Homework: 13s: Critical Thinking Challenge

[karush]

View attachment 8848

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

critical thinking accepted...

 

[topsquark]

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

critical thinking accepted...

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

 

[topsquark]

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

 

[karush]

so you mean this
$S\left [ \begin{matrix} -5x \\ -5y \end{matrix} \right ]
=\left [ \begin{matrix} (-5x) - 2(-5y) \\ 3(-5x) - (-5y) \end{matrix} \right ]
=\left [ \begin{matrix} -5x-10y \\ -15x-5y \end{matrix} \right ]$

 

[karush]

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..

 

[topsquark]

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 - - -

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

 

[topsquark]

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

-Dan

 

[karush]

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