# The Matrices

1. Oct 12, 2008

### 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 $$R_{\theta}R_{-\theta}.$$

3. 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 BT and hence compute ABT.

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

2. Oct 13, 2008

### Dick

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

3. Oct 13, 2008

### 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 $$R_{-\theta}$$?

4. Oct 13, 2008

### Defennder

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

5. Oct 13, 2008

### 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…

6. Oct 13, 2008

### Staff: Mentor

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

7. Oct 13, 2008

### roam

I seriously have no idea as how do you simplify the entries in R.

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.

8. Oct 13, 2008

### Dick

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

9. Oct 13, 2008

### 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 $$-\theta$$?

10. Oct 14, 2008

### roam

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

Is it: sin(-θ) = sin(θ) ?

Does this mean that the entries in R simplifies into:

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

11. Oct 14, 2008

### HallsofIvy

Staff Emeritus
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 $sin(-\theta)= -sin(-\theta)$

12. Oct 14, 2008

### roam

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?

13. Oct 14, 2008

### Staff: Mentor

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.