Determine matrix for reflection followed by rotation

[Richie Smash]

Homework Statement

Hi good morning to all.
The problem at hand states, that the points A (3,0) and B (5,0) are reflected in the mirror line y=x.
Determine the images A' and B' of these points.

I've done that using the reflection in the line y=x which i know to be
\begin{bmatrix}
0 &1 \\
1 & 0
\end{bmatrix}

I got A'(0,3) and B'(0,5). The question then says, A' and B' are rotated 90° counterclockwise about the origin.
Find the images A'' and B'' of A' and B'.

I've also done this using the fact that a 90 degree anti-clockwise rotation looks like this (x,y) = (-y,x)

SO now I'm at this point, where they ask, determine the single matrix which represents the reflection followed by the rotation above.

I'm unsure of how to approach this.

Homework Equations

The Attempt at a Solution

 

[DrClaude]

I've also done this using the fact that a 90 degree anti-clockwise rotation looks like this (x,y) = (-y,x)

Can you write this transformation as a matrix?

After that, think about what applying consecutive transformations means in terms of matrices and vectors.

 

[Richie Smash]

Well, according to my book a 90 degree anti-clockwise rotation is written as
\begin{bmatrix}
0 &-1 \\
1 & 0
\end{bmatrix}

Now what comes to my mind is multiplying both matrices I have and I'm sure it will give me the answer, but the problem is, what order will I multiply them in?
I know matrix multiplication is funny like that, and the order matters.

Do I multiply rotation by reflection, or reflection by rotation?

 

[DrClaude]

Think about it step by step. If $\vec{a}$ is the original vector, after reflection by matrix $\mathbf{M}$, you have
$$
\vec{a}' = \mathbf{M} \vec{a}
$$
Then you need to rotate $\vec{a}'$ with matrix $\mathbf{R}$. What does that look like?

 

[Richie Smash]

vector a'' = Rvec(a')

 

[DrClaude]

vector a'' = Rvec(a')

Yes, so combine both operations into one: how do you go from a to a''?

 

[Richie Smash]

You have to go through M then R which when I do gives me
\begin{bmatrix}
1 &0 \\
0 & -1
\end{bmatrix}

However, when using this new matrix T and multiplying by say the original vector a, it does not give me a''.

Only when I multiply R by M in that order does it give me
\begin{bmatrix}
-1 &0 \\
0 & 1
\end{bmatrix}

And using the same vector a to verify, it does indeed give me a'' which is (-3,0)

 

[DrClaude]

You have to go through M then R

Mathematically, what does this look like? The answer will tell you the order of the matrices to construct the full transformation matrix.

 

[Richie Smash]

I edited my post above.

 

[DrClaude]

You have to go through M then R which when I do gives me
\begin{bmatrix}
1 &0 \\
0 & -1
\end{bmatrix}

However, when using this new matrix T and multiplying by say the original vector a, it does not give me a''.

Only when I multiply R by M in that order does it give me
\begin{bmatrix}
-1 &0 \\
0 & 1
\end{bmatrix}

And using the same vector a to verify, it does indeed give me a'' which is (-3,0)

The first approach is wrong. Again, it would help if you wrote it mathematically. You start with the mirror transform
$$
\vec{a}' = \mathbf{M} \vec{a}
$$
then you apply rotation
$$
\vec{a}'' = \mathbf{R} \vec{a}'
$$
So substituting $\vec{a}'$ in the 2nd using the 1st equation
$$
\begin{align*}
\vec{a}'' &= \mathbf{R} \vec{a}' = \mathbf{R} \mathbf{M} \vec{a} \\
&= \mathbf{T} \vec{a}
\end{align*}
$$
So you see that $\mathbf{T} = \mathbf{R} \mathbf{M}$.

 

[Richie Smash]

OH wow thanks a lot doctor, that really was super clear... wow...