1) Uniform scaling with a factor of 2 in XY plane. 2) Non uniform scaling with a factor of 2 and 1.5 in X and Y directions, while anchoring point (1, 1)

i. Uniform Scaling with a factor of 2 in xy plane and anchoring point $(1,1)$

Step 1: Transformation of Point $\mathrm{A}(1,1)$ to origin $\mathrm{O}(0,0)$ Translation Matrix is given as:

$$[\mathrm{Tr}]=\left[\begin{array}{lll}{1} & {0} & {0} \\ {0} & {1} & {0} \\ {\mathrm{tx}} & {\text { ty }} & {1}\end{array}\right] \quad \mathrm{tx}=-1, \mathrm{ty}=-1$$

$$\therefore[\mathrm{Tr}]=\left[\begin{array}{ccc}{1} & {0} & {0} \\ {0} & {1} & {0} \\ {-1} & {-1} & {1}\end{array}\right]$$

Step 2: Uniform Scaling with a factor of 2 in xy plane

$$[\mathrm{S}]=\left[\begin{array}{ccc}{\mathrm{Sx}} & {0} & {0} \\ {0} & {\mathrm{Sy}} & {0} \\ {0} & {0} & {1}\end{array}\right]=\left[\begin{array}{lll}{2} & {0} & {0} \\ {0} & {2} & {0} \\ {0} & {0} & {1}\end{array}\right]$$

Step 3: Inverse translation of Point P from origin to its original position.

Translation Matrix is:

$$[\mathrm{Tr}]^{-1}=\left[\begin{array}{lll}{1} & {0} & {0} \\ {0} & {1} & {0} \\ {\mathrm{tx}} & {\text { ty }} & {1} \end{array}\right] \quad\{\mathrm{tx}=1, \mathrm{ty}=1\}$$

$$\therefore[\mathrm{Tr}]^{-1}=\left[\begin{array}{lll}{1} & {0} & {0} \\ {0} & {1} & {0} \\ {1} & {1} & {1}\end{array}\right]$$

Now, combine Transformation Matrix is

$$[\mathrm{T}] \quad=\quad[\operatorname{Tr}][\mathrm{S}][\mathrm{Tr}]^{-1}$$

$$=\left[\begin{array}{ccc}{1} & {0} & {0} \\ {0} & {1} & {0} \\ {-1} & {-1} & {1}\end{array}\right]\left[\begin{array}{lll}{2} & {0} & {0} \\ {0} & {2} & {0} \\ {0} & {0} & {1}\end{array}\right]\left[\begin{array}{ccc}{1} & {0} & {0} \\ {0} & {1} & {0} \\ {1} & {1} & {1}\end{array}\right]$$

$$=\left[\begin{array}{ccc}{2} & {0} & {0} \\ {0} & {2} & {0} \\ {-1} & {-1} & {1}\end{array}\right]$$

Now, new resultant coordinates of square are:

$$\left[\mathrm{X}^{\prime}\right]=[\mathrm{X}][\mathrm{T}]$$

$$\left[\begin{array}{l}{\mathrm{A}^{\prime}} \\ {\mathrm{B}^{\prime}} \\ {\mathrm{C}^{\prime}} \\ {\mathrm{D}^{\prime}}\end{array}\right]=\left[\begin{array}{lll}{1} & {1} & {1} \\ {2} & {1} & {1} \\ {2} & {2} & {1} \\ {1} & {2} & {1}\end{array}\right]\left[\begin{array}{ccc}{2} & {0} & {0} \\ {0} & {2} & {0} \\ {-1} & {-1} & {1}\end{array}\right]$$

$$=\left[\begin{array}{lll}{1} & {1} & {1} \\ {3} & {1} & {1} \\ {3} & {3} & {1} \\ {1} & {3} & {1}\end{array}\right]$$

Hence,

$$A^{\prime}=(1,1)$$ $$B^{\prime}=(3,1)$$ $$C^{\prime}=(3,2)$$ $$D^{\prime}=(1,3)$$

ii. Non uniform scaling with a factor of 2 and 1.5 in $\mathrm{x}$ and $\mathrm{y}$ directions, while anchoring Point $(1,1)$

Step 1: Translation of Point P(1,1) to origin O(0,0)

Translation Matrix is given as:

$$\left[\mathrm{T}_{\mathrm{r}}\right] \quad=\left[\begin{array}{lll}{1} & {0} & {0} \\ {0} & {1} & {0} \\ {\mathrm{tx}} & {\text { ty }} & {1}\end{array}\right] \quad\{\mathrm{tx}=-1, \mathrm{ty}=-1\}$$

$$\therefore[\mathrm{Tr}]=\left[\begin{array}{ccc}{1} & {0} & {0} \\ {0} & {1} & {0} \\ {-1} & {-1} & {1}\end{array}\right]$$

Step 2: Non-Uniform Scaling

$$[\mathrm{S}]=\left[\begin{array}{ccc}{\operatorname{Sx}} & {0} & {0} \\ {0} & {\operatorname{Sy}} & {0} \\ {0} & {0} & {1}\end{array}\right]=\left[\begin{array}{ccc}{2} & {0} & {0} \\ {0} & {1.5} & {0} \\ {0} & {0} & {1}\end{array}\right]$$

Step 3: Inverse translation of Point P from origin to its original position.

Translation Matrix is:

$$[\mathrm{Tr}]^{-1}=\left[\begin{array}{lll}{1} & {0} & {0} \\ {0} & {1} & {0} \\ {\mathrm{tx}} & {\text { ty }} & {1}\end{array}\right] \quad\{\mathrm{tx}=1, \mathrm{ty}=1\}$$

$$\therefore[\mathrm{Tr}]^{-1}=\left[\begin{array}{lll}{1} & {0} & {0} \\ {0} & {1} & {0} \\ {1} & {1} & {1}\end{array}\right]$$

Now, combine Transformation Matrix is

$$[\mathrm{T}]=[\operatorname{Tr}][\mathrm{S}][\mathrm{Tr}]^{-1}$$

$$\begin{aligned}=&\left[\begin{array}{ccc}{1} & {0} & {0} \\ {0} & {1} & {0} \\ {-1} & {-1} & {1}\end{array}\right]\left[\begin{array}{ccc}{2} & {0} & {0} \\ {0} & {1.5} & {0} \\ {0} & {0} & {1}\end{array}\right]\left[\begin{array}{ccc}{1} & {0} & {0} \\ {0} & {1} & {0} \\ {1} & {1} & {1}\end{array}\right] \\ & = \left[\begin{array}{ccc}{2} & {0} & {0} \\ {0} & {1.5} & {0} \\ {-1} & {-0.5} & {1}\end{array}\right] \end{aligned}$$

Now, new resultant coordinates of square are:

$$\left[\mathrm{X}^{\prime}\right]=[\mathrm{X}][\mathrm{T}]$$

$$\left[\begin{array}{l}{\mathrm{A}^{\prime}} \\ {\mathrm{B}^{\prime}} \\ {\mathrm{C}^{\prime}} \\ {\mathrm{D}^{\prime}}\end{array}\right]=\left[\begin{array}{lll}{1} & {1} & {1} \\ {2} & {1} & {1} \\ {2} & {2} & {1} \\ {1} & {2} & {1}\end{array}\right]\left[\begin{array}{ccc}{2} & {0} & {0} \\ {0} & {1.5} & {0} \\ {-1} & {-0.5} & {1}\end{array}\right]$$

$$=\left[\begin{array}{ccc}{1} & {1} & {1} \\ {3} & {1} & {1} \\ {3} & {2.5} & {1} \\ {1} & {2.5} & {1}\end{array}\right]$$

Hence,

$$A^{\prime}=(1,1)$$ $$B^{\prime}=(3,1)$$ $$C^{\prime}=(3,2.5)$$ $$D^{\prime}=(1,2.5)$$