(a) The new coordinates geometrically if X represents the point (0, sqrt{2}) m and this point is rotated about the origin 45^{circ} clockwise and then translated 2 units to the right and 3 units upward. (b) The value of Y = ABX, and explain the result. (c) If ABX equal to BAX. Interpret the resul. (d) A matrix that translate Y back to X

(a) The new coordinates geometrically if X represents the point (0, sqrt{2}) m and this point is rotated about the origin 45^{circ} clockwise and then translated 2 units to the right and 3 units upward. (b) The value of Y = ABX, and explain the result. (c) If ABX equal to BAX. Interpret the resul. (d) A matrix that translate Y back to X

Question
Transformation properties
asked 2021-02-19
(a) The new coordinates geometrically if X represents the point \((0,\ \sqrt{2})\) m and this
point is rotated about the origin \(45^{\circ}\) clockwise and then translated 2 units to the right and 3 units upward.
(b) The value of \(Y = ABX,\) and explain the result.
(c) If ABX equal to BAX. Interpret the resul.
(d) A matrix that translate Y back to X

Answers (1)

2021-02-20
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.
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