Q. 1 (a) This looks good

The last row reads

,

but

so this implies that we have no solutions (there is no and such that ).

Q. 1 (b) This also looks good.

Now we do have solutions. Now the difference between the number of variables ( & ) and the number of non-zero rows in reduced form is one so the number of parameters is . There is nothing saying that has to equal anything so we may let be any real number and then solve for .

We did Q. 1 on page 18.

Q. 2 First of all not …

Here we definitely have solutions. The difference between the number of variables (, , and ) and the number of non-zero rows in reduced form is so the number of parameters is . The last row says that must equal . There is nothing that says that has to equal anything in particular so we may let be any real number and then solve for and .

Regards,

J.P.

Just going over the sample test and I am stuck in few questions.

Q 1 I have strange results or I am doing something wrong

Q. 1 (a)

Q. 1 (b)

How do I describe the solution set ?

Q. 2 Why I need “” if I can get the values of

.

Regards,

]]>To be blunt this is something that you are supposed to be able to do before coming into this module.

You can find loads of information by googling ‘factoring quadratics’.

I did it in one line because I assumed ye could too but this is how I do it: if you can rewrite the middle term in terms of factors of the product of the first and last coefficients then you will be able to factor.

In this example, the first coefficient (of ) is one and the last is two. The product of these is two. The only factors of two is one and two and you can rewrite three — the middle coefficient — as two plus one; i.e. :

.

Note that we factored the first two terms and then we took something out of the second pair of terms such that would be left again.

Now is common to both and so we take it out as a common term to get

If this is all a little worrying at this point I have to point out that you won’t need to factor any quadratics in any MATH6038 Assessment.

For MATH6037 however, you will have to be an expert!

Regards,

J.P.

I don’t have an answer sheet but you send me your answers I can check them for you.

We did Q. 1 on P.18 of the notes using Gaussian Elimination. It is only part c that you could do by using the Gauss Jordan Algorithm to find the inverse of the matrix.

Big difference between and and the order in which you multiply matrices is important because, unlike with numbers, in general .

In fact, in Q. 7 it HAS to be . We show this here: suppose and we want . We can undo the action of by multiplying ON THE LEFT on both sides with :

Now undoes the action of so we just have .

Indeed not only is not equal to , isn’t even defined:

,

can’t be done because the number of columns in does not match the number of rows in .

Regards,

J.P.

I was wondering if you had an answer sheet to your sample paper that you attached to your mail. It’s just that I have it done and I was looking to see if the way I was doing them was right because in class we did a question like question 1 only it was with Gauss-Jordan not Gaussian Elimination.

Is there a difference in the answer because in part C you can do Gauss-Jordan to get and values.

Also Question 7 is the only difference between and …is the way you multiply the rows and columns?

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