Find the matrix that represents a certain transformation Algebra I

We know that the following matrix is the counterclockwise rotation of an angle \theta through the origin.

\begin{pmatrix} \cos \theta & -\sin\theta \cr\cr \sin\theta & \cos\theta \end{pmatrix}

Hence for \theta = 90^\circ the matrix is

\begin{pmatrix} \cos 90^\circ & -\sin 90^\circ \cr\cr\sin 90^\circ &\cos 90^\circ \end{pmatrix} = \begin{pmatrix} 0 & -1 \cr\cr 1 & 0 \end{pmatrix}

We know that the following matrix is the counterclockwise rotation of an angle \theta through the origin.

\begin{pmatrix} \cos \theta & -\sin\theta \cr\cr \sin\theta & \cos\theta \end{pmatrix}

Since a clockwise rotation of 45^\circ is needed, we can conclude that \theta = -45^\circ. Therefore, the matrix is:

\begin{pmatrix} \cos -45^\circ & -\sin -45^\circ \cr\cr\sin -45^\circ &\cos -45^\circ \end{pmatrix} = \begin{pmatrix} \dfrac{\sqrt{2}}{2}& \dfrac{\sqrt{2}}{2} \cr\cr -\dfrac{\sqrt{2}}{2} & \dfrac{\sqrt{2}}{2} \end{pmatrix}

A rotation of 360 degrees is a full rotation, which means turning the object around until it is in the same place and the same direction. This rotation makes no changes.

Thus the matrix that represents this transformation is the identity matrix.

We know that the following matrix is the counterclockwise rotation of an angle \theta through the origin.

\begin{pmatrix} \cos \theta & -\sin\theta \cr\cr \sin\theta & \cos\theta \end{pmatrix}

A reflection over the origin is a rotation of 180^\circ and we can write:

\begin{pmatrix} \cos 180^\circ & -\sin 180^\circ \cr\cr\sin 180^\circ &\cos 180^\circ \end{pmatrix} = \begin{pmatrix} -1 & 0 \cr\cr 0 & -1 \end{pmatrix}.

Assume that the following matrix represents the reflection through the x -axis.

\begin{pmatrix}a & b \cr\cr c & d \end{pmatrix}

Since a reflection through the x -axis changes \left(x,y\right) to \left(x,-y\right), we have:

\begin{pmatrix}a & b \cr\cr c & d \end{pmatrix} \begin{pmatrix} x \cr\cr y \end{pmatrix} = \begin{pmatrix} x \cr\cr -y \end{pmatrix}

Therefore:

\begin{pmatrix} ax+by \cr\cr cx+dy \end{pmatrix} = \begin{pmatrix} x \cr\cr -y \end{pmatrix}

Equivalently:

\begin{cases} ax+by=x \cr \cr cx+dy=-y \end{cases}

One obvious solution for this system is:

\begin{cases} a=1 \cr \cr b=0 \cr \cr c=0 \cr \cr d=-1 \end{cases}

Thus the matrix of this transformation is:

\begin{pmatrix}1 & 0 \cr\cr 0 & -1 \end{pmatrix}

Assume that the following matrix represents the reflection through the y -axis.

\begin{pmatrix}a & b \cr\cr c & d \end{pmatrix}

Since a reflection through the x -axis changes \left(x,y\right) to \left(-x,y\right) we have:

\begin{pmatrix}a & b \cr\cr c & d \end{pmatrix} \begin{pmatrix} x \cr\cr y \end{pmatrix} = \begin{pmatrix} -x \cr\cr y \end{pmatrix}

Therefore:

\begin{pmatrix} ax+by \cr\cr cx+dy \end{pmatrix} = \begin{pmatrix} -x \cr\cr y \end{pmatrix}

Equivalently:

\begin{cases} ax+by=-x \cr \cr cx+dy=y \end{cases}

One obvious solution for this system is:

\begin{cases} a=-1 \cr \cr b=0 \cr \cr c=0 \cr \cr d=1 \end{cases}

Thus the matrix of this transformation is:

\begin{pmatrix}-1 & 0 \cr\cr 0 & 1 \end{pmatrix}

Assume that the following matrix represents the reflection through y=x.

\begin{pmatrix}a & b \cr\cr c & d \end{pmatrix}

Since a reflection through y=x changes \left(x,y\right) to \left(y,x\right) we have:

\begin{pmatrix}a & b \cr\cr c & d \end{pmatrix} \begin{pmatrix} x \cr\cr y \end{pmatrix} = \begin{pmatrix} y \cr\cr x \end{pmatrix}

Therefore:

\begin{pmatrix} ax+by \cr\cr cx+dy \end{pmatrix} = \begin{pmatrix} y \cr\cr x \end{pmatrix}

Equivalently:

\begin{cases} ax+by=y \cr \cr cx+dy=x \end{cases}

One obvious solution for this system is:

\begin{cases} a=0 \cr \cr b=1 \cr \cr c=1 \cr \cr d=0 \end{cases}

So the matrix of this transformation is:

\begin{pmatrix}0 & 1 \cr\cr 1 & 0 \end{pmatrix}