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Solution<br />

(a) The matrix that maps ∆P1Q1R1 onto ∆PQR is the inverse of N.<br />

N -1 = 1 3 2<br />

3 2<br />

1<br />

= <br />

2 1<br />

2 1<br />

(b)<br />

Example<br />

P1 Q1 R1 P Q R<br />

3 2<br />

3 2 0<br />

<br />

5 8 2<br />

= <br />

2 1<br />

2 1 1<br />

4 5 1<br />

Therefore, the coordinates of PQR are P(-5, 4), Q(8, -5), R(2, -1)<br />

2<br />

1 <br />

A(2, 1), B(5, 1) and C(5, 3) are vertices of ∆ABC and M = .<br />

3<br />

2<br />

(a) Find the image A1B1C1 of ABC under the transformation represented by<br />

matrix M.<br />

(b) Find M -1 .<br />

(c) Find the image of A1B1C1 under the transformation represented by matrix<br />

M -1 .<br />

(d) Comment on your results in (c).<br />

Solution<br />

A B C A1 B1 C1<br />

2<br />

1 2<br />

5 5 5<br />

11 13 <br />

(a) = Therefore, the coordinates of the<br />

3<br />

2<br />

1<br />

1 3<br />

8<br />

17 21<br />

vertices of the image are: A1(5, 8), B1(11, 17) and C1(13, 21).<br />

Area and the determinant of a matrix.<br />

Under some transformations, the size and shape of objects do not change,<br />

while under other transformations they change. When a matrix represents a<br />

transformation, the ratio of the area of the image to the area of the object is<br />

equal to the determinant of the matrix of transformation.<br />

Example<br />

A unit square with vertices O(0, 0), I(1, 0), K(1, 1) and J(0, 1) is given a<br />

3<br />

0 <br />

transformation represented by matrix N = .<br />

0<br />

2<br />

(a) Draw the square and its image.<br />

(b) Find the determinant of N.<br />

(c) Find the area of the image OI1K1J1.<br />

Solution<br />

O I K J O I1 K1 J1

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