I can help out with matrices.
Question 3.
You just simply forgot to apply the determinant when you used your calculator.
Step 1.
[1 3][/x]=[9]
[2 5][y]= [16]
Step 2.
Here we rearrange the equation to solve for x and y, we do this by taking the inverse of "A" which is the 2x2 matrix. You can simply do this on the calculator by placing the original matrix in and raising it to the power of one. However we can also do it by hand, we start by finding the determinant.
Det (A) = 1x5 - 2x3
= -1
Then we flip a and d and making c and b negative, at this stage the inverse looks like this.
1/-1 x [5 -3]
[-2 1]
Then we sub in the inverse if the determinant by simple multiplication, so the inverse is this.
[-5 3]
[2 -1]
Then we multiply this by [9] and that is how we solve for x and y.
[16]
Question 6
This comes down to the way we actually multiply matrices. If we look at what actually happens when we multiply the matrix, we can work out what each figure represents.
For instance, if we do TxU, we get this in row 1 column one.
[Vincent time photocopier x Alton Photocopiers + Nev time for photocoper x Bolton Photocopiers + Rani time for photocopier x Carlon Photocopier]
This does not show how they long it will take working alone and is also spread across three schools and people, it doesn't really represent anything.
If we do UxT, we get.
[Alton Photocopiers x Vincent time for photocopiers + Alton faxs x Vincent time for faxes + Alton scanners x Vincent time for scanners]
This shows us how long it will take Vincent to work on Alton, which is what the question wants. The easiest way to work this out, is this.
U times T, on the outside is U's rows and T's columns, these form our new matrix.
UT = V N R
[ ] A
[ ] B
[ ] C
This shows us the time it takes for each person, at each school. Which is why it is correct.
Question 8
A matrix is defined when the number of columns of the first matrix, equal the number of rows in the second matrix.
In this question, no matter which way you multiply them, either n or m will be both these values. Therefore, no matter what they are, they will be the same and the matrix will be defined.
Question 9
The matrix tells us what will happen based on what he answered for his last question.
If he chose A, he will choose A again.
If he chose B, he will choose D next time.
If he chose C, he will choose B next time.
If he chose D, he will choose C next time.
Then you just see which statement complies with these new rules.
a. This does not work as if he answers the first question with A he will answer all four with A.
b. This is the correct answer, it works for them all.
A-A-A-A
B-D-C-B
D-C-B-D
C-B-D-C
It will always have the same first and fourth answer.
c. Doesn't work as if you answer C for question two, you will answer B for the third.
d. No, as he will answer A.
e. Once again no, if he gives A it will be 3 As not four. The others will rotate in a cycle of three and the maximum is two.
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