EdExcel Further Pure 1. © MEI, 15/03/10. 1/3. Matrix algebra. Chapter ... EdExcel
FP1 Matrices Assessment solutions. © MEI, 09/06/08. 2/3. Matrix algebra.
EdExcel Further Pure 1 Matrix algebra Chapter Assessment 3 2 1. The matrix represents a transformation, T. The inverse of transformation 2 1 T is W. (i) Find the matrix representing W. [3] (ii) Show that TW = WT = I. [3] 2. A transformation T is given by: x x 2y T: y y x (i) Write down the matrix representing the transformation T. [2] (ii) The transformation S is an anticlockwise rotation through 90° about the origin. Find the matrix representing the transformation S. [2] (iii) Find the single matrix representing the transformation S followed by the transformation T. [3]
6 3 3. The plane is transformed by means of the matrix M = . 4 k (i) In the case where k = 3: (a) find the determinant of M, (b) find the area of the image of a triangle with area 4 square units. (ii) Find the value of k for which the matrix is singular.
[2] [2] [2]
2 1 1 0 4. The matrix A and the matrix B . 3 1 3 2 (i) Find A-1 and B-1. (ii) Hence or otherwise, find (AB)-1.
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2 3 . 5. Matrix A represents a transformation T where A = 4 4 (i) Find the image of the point (2, -1) under the transformation T. (ii) Find the inverse of A. (iii) Find the coordinates of the point that is mapped to (9, 16) under transformation T. (iv) Find A2. d 0 stating the value of d. (v) Show that A3 = 0 d (vi) Give a geometrical description of the matrix A3.
[2] [3] [3] [2] [2] [2]
Total 40 marks
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EdExcel FP1 Matrices Assessment solutions Matrix algebra Solutions to chapter assessment 1. (i)
3 2 det 3 ( 4) 1 2 1 1 2 W T 1 2 3 [3]
3 2 1 2 1 0 (ii) TW I 2 1 2 3 0 1 1 2 3 2 1 0 WT I 2 3 2 1 0 1 [3] 2. (i) The point (1, 0) is mapped to the point (1, -1). The point (0, 1) is mapped to the point (2, 1). 1 2 The matrix representing T is therefore . 1 1 [2] (ii) The point (1, 0) is mapped to the point (0, 1). The point (0, 1) is mapped to the point (-1, 0). 0 1 The matrix representing S is therefore . 1 0 [2] (iii) S followed by T is represented by the matrix product TS. 1 2 0 1 TS 1 1 1 0 2 1 1 1 [3] 3. (i) (a) (b)
det M
6 3 18 12 6 4 3
Area factor = 6, so area of image 4 6 24 square units.
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(ii) If the matrix is singular, then the determinant is zero. 6 3 det M 6k 12 4 k
det M 0 6k 12 0 k 2
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EdExcel FP1 Matrices Assessment solutions 4. (i)
det A (2 1) ( 1 3) 2 3 5 A 1
1 1 3 2
1 5
det B (1 2) (0 3) 2 B 1
1 2 0 1 2 0 2 3 1 2 3 1
(ii) (AB)-1 B 1 A 1
1 2 0 1 1 1 2 3 1 5 3 2 1 2 2 10 6 1
5. (i)
2 3 2 7 4 4 1 12 The image is (7, 12).
2 3 (ii) det 8 12 4 4 4 1 4 3 A-1 = 4 4 2
x 9 (iii) A y 16 9 1 4 3 9 1 12 3 x 1 y A 16 4 2 16 4 1 4 4 The point (3, -1) is mapped to (9, 16). 2 3 2 3 8 6 (iii) A² = . 4 4 4 4 8 4 8 6 2 3 8 0 (iv) A³ = so d = 8 8 4 4 4 0 8 (v) The matrix A³ is an enlargement, centre the origin, scale factor 8.
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