Matrices Questions and Answers

Matrices Questions

1. Given that and find C such that B × C = A (3mks)

2. Use matrix method to solve the (3mks)

3y + 2x = 13

2y – 3x = 0

3. A matrix. Find the values of a and b given that PQ = R. Using matrix method. (3mks)

4. A matrix A is given as

 (i) Determine A² (1mk)

5. Two matrices A and B are such that

A = k 4 B = 1 2

 3 2 3 4

Given that the determinant of AB = 4, find the value of K. (3 mks)

6. Given that A is 3 2 and A 1 2

4 -1 11 11

4 -3

11 11

 Find the value of a and b in the expression: (3 mks)

 3 2 a = 12

 4 -1 b 5

7. Solve for the unknowns given that the following is a singular matrix.

1 2

x
x-3

8. Given that A = and B = and that C = AB, find C^-1

9. B is a matrix 3 2 and C is the matrix 9 -3

2 2 2 1

. If A is a 2 x 2 matrix and A x B = C. determine the matrix A.

10. An object of area 20 cm² undergoes a transformation given by the matrix

-1 -2 followed by 2 3 find the area of the final image.

4 3 -1 2

11. Find the matrix B such that AB = I and A = 3 2 . Hence find the point of intersection of the

-1 3

lines 3x + 2y
= 10 and 3y – 4 = x.

12. Given that P = and Q = find the matrix product PQ. Hence solve the

 simultaneous equations below:-

2x – 3y = 5

– x + 2y = – 3

13. Solve for x and y in the following matrix equation using elimination method

 ½ – ¼ x 2

²/₅
¹/₆ y 6

14. A triangle XYZ , X (-1, -1) , Y (-2, -4) Z (-6 , -9) is reflected in the line X axis followed

by a reflection in line X= Y. Find the image of the final image

15. Triangle ABC is the image of triangle PQR under a transformation M = 2 4 where

P, Q, R map onto A, B, C respectively. 0 2

 Given the points P (5, -1) Q (6, -1) and R(4, – 0.5) draw the triangle ABC on the grid

provided.

b) Triangle ABC in (a) above is to be enlarged by scale factor 2 with centre at (11, – 6) to map

onto A¹B¹ and C¹. Construct and label triangle A¹B¹ and C¹ on the same grid.

c) By construction, find the coordinates of the centre and the angle of rotation which can be used

to rotate triangle A^IB^IC^I onto triangle A^IIB^II C^II whose coordinates are (-3, -2) , (-3, -6) and

(-1, -2) respectively.

16. Triangle ABC with an area of 15 cm² is mapped onto triangle A^IB^IC^I using matrix

M = 2 -3 . Find the area of triangle A^IB^IC^I.

1 1

17. T is a transformation represented by the matrix under T a square whose area

is 10cm² is mapped onto a square of area 110cm². Find the possible values of X

18. Triangle A¹B¹C¹ is the image of  ABC under a transformation represented by the matrix

M = 3 2

9 5

 If the area of triangle A¹B¹C¹ is 54cm². Determine the area of triangle ABC

19. Find the matrix B such that AB = I and A = 3 2 . Hence find the point of intersection of the

-1 3

lines 3x + 2y
= 10 and 3y – 4 = x.

Matrices Answers

1.

3 2 a = 12

4 -1 b 5

1
2 3 2 a 1
2 12

11 11 = 11 11

4
-3 4 -1 b 4
-3 5

11 11 11 11

1 0 a = 2

0 1 b 3

a = 2

b 3

a = 2 √ and b = 3 √

2. (x-3) – (2x) = 0

x-3-2x = 0

-2x + x – 3 = 0

-x -3 = 0

x = 3

3. 1 5 7 3 = -13 -7

3 7 -4 -2 -4 -2

Determinant = + 65 – 49 = 16

C^-1 = 1 -5 7

7 -13

4.

3 2 a b = 9 -3

2 2 c d 2 1

3a + 2c = 9

2a + 2c = 2

a = 7

c = -6

3b + 2d = -3

2b + 2d = 1

b = -4

d = 4.5

A = 7 – 4

-6 4.5

5. 20x (-3 – ^–8)

 100 area of 1^st image.

 100 x (4 – ^–3)

 700 area of 2^nd image

6. Det. 9 + 2 = 11

A¹ = 1 3 -2

11 1 3

3 2 x = 10

3 -1 y 4

x = 1 3 -2 10

y 11 1 3 4

x = 1 22

y 11 22

x = 2

y 2

 P (2, 2)

7 PQ = 1 0

 0 1

 2 -3 x = 5

-1 2 y -3

2 3 2 -3 x = 2 3 5

1 2 -1 2 y 1 q -3

x = 1

y -1

x =1 y= -2

8. ½ x – ¼ y = 2

²/₅ + ¹/₆ = 6

2x _ y = 8

12x + 5y = 180

10x – 5y = 40 +

22x = 220

x = 10

¼ y = ½ (10) ^-2

¼ y = 5-2 = 3

Y = 12

9.

=

=

Final image X¹¹ Y¹¹Z¹¹

X¹¹(1, -1) Y¹¹(4, -2), Z¹¹(9, -6)

10. P Q R A B C

a|: 2 2 5 6 4 = 6 8 6

0 4 -1 -1 – ½ -2 2 -1

(c) Centre (-3,2)

Angle + 90^o

A B C

a) 2 4 5 6 4 = 6 8 6

0 2 -1 -1 – ½ 2 2 -1

11. Det 2 – -3 = 5

Area of A^IB^IC^I = 5 x 15

= 75 cm²

12. A.S.F = 110 = 11

 10

 5X (X) – -6 = 11

 5X² + 6 = 11

 5x² = 5

 X² = 1

X =  1

13. Area of the image = Area of the object x Det.

Det. (∆) = 15 – 18 = -3

54 cm² = A x -3

54 cm² = A

3

Area of ∆ ABC = 18 cm²

14. Det. 9 + 2 = 11

A¹ = 1 3 -2

11 1 3

3 2 x = 10

3 -1 y 4

x = 1 3 -2 10

y 11 1 3 4

x = 1 22

y 11 22

x = 2

y 2

 P (2, 2)