VCE Methods Question Thread!

[keltingmeith]

We start by putting this into matrix form:



Now, we know that this matrix equation cannot be solved if the inverse of the first matrix doesn't exist. For an inverse to exist, the matrix cannot be singular - this means that the determinant must not equal zero. So, if we find the determinant of the first matrix when it equals zero, this will give us .

However, we're not done yet - graphically, we know that there are two situations when we cannot find a solution. When there are no solutions (parallel lines), or infinite solutions (they're actually the same line). The question wants us to find when there are no solutions, so we check which situation we have by picking values for a and b in the ratio given above. So, if we let b = 4, then a = -5, giving us:



So, we see there are no solutions as the lines are parallel. So, the given simultaneous equations have no solution when , so the answer is C.

EDIT: upon re-reading the question, I realised that I actually went too far into it. Since the question only asks for when there are no unique solutions, you don't have to double check, essentially stopping once you found that . However, if they ask for when there's no solutions or infinite solutions, you must confirm as above.

[LiquidPaperz]

Thanks guys!!

I also had this one, last one. Why have they put the values across and not down the column to create a transition matrix?

[JHardwickVCE]

Thanks guys!!

I also had this one, last one. Why have they put the values across and not down the column to create a transition matrix?

I don't think this can be categorised as a transition matrix, as there are no further iterations. The values are filled in horizontally because when they are multiplied by a 3*1 matrix, each percentage will multiply through and correspond to the price of a kind of sweet.

[lzxnl]

Transition matrices have a few requirements:
1. Square
2. All entries are positive
3. Sum of entries in any column equals 1

[hyunah]

help please

a survey showed that of a population 26% read Advertiser 25% read Bulletin and 14% read Chaser. Both the advertiser and bulletin is read by 11%. Both Bulletin and Chaser read by 10% and both Chaser and Advertiser read by 9%. None read Chaser only.

a) draw a venn diagram of the situation
b)% of population who read at least one of the three newspapers
c) % of population who read all 3

thank you

[kinslayer]

Thanks guys!!

I also had this one, last one. Why have they put the values across and not down the column to create a transition matrix?

If they put the sweets as columns rather than rows, then calculating the price of a mix of sweets would require multiplying on the left by a row vector (1x3), as opposed to multiplying on the right by a column vector (3x1). I can't see any other important difference.

Whether or not it is (or could be) a transition matrix isn't relevant for this question because we're not given a state space for it to represent.

In my opinion, the matrix transpose of M (rows & columns switched) would be a valid answer too.

[knightrider]

How would you solve
sin(5pi/8)

 
 


[JHardwickVCE]

help please

a survey showed that of a population 26% read Advertiser 25% read Bulletin and 14% read Chaser. Both the advertiser and bulletin is read by 11%. Both Bulletin and Chaser read by 10% and both Chaser and Advertiser read by 9%. None read Chaser only.

a) draw a venn diagram of the situation
b)% of population who read at least one of the three newspapers
c) % of population who read all 3

thank you

[Phy124]

[soNasty]

how would i input 'union' into a tinspire when plotting/defining a function?

[keltingmeith]

how would i input 'union' into a tinspire when plotting/defining a function?

There is a symbol for it in the ti-nspire symbol library, but I'm not sure if it works for plotting a function...? Can you give us the situation, you're probably better off just figuring out what the union would become and entering in that interval instead.

[soNasty]

entering this in to be plotted

[LiquidPaperz]

help please

[keltingmeith]

(Image removed from quote.)

entering this in to be plotted

The points they haven't included, the function isn't actually defined. I can go through the algebra if you want, but it's unnecessary and I wouldn't expect them to ask you about that.

Otherwise, don't graph the function once - graph it three times. The first time, use the domain [0, 3pi/2), the second time use (7pi/2, 11pi/2) and the third time use (a, b).

help please

A has an inverse such that . If , then . This gives us:

[alchemy]

help please

determinant=-9+4k.

just multiply the top column only to perhaps save time and equate it to 3 since 3 is what's on the top right column of the matrix it's meant to be the same as.
-27+12k=-3
so k=2.
edit: beaten..
edit2: thanks eulerfan for spotting the error.