Triangle PQR has vertices P(2,6), Q(2,2), and R(4,6). Find the image of triangle PQR under the transformations given by matrices a) [[0, 1], [1, 0]] and b) [[1, 2], [-1, 1]].

Here are the images of triangle PQR under the given transformations. The vertices of triangle PQR are $P(2,6)$, $Q(2,2)$, and $R(4,6)$.

a) For the matrix $\begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}$:

Step 1: Calculate the image of point P.

$$P' = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} \begin{pmatrix} 2 \\ 6 \end{pmatrix} = \begin{pmatrix} (0)(2) + (1)(6) \\ (1)(2) + (0)(6) \end{pmatrix} = \begin{pmatrix} 6 \\ 2 \end{pmatrix}$$

Step 2: Calculate the image of point Q.

$$Q' = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} \begin{pmatrix} 2 \\ 2 \end{pmatrix} = \begin{pmatrix} (0)(2) + (1)(2) \\ (1)(2) + (0)(2) \end{pmatrix} = \begin{pmatrix} 2 \\ 2 \end{pmatrix}$$

Step 3: Calculate the image of point R.

$$R' = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} \begin{pmatrix} 4 \\ 6 \end{pmatrix} = \begin{pmatrix} (0)(4) + (1)(6) \\ (1)(4) + (0)(6) \end{pmatrix} = \begin{pmatrix} 6 \\ 4 \end{pmatrix}$$

The image of triangle PQR is $\boxed{P'(6,2), Q'(2,2), R'(6,4)}$.

b) For the matrix $\begin{pmatrix} 1 & 2 \\ -1 & 1 \end{pmatrix}$:

Step 1: Calculate the image of point P.

$$P' = \begin{pmatrix} 1 & 2 \\ -1 & 1 \end{pmatrix} \begin{pmatrix} 2 \\ 6 \end{pmatrix} = \begin{pmatrix} (1)(2) + (2)(6) \\ (-1)(2) + (1)(6) \end{pmatrix} = \begin{pmatrix} 2 + 12 \\ -2 + 6 \end{pmatrix} = \begin{pmatrix} 14 \\ 4 \end{pmatrix}$$

Step 2: Calculate the image of point Q.

$$Q' = \begin{pmatrix} 1 & 2 \\ -1 & 1 \end{pmatrix} \begin{pmatrix} 2 \\ 2 \end{pmatrix} = \begin{pmatrix} (1)(2) + (2)(2) \\ (-1)(2) + (1)(2) \end{pmatrix} = \begin{pmatrix} 2 + 4 \\ -2 + 2 \end{pmatrix} = \begin{pmatrix} 6 \\ 0 \end{pmatrix}$$

Step 3: Calculate the image of point R.

$$R' = \begin{pmatrix} 1 & 2 \\ -1 & 1 \end{pmatrix} \begin{pmatrix} 4 \\ 6 \end{pmatrix} = \begin{pmatrix} (1)(4) + (2)(6) \\ (-1)(4) + (1)(6) \end{pmatrix} = \begin{pmatrix} 4 + 12 \\ -4 + 6 \end{pmatrix} = \begin{pmatrix} 16 \\ 2 \end{pmatrix}$$

The image of triangle PQR is $\boxed{P'(14,4), Q'(6,0), R'(16,2)}$.

c) For the matrix $\begin{pmatrix} -2 & 0 \\ 0 & -3 \end{pmatrix}$:

Step 1: Calculate the image of point P.

$$P' = \begin{pmatrix} -2 & 0 \\ 0 & -3 \end{pmatrix} \begin{pmatrix} 2 \\ 6 \end{pmatrix} = \begin{pmatrix} (-2)(2) + (0)(6) \\ (0)(2) + (-3)(6) \end{pmatrix} = \begin{pmatrix} -4 \\ -18 \end{pmatrix}$$

Step 2: Calculate the image of point Q.

$$Q' = \begin{pmatrix} -2 & 0 \\ 0 & -3 \end{pmatrix} \begin{pmatrix} 2 \\ 2 \end{pmatrix} = \begin{pmatrix} (-2)(2) + (0)(2) \\ (0)(2) + (-3)(2) \end{pmatrix} = \begin{pmatrix} -4 \\ -6 \end{pmatrix}$$

Step 3: Calculate the image of point R.

$$R' = \begin{pmatrix} -2 & 0 \\ 0 & -3 \end{pmatrix} \begin{pmatrix} 4 \\ 6 \end{pmatrix} = \begin{pmatrix} (-2)(4) + (0)(6) \\ (0)(4) + (-3)(6) \end{pmatrix} = \begin{pmatrix} -8 \\ -18 \end{pmatrix}$$

The image of triangle PQR is $\boxed{P'(-4,-18), Q'(-4,-6), R'(-8,-18)}$.

d) For the matrix $\begin{pmatrix} \frac{1}{2} & \frac{1}{4} \\ \frac{1}{2} & \frac{1}{2} \end{pmatrix}$:

Step 1: Calculate the image of point P.

$$P' = \begin{pmatrix} \frac{1}{2} & \frac{1}{4} \\ \frac{1}{2} & \frac{1}{2} \end{pmatrix} \begin{pmatrix} 2 \\ 6 \end{pmatrix} = \begin{pmatrix} (\frac{1}{2})(2) + (\frac{1}{4})(6) \\ (\frac{1}{2})(2) + (\frac{1}{2})(6) \end{pmatrix} = \begin{pmatrix} 1 + \frac{6}{4} \\ 1 + 3 \end{pmatrix} = \begin{pmatrix} 1 + \frac{3}{2} \\ 4 \end{pmatrix} = \begin{pmatrix} \frac{2+3}{2} \\ 4 \end{pmatrix} = \begin{pmatrix} \frac{5}{2} \\ 4 \end{pmatrix}$$

Step 2: Calculate the image of point Q.

$$Q' = \begin{pmatrix} \frac{1}{2} & \frac{1}{4} \\ \frac{1}{2} & \frac{1}{2} \end{pmatrix} \begin{pmatrix} 2 \\ 2 \end{pmatrix} = \begin{pmatrix} (\frac{1}{2})(2) + (\frac{1}{4})(2) \\ (\frac{1}{2})(2) + (\frac{1}{2})(2) \end{pmatrix} = \begin{pmatrix} 1 + \frac{2}{4} \\ 1 + 1 \end{pmatrix} = \begin{pmatrix} 1 + \frac{1}{2} \\ 2 \end{pmatrix} = \begin{pmatrix} \frac{2+1}{2} \\ 2 \end{pmatrix} = \begin{pmatrix} \frac{3}{2} \\ 2 \end{pmatrix}$$

Step 3: Calculate the image of point R.

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