Dekoyl's Questions

[dekoyl]

Accidentally deleted post

[enwiabe]

Row-echelon form means leading ones only, an upper triangle matrix.


So they want, (forgive the lack of latex)

[1  a  b |  c]
[0  1  d |  e]
[0  0  1 |  f]

meaning you have

z = f
y + dz = e
x + ay + bz = c

where a, b, c, d, e, and f are all arbitrary constants in the real set.

Back substitution then means subbing z = f into y + dz = e to obtain y = e - df

and then subbing in z = f and y = e - df into x + ay + bz = c to obtain x = c - bf - ae + adf

[enwiabe]

P.S. Linear Algebra is finicky as hell, be very careful what you're multiplying/dividing/adding/subtracting in your row operations. One mistake and it will all go horrrrribly wrong. Check, double check, triple check.

[squance]

P.S. Linear Algebra is finicky as hell, be very careful what you're multiplying/dividing/adding/subtracting in your row operations. One mistake and it will all go horrrrribly wrong. Check, double check, triple check.

Im currently doing Linear Algebra as a summer subject and yeah...be very careful with your row operations.
Also remember to use correct notation.

[dekoyl]

Thanks enwiabe and squance. I think I've got it =D

[dekoyl]

Just a quick one:
I'm given three linear equations. Using matrices, I find that it's in fact 2 equations and that there are infinite solutions. And so to show the solution to the question, I display it with matrices (don't know how to use LaTeX for this but it's something like):
[x|  [1]   [4]
[y|=[2]+[5]t
[z|  [3]   [6]
Should I say or can it be anything? If so, how do I denote that?

Thanks

[squance]

hmm...I'm not quite sure what you are asking here:
I have an example in my workbook that might help you:

Solve the linear system in the variables x, y ,z which has the row echelon form:

[1 0 1 -2
 0 2 2 4
 0 0 0 0]

Last row reads 0 = 0. z is not specified, so let z =t.

Second equation reads
2y + 2z = 4 ---> y + z =2
But z=t, so y = 2-t

First equation reads x + z = -2
but z=t, so x = -2-t

Hence solution is {(x,y,z): x=2-t, y =2-t,z=t, t for all real numbers}
or
{(x,y,z) = (2, 2, 0) + t(-1,-1,1), t for all real numbers}

Thats how I've taught to show the solution of a linear system.

ANd yes, in general you should say t for all real numbers...or if you have more than one parameter, eg: s and t, then you should write, "s,t for all real numbers"

Hope I've been some help

[dekoyl]

ANd yes, in general you should say t for all real numbers...or if you have more than one parameter, eg: s and t, then you should write, "s,t for all real numbers"

Thank you! Sorry I wasn't very clear but that was exactly what I was asking.

And thanks for showing how you set out the solution. Instead of doing (x,y,z) we were shown in column form.

Oh and forgot to add, that Gaussian elimination method you posted in your thread was really helpful =] Thanks squance.

[enwiabe]

Column/row form are both acceptable. Generally when giving a solution set, however, column form is the convention. Row form is the convention for describing a line in 3-space (generally).

[dekoyl]

^Thanks enwiabe.

Another quick one:
When questions (related to matrices I did above) are like: "Find the values of k for which the system of equations has:
(i)one solution (ii)no solution" etc., when you work out the value of k that does satisfy the question, is the only way to show the solution by substituting into the matrix again to show that it satisfies the requirement?

Eg (my setting out for the answer):

If k = 3:

Infinitely many solutions

Edit: Sorry - I copy-pasted the LaTeX from my first post so the matrix didn't make sense

[enwiabe]

Well, that's wrong, because R3 in and of itself does not imply infinitely many solutions.

I'd say "more unknowns than equations, therefore infinitely many solutions"

[dekoyl]

Ah okay thanks - bad example I did there
So I have to draw up another matrix to prove my solution, right? Because I think my teacher said there was another way of setting out the answer.

[Mao]

that is the way, though your reasoning is wrong. Consider these two matrices:

and

In both cases, the last row is consisted of zeroes, but that is insufficient to say there are infinite number of solutions [in the second matrix it clearly had a distinct solution]. The key is leading entries, when a column does not have a leading entry, it is unbounded and hence there are infinite number of solutions.


but yes, to show that your 'k' value is correct, substitute back in and show the ref with (i) columns w/o leading entry or (ii) [0 0 0 ... | *]

[enwiabe]

Mao's answer is more rigorous, re: terminology.

[dekoyl]

Is this right for linear mapping? I might've copied the notes wrong.
If , the line transforms to:






Should it be instead of the I did above?




Also:
In 2D, is a dilation by a factor of   from the y-axis or from the y-axix or by a factor of from the x-axis?

Yes I'm still trying to understand some concepts :p

Thanks