### Composition and Matrix Multiplication

Exercise Let

**A** = [ |
2 |
0 |
2 |
1 |
] |

1 |
1 |
-1 |
0 |

0 |
1 |
-2 |
1 |

For each of the following matrix **B**,
find a matrix **X** such that `AX` = **B**.

1)

Answer **X** must be a vertical vector of dimension 4.
The equation `AX` = **B** is the usual system of 3 linear equations in 4 variables.
It is in fact this
system.
By the earlier computation, the augmented matrix [`A B`]
has the following row echelon form.

[ |
1 |
1 |
-1 |
0 |
-1 |
] |

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

0 |
0 |
0 |
3 |
-1 |

From this we find one solution

2)

Answer Let

**b**_{1} = [ |
1 |
],
**b**_{2} = [ |
1 |
] |

-1 |
0 |

-2 |
1 |

be the two columns of **B**. Let **x**, **y** be the two columns of
**X**. By `AX` = **A**[`x y`] = [`Ax Ay`],
`AX` = **B** is equivalent to two separate systems
`Ax` = **b**_{1}, `Ay` = **b**_{2}.
The first system has been studied in the problem above and a solution has been obtained.
The second system has an obvious solution

Combining the solutions together, we get a solution

**X** = [ |
1 |
0 |
] |

-2 |
0 |

0 |
0 |

-1/3 |
1 |

of `AX` = **B**.

3)

**B** = [ |
1 |
0 |
0 |
] |

0 |
1 |
0 |

0 |
0 |
1 |

Answer The columns of **B** are the standard basis vectors **e**_{1},
**e**_{2}, **e**3. By a similar argument as before,
the columns of **X** are the solutions of the systems
`Ax` = **e**_{i}.
Combining the augmented matrices of the three systems, we get

[**A** **e**_{1} **e**_{2} **e**_{3}]
= [`A I`] = [ |
2 |
0 |
2 |
1 |
1 |
0 |
0 |
] |

1 |
1 |
-1 |
0 |
0 |
1 |
0 |

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

Row operations (e.g., as in this
exercise)
change the matrix into

[ |
1 |
1 |
-1 |
0 |
0 |
1 |
0 |
] |

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

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

By taking [col 1-5], we get a solution (1/3, -1/3, 0, 1/3) of
`Ax` = **e**_{1}.
By taking [col 1-4] and [col 6], we get a solution (1/3, 2/3, 0, -2/3) of
`Ax` = **e**_{2}.
By taking [col 1-4] and [col 7], we get a solution (-1/3, 1/3, 0, 2/3) of
`Ax` = **e**_{3}.
Putting the three columns together, we get a solution

**X** = [ |
1/3 |
1/3 |
-1/3 |
] |

-1/3 |
2/3 |
1/3 |

0 |
0 |
0 |

1/3 |
-2/3 |
2/3 |

of `AX` = **B**.

Remark From this sequence of exercises,
you can see the relation between solving equations like `AX` = **B**
and our usual systems `Ax` = **b** of linear equations.
Apply your new insight to this exercise
and this exercise.
Moreover, study the existence/uniqueness/general solution problem for
`AX` = **B**,
just as we have done for systems of linear equations.