Finding AB & BA: Can It Be Done?

[roam]

(a) Let A = \left(\begin{array}{ccc}1\\2\\-2\end{array}\right), B = (0 3 -1)
Find AB and BA or else explain why it cannot be done.


(b) Let R_{\theta} = \left(\begin{array}{ccc}cos\theta&-sin\theta\\sin\theta&cos\theta\end{array}\right), Find R_{\theta}R_{-\theta}.

The Attempt at a Solution

(a) I believe that AB & BA cannot be computed because one of them is a row while the other one is a column. A column represents a vector whilst a row represents a point.

The only way to find AB is to find B transpose aka B^T and hence compute AB^T.

Is this the right answer for part (a) ? Thanks.

 

[Dick]

No, both AB and BA can be computed. One is a number and the other is a 3x3 matrix. Why?

 

[roam]

Oh I realized that! Thanks a lot. I managed to compute both AB and BA.


But I don't understand part (b), we are required to find R_{\theta}R_{-\theta}.
I reckon we need to multiply the matrix R_{\theta} by R_{-\theta}.

I don't understand, what is R_{-\theta}?

 

[Defennder]

R_{-\theta} is what you get when you replace all the \theta in the matrix with -\theta.

 

[roam]

R_{\theta} = \left(\begin{array}{ccc}cos\theta&-sin\theta\\sin\theta&cos\theta\end{array}\right)

R_{-\theta} = \left(\begin{array}{ccc}cos(-\theta)&-sin(-\theta)\\sin(-\theta)&cos(-\theta)\end{array}\right)

Now I need multiply the two in order to find R_{\theta}R_{-\theta}

For example to find the entry of the 1st row/1st column we are required to do the following;

cos(θ) . cos(-θ) + -sin(θ) . sin(-θ)

Is there a simplification for this? If so, how should I simplify it because each time I get a wrong answer…

 

[Mark44]

Before you multiply, simplify the entries in R_-theta. For example, what is cos (-theta)? sin(-theta)?

 

[roam]

Before you multiply, simplify the entries in R_-theta. For example, what is cos (-theta)? sin(-theta)?

I seriously have no idea as how do you simplify the entries in R_-θ.

But if I had to guess I'd say cos(θ) . cos(-θ) simplifies to cos²(-θ)

But the problem is that one of the θ's is positive while the other one is negative.

 

[Dick]

cos(-theta)=cos(theta). Look it up. What about sin(-theta)?

 

[jjou]

To simplify the entries of R_{-\theta}, think about a point on the unit circle.

First, how do the x and y coordinates relate to the sine and cosine functions?

Secondly, pick a \theta. Look at the x and y values corresponding to that \theta. How do they relate to the x and y values corresponding to -\theta?

 

[roam]

To simplify the entries of R_{-\theta}, think about a point on the unit circle.

First, how do the x and y coordinates relate to the sine and cosine functions?

Secondly, pick a \theta. Look at the x and y values corresponding to that \theta. How do they relate to the x and y values corresponding to -\theta?

P(x,y) = (cos(\theta), sin(\theta))

cos(-theta)=cos(theta). Look it up. What about sin(-theta)?

Is it: sin(-θ) = sin(θ) ?

Does this mean that the entries in R_-θ simplifies into:

\left(\begin{array}{ccc}cos\theta&-sin\theta\\sin\theta&cos\theta\end{array}\right) ?

 

[HallsofIvy]

No, it doesn't! You had before
\left(\begin{array}{cc} cos(-\theta) & -sin(-\theta) \\ sin(-\theta) & cos(-theta)\end{array}\right)
and you were told that cos(-\theta)= cos(\theta) and sin(-\theta)= -sin(-\theta)
Put those into your matrix.

 

[roam]

No, it doesn't! You had before
\left(\begin{array}{cc} cos(-\theta) & -sin(-\theta) \\ sin(-\theta) & cos(-theta)\end{array}\right)
and you were told that cos(-\theta)= cos(\theta) and sin(-\theta)= -sin(-\theta)
Put those into your matrix.

Yes;

R_{-\theta} = \left(\begin{array}{ccc}cos(\theta)&sin(-\theta)\\-sin(-\theta)&cos(\theta)\end{array}\right)

And to find R_{\theta}R_{-\theta} I multiply the two:

\left(\begin{array}{ccc}cos\theta&-sin\theta\\sin\theta&cos\theta\end{array}\right) . \left(\begin{array}{ccc}cos(\theta)&sin(-\theta)\\-sin(-\theta)&cos(\theta)\end{array}\right)


It's a 2x2 matrix;\left(\begin{array}{ccc}a_{1 1}&a_{12}\\a_{21}&a_{22}\end{array}\right) so;

a₁₁ = cos(\theta) . cos(\theta) + (-sin(-\theta) . (-sin(-\theta) => cos²θ + sin²(-θ)


a₁₂ = cos(θ) . (sin(-θ)) + (-sin(θ)) . cos(θ)

Am I on the right track?

 

[Mark44]

Yes;

R_{-\theta} = \left(\begin{array}{ccc}cos(\theta)&sin(-\theta)\\-sin(-\theta)&cos(\theta)\end{array}\right)

And to find R_{\theta}R_{-\theta} I multiply the two:

\left(\begin{array}{ccc}cos\theta&-sin\theta\\sin\theta&cos\theta\end{array}\right) . \left(\begin{array}{ccc}cos(\theta)&sin(-\theta)\\-sin(-\theta)&cos(\theta)\end{array}\right)


R_{-\theta} = \left(\begin{array}{ccc}a_{1 1}&a_{12}\\a_{21}&a_{22}\end{array}\right)

a₁₁ = cos(\theta) . cos(\theta) + (-sin(-\theta) . (-sin(-\theta) => cos²θ + sin²(-θ)


a₁₂ = cos(θ) . (sin(-θ)) + (-sin(θ)) . cos(θ)

Am I on the right track?

There was a typo in HallsofIvy's post. sin(-theta) is not equal to -sin(-theta); it should have been sin(-theta) = -sin(theta).

You've received a lot of help with this problem. I hope that you will be able to finish this one without needing any more clues.