Congratulations Andrei Lazanu, age 14, of School No. 205, Bucharest, Romania for this excellent solution.

I looked at the images of points (1, 0) and (0, 1) respectively, under the transformation by the matrix

(\begin{matrix} a & b c & d \end{matrix})

(\begin{matrix} a & b c & d \end{matrix}) (\begin{matrix} 10 \end{matrix}) = (\begin{matrix} a c \end{matrix})

This means that under this transformation, the point (1, 0) maps to the point

(a, c)

(\begin{matrix} a & b c & d \end{matrix}) (\begin{matrix} 01 \end{matrix}) = (\begin{matrix} b d \end{matrix})

Under this transformation, the point (0,1) maps to the point

(b, d)

.

Now, I look at the effect on the plane of the four transformations:

1.

b = c = 0, a = d = - 1

(\begin{matrix} - 1 & 0 0 & - 1 \end{matrix}) (\begin{matrix} x y \end{matrix}) = (\begin{matrix} - x - y \end{matrix}) .

This means each point (x, y) transforms into its symmetrical image with respect to the origin, i.e. into the point (-x, -y). This is a rotation of 180 degrees about the origin.

2. $b = c = 0, a = - 1, d = 1$

$(\begin{matrix} - 1 & 0 0 & 1 \end{matrix}) (\begin{matrix} x y \end{matrix}) = (\begin{matrix} - x y \end{matrix}) .$

Here, it is a reflection in the y-axis.

3. $b = c = 0, a = d = 1$

$(\begin{matrix} 1 & 0 0 & 1 \end{matrix}) (\begin{matrix} x y \end{matrix}) = (\begin{matrix} x y \end{matrix}) .$

In this situation each point transforms into itself.

4. $b = c = 0, a = 1, d = - 1$

$(\begin{matrix} 1 & 0 0 & - 1 \end{matrix}) (\begin{matrix} x y \end{matrix}) = (\begin{matrix} x - y \end{matrix}) .$

This transformation leaves the abscissa unchanged and modifies the sign of the ordinate, being a reflection in the x-axis.

Now, I look at the next set of transformations.

5. $a = d = 0, b = c = 1$

$(\begin{matrix} 0 & 1 1 & 0 \end{matrix}) (\begin{matrix} x y \end{matrix}) = (\begin{matrix} y x \end{matrix}) .$

In this transformation, the abscissa and the ordinate are interchanged, the transformation being a reflection in respect to the line $y = x$ , the angle bisector of the first quadrant.

6. $a = d = 0, b = 1, c = - 1$

$(\begin{matrix} 0 & 1 - 1 & 0 \end{matrix}) (\begin{matrix} x y \end{matrix}) = (\begin{matrix} y - x \end{matrix}) .$

Looking to the solution to "Complex Rotations" from the NRICH +15, July 2003, I see that this comes to the transformation of point C into point G: here the plane is complex, on the abscissa there is the real part, and the ordinate the imaginary one, but in fact the things are completely similar. This is a rotation of 270 degrees anti-clockwise about the origin.
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7. $a = d = 0, b = c = - 1$

$(\begin{matrix} 0 & - 1 - 1 & 0 \end{matrix}) (\begin{matrix} x y \end{matrix}) = (\begin{matrix} - y - x \end{matrix}) .$

This corresponds to a reflection in the line $y = - x$ .

8. $a = d = 0, b = - 1, c = 1$

$(\begin{matrix} 0 & - 1 1 & 0 \end{matrix}) (\begin{matrix} x y \end{matrix}) = (\begin{matrix} - y x \end{matrix}) .$

This corresponds to a rotation of the point by 90 degrees anti-clockwise about the origin.

The difference between the transformations (5 - 8) and the transformations (1 - 4) is that not only the signs are changed, but a reversal of the x and y coordinates takes place simultaneously.

I start to analyse the effect of these transformations on the unit square with vertices (0,0), (0,1), (1,1), (1,0). By observing what happens to the two points A(1,0) and B(0,1) we can see what happens to the whole square which moves rigidly with the edges OA and OB. All the points in the plane are transformed in the same way as the unit square.

1) The matrix $(\begin{matrix} - 1 & 0 0 & - 1 \end{matrix})$ maps (1,0) to (-1,0) and (0,1) to (0,-1), a rotation of 180 degrees about the origin.

2) The matrix $(\begin{matrix} - 1 & 0 0 & 1 \end{matrix})$ maps (1,0) to (-1,0) and (0,1) to (0,1), a reflection in the y-axis.

3) The matrix $(\begin{matrix} 1 & 0 0 & 1 \end{matrix})$ maps (1,0) to (1,0) and (0,1) to (0,1), the identity transformation which leaves all points unchanged.

4) The matrix $(\begin{matrix} 1 & 0 0 & - 1 \end{matrix})$ maps (1,0) to (1,0) and (0,1) to (0,-1), a reflection in the x-axis.

5) The matrix $(\begin{matrix} 0 & 1 1 & 0 \end{matrix})$ maps (1,0) to (0,1) and (0,1) to (1,0), a reflection in the line $y = x$ .

6) The matrix $(\begin{matrix} 0 & 1 - 1 & 0 \end{matrix})$ maps (1,0) to (0-1) and (0,1) to (1,0), a rotation of 90 degrees clockwise or 270 degrees anti-clockwise about the origin.

7) The matrix $(\begin{matrix} 0 & - 1 - 1 & 0 \end{matrix})$ maps (1,0) to (0,-1) and (0,1) to (-1,0), a reflection in the line $y = - x$ .

8) The matrix $(\begin{matrix} 0 & - 1 1 & 0 \end{matrix})$ maps (1,0) to (0,1) and (0,1) to -1,0), a rotation of 90 degrees anti-clockwise about the origin.

Each point transforms into its symmetrical in respect to the origin, so that the result is as follows: the square transforms into another square, the image of the red square under the transformation being the blue one.
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%Insert figure 2 \end 2) This in the reflection in respect to the y axis. Figure 2 3) This is the identity transformation, each point being left as is was: Figure 3 4) Each point of the plane, i.e. the points inside or on the edges and vertices of the unit square transform into points with the same abscissa and with opposite sign ordinate: Figure 4 5. ' The same as Figure 3, but only if the identity of the vertices is lost. Being a reflection in respect to the line y = x, and starting from a square with two vertices on it, we obtained in fact a rotated square.. 6. ' The effect on the square is the same as the one described by transformation 4, see Fig. 4. 7. ' The same as Figure 1. 8. ' The same as Figure 2. One of the ways to work with transformations is to use a matrix. If you have not met matrices before don't be put off, they are very easy. In this question you will use some simple matrices for rotations and reflections and see how they work.

First you need to know how to multiply a matrix like $(\begin{matrix} a & b c & d \end{matrix})$ by the vector $(\begin{matrix} x y \end{matrix})$ to give the image of the point $(x, y)$ . This multiplication is defined as follows:

$(\begin{matrix} a & b c & d \end{matrix}) (\begin{matrix} x y \end{matrix}) = (\begin{matrix} ax + by cx + dy \end{matrix}) .$

Find the images of the points $(1, 0)$ and $(0, 1)$ under the transformation given by the matrix $(\begin{matrix} a & b c & d \end{matrix})$ .

Describe the effect on the plane of the four transformations where $b = c = 0$ and $a$ and $d$ take all possible combinations of the values $\pm 1$ . Now describe the effect on the plane of the four transformations where $a = d = 0$ and $b$ and $c$ take all possible combinations of the values $\pm 1$

Explain why transformations have the same effect on the whole plane as on the unit square with vertices $(0, 0), (0, 1), (1, 1), (1, 0)$ .