** AX** =

`x` - 2`u` = 1, - 2`x` + 3`u` = 0, 3`x` - 4`u` = 0,

`y` - 2`v` = 0, - 2`y` + 3`v` = 1, 3`y` - 4`v` = 0,

`z` - 2`w` = 0, - 2`z` + 3`w` = 0, 3`z` - 4`w` = 1.

Note that we actually have three systems of linear equations, with different variables
((`x`, `u`) for the first system,
(`y`, `v`) for the second, (`z`, `w`) for the third),
but with the same matrix ` A` as the coefficient matrix.
Since the system in (