# (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

(a) The new coordinates geometrically if X represents the point 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
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d2saint0
Given:
The matrices

Concept used:
When a point is rotated an angle theta clockwise about the origin, the transforming matrix is
$\left[\begin{array}{ccc}\mathrm{cos}\theta & \mathrm{sin}\theta & 0\\ -\mathrm{sin}\theta & \mathrm{cos}\theta & 0\\ 0& 0& 1\end{array}\right]$
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 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

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
$Y=ABX$