(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

Given:
The matrices
$A=\left[\begin{array}{ccc}1& 0& 2\\ 0& 1& 3\\ 0& 0& 1\end{array}\right],B=\left[\begin{array}{ccc}\frac{1}{\sqrt{2}}& \frac{1}{\sqrt{2}}& 0\\ -\frac{1}{\sqrt{2}}& \frac{1}{\sqrt{2}}& 0\\ 0& 0& 1\end{array}\right]$
Concept used:
When a point is rotated an angle theta clockwise about the origin, the transforming matrix is
$\left[\begin{array}{ccc}\cos \theta & \sin \theta & 0\\ -\sin \theta & \cos \theta & 0\\ 0& 0& 1\end{array}\right]$
To translate a point (x, y) horizontallyh units and vertically k units, we use the transformation matrix
$\left[\begin{array}{ccc}1& 0& h\\ 0& 1& k\\ 0& 0& 1\end{array}\right]$
Calculation:
(a) Let the point $(0,\sqrt{2})$ be represented by column matrix
$X=\left[\begin{array}{c}1\\ \sqrt{2}\\ 1\end{array}\right]$
When it is rotated ${45}^{\circ }$ about the origin, then we have
$\left[\begin{array}{ccc}\frac{1}{\sqrt{2}}& \frac{1}{\sqrt{2}}& 0\\ -\frac{1}{\sqrt{2}}& \frac{1}{\sqrt{2}}& 0\\ 0& 0& 1\end{array}\right]\left[\begin{array}{c}0\\ \sqrt{2}\\ 1\end{array}\right]=\left[\begin{array}{c}1\\ 1\\ 1\end{array}\right]$
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
$\left[\begin{array}{ccc}1& 0& 2\\ 0& 1& 3\\ 0& 0& 1\end{array}\right]\left[\begin{array}{c}1\\ 1\\ 1\end{array}\right]=\left[\begin{array}{c}3\\ 4\\ 1\end{array}\right]$
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
$Y=ABX=\left[\begin{array}{ccc}1& 0& 2\\ 0& 1& 3\\ 0& 0& 1\end{array}\right]\left[\begin{array}{ccc}\frac{1}{\sqrt{2}}& \frac{1}{\sqrt{2}}& 0\\ -\frac{1}{\sqrt{2}}& \frac{1}{\sqrt{2}}& 0\\ 1& 0& 1\end{array}\right]\left[\begin{array}{c}0\\ \sqrt{2}\\ 1\end{array}\right]$
$=\left[\begin{array}{ccc}1& 0& 2\\ 0& 1& 3\\ 0& 0& 1\end{array}\right]\left[\begin{array}{c}1\\ 1\\ 1\end{array}\right]$
$=\left[\begin{array}{c}3\\ 4\\ 1\end{array}\right]$
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
$BAX=\left[\begin{array}{ccc}\frac{1}{\sqrt{1}}& \frac{1}{\sqrt{2}}& 0\\ -\frac{1}{\sqrt{2}}& \frac{1}{\sqrt{2}}& 0\\ 0& 0& 1\end{array}\right]\left[\begin{array}{c}2\\ 3+\sqrt{2}\\ 1\end{array}\right]\left[\begin{array}{c}\frac{5+\sqrt{2}}{\sqrt{2}}\\ \frac{1+\sqrt{2}}{\sqrt{2}}\\ 1\end{array}\right]$
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
$Y=ABX$

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