Given:

The matrices

\(A=\begin{bmatrix}1 & 0 & 2 \\ 0 & 1 & 3 \\ 0 & 0 & 1 \end{bmatrix},\ B=\begin{bmatrix}\frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} & 0 \\ -\frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} & 0 \\ 0 & 0 & 1 \end{bmatrix}\)

Concept used:

When a point is rotated an angle theta clockwise about the origin, the transforming matrix is

\(\begin{bmatrix} \cos\theta & \sin\theta & 0 \\ -\sin\theta & \cos\theta & 0 \\ 0 & 0 & 1 \end{bmatrix}\)

To translate a point (x, y) horizontallyh units and vertically k units, we use the transformation matrix

\(\begin{bmatrix} 1 & 0 & h \\ 0 & 1 & k \\ 0 & 0 & 1 \end{bmatrix}\)

Calculation:

(a) Let the point \((0,\ \sqrt{2})\) be represented by column matrix

\(X=\begin{bmatrix} 1 \\ \sqrt{2} \\ 1 \end{bmatrix}\)

When it is rotated \(45^{\circ}\) about the origin, then we have

\(\begin{bmatrix}\frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} & 0 \\ -\frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} & 0 \\ 0 & 0 & 1 \end{bmatrix}\begin{bmatrix} 0 \\ \sqrt{2} \\ 1 \end{bmatrix}=\begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix}\)

So after rotating \(45^{\circ}\) about the origin, the new coordinates are (1, 1).

Now we need to translate this point 2 units to the right and 3 units upward, so using above concept we have

\(\begin{bmatrix}1 & 0 & 2 \\ 0 & 1 & 3 \\ 0 & 0 & 1 \end{bmatrix}\begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix}=\begin{bmatrix} 3 \\ 4 \\ 1 \end{bmatrix}\)

Hence after rotating \(45^{\circ}\) about origin and then translating 2 units to the right and 3 units upward, we reach to the point (3, 4).

(b) Now to compure \(Y = ABX,\) we have

\(Y=ABX=\begin{bmatrix}1 & 0 & 2 \\ 0 & 1 & 3 \\ 0 & 0 & 1 \end{bmatrix}\begin{bmatrix} \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} & 0 \\ -\frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} & 0\\ 1 & 0 & 1 \end{bmatrix}\begin{bmatrix} 0 \\ \sqrt{2} \\ 1 \end{bmatrix}\)

\(=\begin{bmatrix}1 & 0 & 2 \\ 0 & 1 & 3 \\ 0 & 0 & 1 \end{bmatrix}\begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix}\)

\(=\begin{bmatrix} 3 \\ 4 \\ 1 \end{bmatrix}\)

It is same as the above transformations.

Hence the first matrix A represents the translation of 2 units the right and 3 units upward.

And the matrix B represent the rotation of \(45^{\circ}\) clockwise about the origin of the point X.

(c) Now to check if ABX equal to BAX, we compute BAX first. So we have

\(BAX=\begin{bmatrix}\frac{1}{\sqrt{1}} & \frac{1}{\sqrt{2}} & 0 \\ -\frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} & 0 \\ 0 & 0 & 1 \end{bmatrix}\begin{bmatrix}2 \\ 3\ +\ \sqrt{2} \\ 1 \end{bmatrix}\begin{bmatrix}\frac{5\ +\ \sqrt{2}}{\sqrt{2}} \\ \frac{1\ +\ \sqrt{2}}{\sqrt{2}} \\ 1 \end{bmatrix}\)

We can see that ABX is not equal to BAX. It implies that if we translate the point X, 2 units to the right and 3 units upward first and then rotate about the origin \(45^{\circ}\) clockwise we will not reach to the same point as we get by first rotating about the origin and then translating.

(d) To find the matrix that translate Y back to X, we use the concept

\(Y = ABX\)

\(\Rightarrow\ (AB)^{-1} Y = X\)

Now we first find \((AB)^{-1} = B^{-1} A^{-1},\) so we get

\((AB)^{-1} = B^{-1} A^{-1},\)

\(\begin{bmatrix}\frac{1}{\sqrt{1}} & -\frac{1}{\sqrt{2}} & 0 \\ \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} & 0 \\ 0 & 0 & 1 \end{bmatrix}\begin{bmatrix}1 & 0 & -2 \\ 0 & 1 & -3 \\ 0 & 0 & 1 \end{bmatrix}\)

Thus, the matrix that translate Y back

\(=\begin{bmatrix}\frac{1}{\sqrt{1}} & -\frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} & -\frac{5}{\sqrt{2}} \\ 0 & 0 & 1 \end{bmatrix}\)

to X, is given by \(\begin{bmatrix}\frac{1}{\sqrt{1}} & -\frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} & -\frac{5}{\sqrt{2}} \\ 0 & 0 & 1 \end{bmatrix}\)

Conclusion:

Using the properties of matrix multiplication, we found the rotation and translation of the point X.

The matrices

\(A=\begin{bmatrix}1 & 0 & 2 \\ 0 & 1 & 3 \\ 0 & 0 & 1 \end{bmatrix},\ B=\begin{bmatrix}\frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} & 0 \\ -\frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} & 0 \\ 0 & 0 & 1 \end{bmatrix}\)

Concept used:

When a point is rotated an angle theta clockwise about the origin, the transforming matrix is

\(\begin{bmatrix} \cos\theta & \sin\theta & 0 \\ -\sin\theta & \cos\theta & 0 \\ 0 & 0 & 1 \end{bmatrix}\)

To translate a point (x, y) horizontallyh units and vertically k units, we use the transformation matrix

\(\begin{bmatrix} 1 & 0 & h \\ 0 & 1 & k \\ 0 & 0 & 1 \end{bmatrix}\)

Calculation:

(a) Let the point \((0,\ \sqrt{2})\) be represented by column matrix

\(X=\begin{bmatrix} 1 \\ \sqrt{2} \\ 1 \end{bmatrix}\)

When it is rotated \(45^{\circ}\) about the origin, then we have

\(\begin{bmatrix}\frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} & 0 \\ -\frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} & 0 \\ 0 & 0 & 1 \end{bmatrix}\begin{bmatrix} 0 \\ \sqrt{2} \\ 1 \end{bmatrix}=\begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix}\)

So after rotating \(45^{\circ}\) about the origin, the new coordinates are (1, 1).

Now we need to translate this point 2 units to the right and 3 units upward, so using above concept we have

\(\begin{bmatrix}1 & 0 & 2 \\ 0 & 1 & 3 \\ 0 & 0 & 1 \end{bmatrix}\begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix}=\begin{bmatrix} 3 \\ 4 \\ 1 \end{bmatrix}\)

Hence after rotating \(45^{\circ}\) about origin and then translating 2 units to the right and 3 units upward, we reach to the point (3, 4).

(b) Now to compure \(Y = ABX,\) we have

\(Y=ABX=\begin{bmatrix}1 & 0 & 2 \\ 0 & 1 & 3 \\ 0 & 0 & 1 \end{bmatrix}\begin{bmatrix} \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} & 0 \\ -\frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} & 0\\ 1 & 0 & 1 \end{bmatrix}\begin{bmatrix} 0 \\ \sqrt{2} \\ 1 \end{bmatrix}\)

\(=\begin{bmatrix}1 & 0 & 2 \\ 0 & 1 & 3 \\ 0 & 0 & 1 \end{bmatrix}\begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix}\)

\(=\begin{bmatrix} 3 \\ 4 \\ 1 \end{bmatrix}\)

It is same as the above transformations.

Hence the first matrix A represents the translation of 2 units the right and 3 units upward.

And the matrix B represent the rotation of \(45^{\circ}\) clockwise about the origin of the point X.

(c) Now to check if ABX equal to BAX, we compute BAX first. So we have

\(BAX=\begin{bmatrix}\frac{1}{\sqrt{1}} & \frac{1}{\sqrt{2}} & 0 \\ -\frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} & 0 \\ 0 & 0 & 1 \end{bmatrix}\begin{bmatrix}2 \\ 3\ +\ \sqrt{2} \\ 1 \end{bmatrix}\begin{bmatrix}\frac{5\ +\ \sqrt{2}}{\sqrt{2}} \\ \frac{1\ +\ \sqrt{2}}{\sqrt{2}} \\ 1 \end{bmatrix}\)

We can see that ABX is not equal to BAX. It implies that if we translate the point X, 2 units to the right and 3 units upward first and then rotate about the origin \(45^{\circ}\) clockwise we will not reach to the same point as we get by first rotating about the origin and then translating.

(d) To find the matrix that translate Y back to X, we use the concept

\(Y = ABX\)

\(\Rightarrow\ (AB)^{-1} Y = X\)

Now we first find \((AB)^{-1} = B^{-1} A^{-1},\) so we get

\((AB)^{-1} = B^{-1} A^{-1},\)

\(\begin{bmatrix}\frac{1}{\sqrt{1}} & -\frac{1}{\sqrt{2}} & 0 \\ \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} & 0 \\ 0 & 0 & 1 \end{bmatrix}\begin{bmatrix}1 & 0 & -2 \\ 0 & 1 & -3 \\ 0 & 0 & 1 \end{bmatrix}\)

Thus, the matrix that translate Y back

\(=\begin{bmatrix}\frac{1}{\sqrt{1}} & -\frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} & -\frac{5}{\sqrt{2}} \\ 0 & 0 & 1 \end{bmatrix}\)

to X, is given by \(\begin{bmatrix}\frac{1}{\sqrt{1}} & -\frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} & -\frac{5}{\sqrt{2}} \\ 0 & 0 & 1 \end{bmatrix}\)

Conclusion:

Using the properties of matrix multiplication, we found the rotation and translation of the point X.