### Composition and Matrix Multiplication

Exercise Let

**A** = [ |
1 |
-1 |
1 |
0 |
2 |
],
**B** = [ |
1 |
3 |
1 |
] |

2 |
-2 |
0 |
2 |
2 |
0 |
2 |
0 |

-1 |
1 |
2 |
-3 |
1 |
0 |
3 |
2 |

-2 |
2 |
1 |
-3 |
-1 |
0 |
0 |
1 |

Can you find a matrix **X** such that `AX` = **B**?
In general, what is the criterion for the existence of **X**
such that `AX` = **B**? What about the uniqueness of **X**?

Answer As we have seen in the previous
exercise,
the equation `AX` = **B** is equivalent to solving three separate systems
`Ax` = **b**_{1},
`Ax` = **b**_{2},
`Ax` = **b**_{3},
where **b**_{1}, **b**_{2}, **b**_{3} are the
three columns of **b**. In an earlier exercise, we have argued that the system `Ax` = **b**_{1}
has no solution. Therefore the equation `AX` = **B** has no solution.

In general, `AX` = **B** has solutions if and only if no column of **B**
is a pivot column of the "augmented matrix" [`A B`].
The solution is unique if and only if all columns of **A** are pivot.