J20

1 1 1
0
1
2
5 5
P Q R
= 2 1 6
1
0
2
1 6
2 5 5
Therefore, the coordinates of the image are P 1 (-2, 2), Q 1 (-1, 5) and R 1 (-6, 5).
We can find the image of any point in the plane by pre-multiplying the position vector of
that point by the matrix of the transformation.
Example
Find the images of the points with the following position vectors under the
2
1
transformation whose matrix is
1
2
1
0
(a)
(b)
0
1
2
1
(c)
(d)
5
1
Solution
(a)
(b)
(c)
(d)
2
1 1
21
1
0 20
2
= = =
1
2
0
11
2
0
1
0
1
1
2
So the image of is
0
1
2
1 0 01
1
= = .
1
2
1
0
2
2
0
1
So the image of is .
1
2
2
1 2 4
5 9
= =
1
2
5
2
10
12
2
9
So the image of is .
5
12
2
1 1 2
1 3
1
3
= = . So the image of is .
1
2
1
1
2
3
1
3
1
Parts (a) and (b) above are especially interesting as the images of and
0
columns of the transformation matrix.
In general,
0
are the
1