## 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

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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 AIBICI onto triangle AIIBII CII whose coordinates are (-3, -2) , (-3, -6) and

(-1, -2) respectively.

16.  Triangle ABC with an area of 15 cm2 is mapped onto triangle AIBICI using matrix

M = 2 -3 . Find the area of triangle AIBICI.

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.

 1 Let M1 M1   A1  3 AlternativeC = B-1A equations B-1 allow any two solving of equations or equivalent 2. M1   M1    A1 03

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 1st image.

100 x (4 – 3)

700 area of 2nd 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 + 90o

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 AIBICI = 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)

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