I am emailing a link of this to everyone on the class list every week. If you are not receiving these emails or want to have them sent to another email address feel free to email me at jippo@campus.ie and I will add you to the mailing list.

This Week

We looked at different types of data and how to present them. Notes here. When we look at histograms next week this section will be complete. In Maple we continued to boost our basic skills and looked at Maple’s LinearAlgebra package.

Next Week

I hope to finish looking at histograms. Then we will begin our study of probability, asking the question what is a random variable and when are random variables independent?

Reminder

The test will be held at 7.05 p.m. sharp on Wednesday 14 March.

Sample Test Answers

Here I give you links to how I checked answers quickly using Wolfram Alpha. It takes Mathematica code (they are the same company) and will probably decipher Maple code too (moreover Maple code with ‘adjustments’). Note that Wolfram Alpha gives us a lot more information than we need but that is the beauty of the thing really.

Question 1 (has a slight typo: should be -6z in the first equation)

Question 2

Question 3 — this implies that there are non-trivial solutions. Now that we know that it has non-trivial (non-zero solutions) so we can use Wolfram Alpha to solve for these (in terms of a parameter t.). Now let z=t so we have (x,y,z)=(t,t/2,t).

Bonus Questions

You could do without all the notation. The underlined would do.

1. Firstly the linear system can be written as a set of simultaneous equations:

a_{11}x_1+\cdots+a_{1n}x_n=b_1

a_{21}x_1+\cdots+a_{2n}x_n=b_2

\vdots+\cdots+\vdots=\vdots

a_{m1}x_1+a_{mn}x_n=b_m

The solution set is not changed by E_1 as this is simply writes the equations in a different order. Neither will E_2 change the solution (well if k\neq0) as multiplying one of the equations by a constant on both sides will not change the solutionFinally E_3 doesn’t change the solution set as we add the LHS_i (left-hand-side) of one equation to the LHS_j of another, while we add the constant b_i to b_j. However the LHS_i=b_i so we have just added the same thing to both sides and this does not change the solution.

2. A\mathbf{x}=\mathbf{b}\Rightarrow A^{-1}A\mathbf{x}=A^{-1}\mathbf{b}\Rightarrow I\mathbf{x}=A^{-1}\mathbf{b}\Rightarrow \mathbf{x}=A^{-1}\mathbf{b}.

3. You can solve the system for one rather than all variables.

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