MHB Homework: 13s: Critical Thinking Challenge

  • Thread starter Thread starter karush
  • Start date Start date
karush
Gold Member
MHB
Messages
3,240
Reaction score
5
View attachment 8848

ok just sent this homework in (maybe typos} :cool:
on TS i think this is as far you can go with the current dimensions

critical thinking accepted...:cool:
 
Physics news on Phys.org
karush said:
ok just sent this homework in (maybe typos} :cool:
on TS i think this is as far you can go with the current dimensions

critical thinking accepted...:cool:
The first two are right, though I'm really not fond of your text trying to add a 2x1 matrix with a 3x1 in the second one. That's rather illegal, but not your problem.

As for [math]S^3[/math], take it step by step.
[math]S \left [ \begin{matrix} x \\ y \end{matrix} \right ] = \left [ \begin{matrix} x - 2y \\ 3x - y \end{matrix}\right ] [/math]

[math]S^2 \left [ \begin{matrix} x \\ y \end{matrix} \right ] = S \left ( S \left [ \begin{matrix} x \\ y \end{matrix} \right ] \right )[/math]

[math] = S \left [ \begin{matrix} x - 2y \\ 3x - y \end{matrix} \right ] = \left [ \begin{matrix} (x - 2y) - 2(3x - y) \\ 3(x - 2y) - (3x -y) \end{matrix} \right ] = \left [ \begin{matrix} -5x \\ -5y \end{matrix} \right ] [/math]

Now you finish the rest.

-Dan
 
karush said:
$\displaystyle = S \left [ \begin{matrix} x - 2y \\ 3x - y \end{matrix} \right ] = \left [ \begin{matrix} (x - 2y) - 2(3x - y) \\ 3(x - 2y) - (3x -y) \end{matrix} \right ] = \left [ \begin{matrix} -5x \\ -5y \end{matrix} \right ]$so then kinda maybe
$S\left [ \begin{matrix} -5x \\ -5y \end{matrix} \right ]
=\left [ \begin{matrix} -5(x-2y) \\ -5(3x-y) \end{matrix} \right ]
=\left [ \begin{matrix} -5x-10y \\ -15x-5y \end{matrix} \right ]$
The transformation S takes a value [math]x \to x - 2y[/math] and [math]y \to 3x - y[/math], so your new x value will be (-5x) - 2(-5y) and your new y value will be 3(-5x) - (-5y).

-Dan
 
so you mean this
$S\left [ \begin{matrix} -5x \\ -5y \end{matrix} \right ]
=\left [ \begin{matrix} (-5x) - 2(-5y) \\ 3(-5x) - (-5y) \end{matrix} \right ]
=\left [ \begin{matrix} -5x-10y \\ -15x-5y \end{matrix} \right ]$
 
so for ST if
$S\left(\left[\begin{array}{c}x \\ y \end{array}\right]\right)
=\left[\begin{array}{c}x-2y \\ 3x-y \end{array}\right], \quad
T\left(\left[\begin{array}{c}x \\ y \end{array}\right]\right)
=\left[\begin{array}{c}x+y \\ x-y\\2x+3y \end{array}\right]$

then

$S\left(\left[\begin{array}{c}x-2y \\ 3x-y \end{array}\right]\right)
=\left[\begin{array}{c}(x-2y)+(3x-y) \\ (x-2y)-(3x-y)\\2(x-2y)+3(3x-y) \end{array}\right]$
if ok then simplify..
 
Last edited:
karush said:
so you mean this
$S\left [ \begin{matrix} -5x \\ -5y \end{matrix} \right ]
=\left [ \begin{matrix} (-5x) - 2(-5y) \\ 3(-5x) - (-5y) \end{matrix} \right ]
=\left [ \begin{matrix} -5x-10y \\ -15x-5y \end{matrix} \right ]$
Watch the double negatives!

-Dan

- - - Updated - - -

karush said:
so for ST if
$S\left(\left[\begin{array}{c}x \\ y \end{array}\right]\right)
=\left[\begin{array}{c}x-2y \\ 3x-y \end{array}\right], \quad
T\left(\left[\begin{array}{c}x \\ y \end{array}\right]\right)
=\left[\begin{array}{c}x+y \\ x-y\\2x+3y \end{array}\right]$

then

$S\left(\left[\begin{array}{c}x-2y \\ 3x-y \end{array}\right]\right)
=\left[\begin{array}{c}(x-2y)+(3x-y) \\ (x-2y)-(3x-y)\\2(x-2y)+3(3x-y) \end{array}\right]$
if ok then simplify..
I think you mean
[math]T \left ( \left [ \begin{array}{c}x-2y \\ 3x-y \end{array} \right ] \right ) [/math]
in that last line. Otherwise, yes, good!

-Dan
 
A random thought. When you were doing the TS part you do realize you do S and then T, right?

-Dan
 
topsquark said:
A random thought. When you were doing the TS part you do realize you do S and then T, right?

-Dan
Yeah I thot it was a product
Not a composite
 

Similar threads

Replies
9
Views
1K
Replies
1
Views
1K
Replies
6
Views
2K
Replies
6
Views
3K
Replies
6
Views
1K
Replies
4
Views
2K
Back
Top