Exercise 1.5.1
We have so
and
Exercise 1.5.2
We haveAnd
Thus
And
Comparing () and () we see both calculations result in the same matrix.
Exercise 1.5.3
For example, are two such matrices.
Exercise 1.5.4
Exercise 1.5.5
To find such a we must solve the equation
This gives a system of equations
We row-reduce the augmented coefficient matrix
Setting gives the solution
Checking:
Exercise 1.5.6
The -th entry of is . Since the term is independent of , we can view the sum independent of as where is the -th column of . I’m not sure what more to say, this is pretty immediately obvious from the definition of matrix multiplication.
Exercise 1.5.7
Suppose
Then implies the following system in has a solution
because is one such solution. The augmented coefficient matrix of this system is
As long as this system row-reduces to the following row-reduced echelon form
Thus we see that necessarily , , , . Thus
Now it’s a simple matter to check that
The loose end is that we assumed . To tie up this loose end we must show that if then necessarily . Suppose that . We will show there is no solution to (), which contradicts the fact that is a solution. If then obviously . So suppose WOLOG that (because by elementary row operations we can move any of the four elements to be the top left entry). Subtracting times the 3rd row from the 4th row of () gives
Now and since also . Thus we get
and it follows that () has no solution.
Exercise 1.5.8
We want to know when we can solve for such that
The right hand side is equal to
Thus the question is equivalent to asking: when can we choose so that the following system has a solution for
The augmented coefficient matrix for this system is
This matrix is row-equivalent to
from which we see that necessarily .
Suppose conversely that . We want to show such that .
We first handle the case when . We know so also . So is in the form
In this case letThenSo we can assume going forward that . We want to show the system () can be solved. In other words we have to find such that the system has a solution in . If we assume and then this matrix row-reduces to the following row-reduced echelon form
We see that necessarily
Since , we can set , and . Then the system can be solved and we get a solution for any choice of and . Setting we get and .
Summarizing, if then:
For example if then and . Checking:
From http://greggrant.org