VCE Methods Question Thread!

[Gentoo]

Can someone give me a quick run down on solving linear equations for unique (infinite?) and no solutions? I have no idea how to do it and what it means.

thank you!

So, what you're talking about is where you have two linear equations, with a particular unknown constant in (typically) both of the equations. What these questions ask you to do is find out the particular value(s) of this constant (they might use k, m, etc.) that either causes both lines to be exactly the same (i.e. have the same equation; an infinite number of solutions), both lines to be parallel (i.e. have the same gradient but a different y-intercept; no solutions) or have one unique solution (the two lines have different gradients and intersect only once).

So how do we figure that out? By setting out the two equations in matrix form and figuring out the determinant of the 2x2 matrix (ad - bc) that's multiplying the x and y matrix (to equal the other matrix) and equating it to 0, you can find the values of the constant (k, m, etc.) that cause the equations to either have no or infinite solutions (you can tell which one by substituting that value of your constant back into the original simultaneous equations). THEN, all values of your constant OTHER than those (usually 2) values that give you no/infinite solutions respectively will give you one unique solution.

Say we had: (and don't be put off by the form the linear equations are put in, they'll usually be expressed as ax + by = c rather than the other way around to make it easier to turn into matrix form)

kx - 3y = k + 3    (equation 1)
4x + (k + 7)y = 1    (equation 2)

That would convert to matrices like this:

    = 

So now we'd go and find the determinant of the 2x2 matrix that's multiplying x and y (ad - bc), which is k*(k + 7) - 4*-3 = k^2 +7k + 12 = (k + 4)(k + 3)
So a k value of either -3 or -4 would make the determinant equal to 0, telling us that these k values will produce either no or an infinite number of solutions. To figure out which is which, let's sub back in to find out:

k = -4 gives us 

-4x - 3y = -1   (equation 1)
4x + 3y = 1   ----> (is equivalent to) -4x - 3y = -1  (equation 2)

We can see that they're both actually the same (because if you multiply either one (I've done it to equation 2) by -1 they're both identical), so k=-4 gives us an infinite number of solutions.

Subbing in k = -3 gives us:

-3x - 3y = 0  (equation 1)
4x + 4y = 1  -----> -3x - 3y = -3/4  (equation 2)

We can see that these 2 lines will never intersect because they have the same gradient (the same ratio of x and y coefficients) but different intercepts, thus k=-3 gives us no solutions.

Thus, the values of k that give us one unique solution are R/{-3, -4} as all other k values will just give two lines that have different gradients and intersect once.

Hi!
I need help with this question. I have a SAC tomorrow, and I am freaking out

The matrix that describes the composition of mappings
- dilation of factor 5 from the x-axis
- reflection in the line y=x
-reflection in the y-axis

In general, I am confused about questions like this (using matrices!)
Please help

Okay, so a dilation by factor 5 from the x-axis will be represented by the matrix transformation:

=  

A reflection in the line y=x will be represented by:

=  

A reflection in the y-axis will be represented by:

=  

That's how to do each individual translation. Or did you want a single matrix transformation that performs ALL of the transformations at the same time? That's a bit trickier. EDIT: I just posted an example meant to demonstrate the fact that you need to be careful working back to establish a particular matrix transformation that produces MULTIPLE graph transformations at once, which was totally wrong so I subsequently deleted it, ignore that if you read it. >.<

Is a 50 still plausible if I lose about 5 marks on a sac out of 60? Because according to some resources, you can only lose 'a few' marks throughout the year. Worried :3

The amount of marks you lose technically doesn't matter, it's just your rank and how well your cohort does on the exam (and most importantly, how well YOU do on the exam). I made quite a lot of silly mistakes on SACs throughout the year, for instance.

[keltingmeith]

(what's the best way to type matrices into forums? If the formatting keeps stuffing up my explanation won't make a whole lot of sense)

\begin{bmatrix} something&cool\\ blah&olol\end{bmatrix} comes out as:



You can also add more rows/columns as necessary by putting in more & and \\s.

[Gentoo]

Ah, thanks, will do.

EDIT: Wait, how do you format a 2x1 matrix?

[keltingmeith]

Ah, thanks, will do.

EDIT: Wait, how do you format a 2x1 matrix?

Just remove &s and \\s as necessary.

[Gentoo]

Yeah I just figured that one out haha oh boy.

[Individu]

EDIT: Okay, so a dilation by factor 5 from the x-axis will be represented by the matrix transformation (again, you'll just have to imagine the matrice brackets, sorry if it's a bit hard to follow):

=  

Shouldn't that be a 5 and not 1/5?

[knightrider]

Using our exponent laws:

Thanks psyxwar

[StupidProdigy]

Shouldn't that be a 5 and not 1/5?

It's the same thing. Dilation by a fac of 1/5 from the y ax is the same as a dilation by a fac of 5 from the x ax

[dankfrank420]

It's the same thing. Dilation by a fac of 1/5 from the y ax is the same as a dilation by a fac of 5 from the x ax

But the matrix has dilated it by 1/5 in the x axis?

[grannysmith]

It's the same thing. Dilation by a fac of 1/5 from the y ax is the same as a dilation by a fac of 5 from the x ax

That's actually a dilation of factor 1/5 from the x axis. In some instances it can be the same thing (e.g. y=x) but for most other functions (e.g. y=x^2) it's not.

[StupidProdigy]

That's actually a dilation of factor 1/5 from the x axis. In some instances it can be the same thing (e.g. y=x) but for most other functions (e.g. y=x^2) it's not.

Okay my bad! hahaha sorry! 
Edit: stuffed up with my basic matrix multiplication 

[Gentoo]

Sorry guys, you're right!!!!!!!

=  

would mean that y' = (1/5)y, rearranging to get y in terms of y' would give y=5y', which after substituting back into the original equation would ultimately result in a dilation of factor 1/5 from the x-axis, NOT a dilation of factor 5 like we wanted. The moral of the story: the bottom-right hand corner value (d) represents the dilation factor from the x-axis.

It should be 5, not 1/5, my bad (I was trying to type this up in such a rush I stuffed up, just edited then).

[knightrider]

Could someone please explain the change of base rule?

Thanks 

[keltingmeith]

Could someone please explain the change of base rule?

Thanks 

The way I think about it, is that the RATIO of two logarithms is always the same, regardless of the base. In terms of equations:



So, if we have a single logarithm by itself, say , this is the same as , since . So, this means that if we were to change to base e:



If you're looking for a proper proof, see here.

[kinslayer]

-4(-3+k)>0
12-4k>0

then -4k > -12
k< 3, how is this wrong?

Sorry, I meant you are right and the question is wrong, sorry for the confusion.