Module 12- Matrices

MODULE 12 MATEMATIK SPM “ENRICHMENT” TOPIC : MATRICES TIME : 2 HOURS

1.

(a)

3  2  is m 5  4

The inverse matrix of 

 4   5

n  3 

Find the value of m and of n. (b)

Answer : (a)

(b)

Hence, using matrices, solve the following simultaneous equations : 3x – 2y = 8 5x – 4y = 13

2.

(a)

m 3  2 n  

Given that G =  

and the inverse matrix of G is

1  4  3  , 14   2 m 

find the value of m and of n. (b)

Hence, using matrices, calculate the value of p and of q that satisfies the following equation :

 p  1  G      q    8 Answer : (a)

(b)

3.

 1 2 1 0    , A  3 5 0 1    

(a)

Given that

(b)

Hence, using the matrix method, find the value of r and s which satisfy the simultaneous equations below. -r + 2s = -4 -3r + 5s = -9

Answer : (a)

(b)

find matrix A.

4.

 4 5  and matrix PQ = 6 8  

Given matrix P =   (a) (b)

Answer : (a)

(b)

1 0   0 1  

Find the matrix Q. Hence, calculate by using the matrix method, the values of m and n that satisfy the following simultaneous linear equations : 4m + 5n = 7 6m + 8n = 10

5.

 4  3  , 8  5  

Given the matrix P is  

1 0  0 1  

(a)

Find the matrix Q so that PQ = 

(b)

Hence, calculate the values of h and k, which satisfy the matrix equation:

 4  3  h    7        8  5 k  11      Answer : (a)

(b)

6.

 k 6 , find the value of k if matrix M has no inverse.  4 2  

(a)

Given matrix M =  

(b)

Given the matrix equations

 7  6  x    4          5 8  y   1  (i) (ii) Answer : (a)

(b)

 x

1  8 6   4 

and      y  h  5 7  1      

Find the value of h Hence, find the value of x and y.

7.

2 5   does not have an inverse matrix. k  2  

It is given that matrix P =   (a) (b)

Find the value of k. If k = 1, find the inverse matrix of P and hence, using matrices, find the values of x and y that satisfy the following simultaneous linear equations. 2x + 5y = 13 x - 2y = -7

Answer : (a)

(b)

8.

 2 4  2 4  M    1 3 1 3    

(a)

Find matrix M such that  

(b)

Using matrices, calculate the values of x and y that satisfy the following matrix equation.

 2 4  x   6         1 3  y   5 

Answer : (a)

(b)

9.

3 1  . 5  2  

(a)

Find the inverse of matrix  

(b)

Hence, using matrices, calculate the values of d and e that satisfy the following simultaneous equations : 2d – e = 7 5d – e = 16

Answer : (a)

(b)

10.

1  2  , find 2 5  

Given matrix M =   (a) (b)

the inverse matrix of M hence, using matrices, the values of u and v that satisfy the following simultaneous equations : u – 2v = 8 2u + 5v = 7

Answer : (a)

(b)

MODULE 12 - ANSWERS TOPIC : MATRICES

1.

m= 

(a)

1 2

1m

n =2 (b)  3

 5

1m

 2  x  8    =    4   y  13   x  1  4      y 2  5 x=3 y= 

2.

(a)

 2

2  8    3 13 

1 2

1m

1m 1m

3  p     4  q 

 1      8

 p  1  4  3  1         q  14   2 5   8  p=2 q = -3

3.

(a)

(b)

4.

(a)

5 3 

A=

1m

1m

n =4 m=5

(b)  5

1m

1m

1m

1m 1m

 2   1 

2m

  1 2  r    4         3 5  s    9

1m

 r  1  5  2   4         s  1  3  1   9 

1m

P=

r = -2

1m

s = -3

1m 1m

1  8  5   32  30   6 4 

(b)

5.

(a)

(b)

1  8  5   = 2   6 4 

1m

 4 5  m   7         6 8  n  10 

1m

 m  1  8  5  7         n  2   6 4 10 

1m

m=3

1m

n = -1

1m

  5 3  1m 1    20  (24)  8 4  1   5 3 1m    4  8 4 

P 

 4  3  h    7         8  5  k    11  h  1   5 3   7         k  2  8 4   11 

6.

1 2    2   100 

1m

1m

h=1 k = -50

1m 1m

(a)

k = -12

1m

(b)

(i)

h = 26

 x 1  8 6   4         y  26  5 7  1  1   26     26   13 

1m

(ii)

1m

1m

x = -1 y= 

7.

(a)

1m

1 2

- 4 – 5k = 0

1m

1m

5k = -4 k=  (b)

8.

4 5

1m

 2 5  x   13         1  2  y    7 

1m

 x 1   2  5  13         y  1 2 9      7 

1m

x = -1 y=3

1m 1m

1 0  0 1

(a)

M=  

2m

(b)

 x 1  3  4  6         y  6  4   1 2  5 

1m

1  3  4  6     2   1 2  5  1   2    2 4  

x = -1 y=2

1m

1m 1m

9.

(a)

1   2 1    6  5   5 3 

1m

1   2 1    1   5 3 

1m



 1  d   7         5  3  e  16   d  1   3 1  7         e   1   5 2 16 

(b)  2



1m

1m

1   5    1   3 

 5     3 d=5 e=3

10.

(a)

 5 2 1   5  (4)   2 1  1  5 2    9   2 1   2  u   8   2 5  v    7        u  1  5 2  8         v  9   2 1  7 

(b)  1

1m 1m

1m

1m

1m

1m

1  54     9   9 6      1 u6 v  1

1m 1m