Matrices Module 12 by Rashidah Iman

HIKMAH MODULE FROM YAYASAN PELAJARAN JOHOR

TOPIC : MATRICES

 2 6  . 1 4

(a) Find inverse of the matrix 

1  4  6 ] 2   1 2 

[ 

(b) Hence, using matrices, calculate the value of x and of y that satisfy the following simultaneous linear equations:

 2 6  k   0         1 4  m   1  [k = –3, m = 1] Answer :

169

SMK SEKSYEN 24(2)

4 3 . k 

Given that matrix G   1 (i)

Find the value of k, if G not have inverse matrix. [k =

(ii)

3 ] 4

Given that k = 2, (a) Find inverse matrix of G. (b) Hence, using matrices, calculate the value of d and of e that satisfy the following matrix equations:

 4 3  d   21        1 k  e   4  (a)

1  2  3   5   1 4 

(b) d = 6, e = –1 Answer :

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HIKMAH MODULE FROM YAYASAN PELAJARAN JOHOR

3  2

1 0

    . M is a 2 x 2 matrix where M  5  4  0 1 (a) Find the matrix M. [

1   4 2  ]  2   5 3 

(b) Write the following simultaneous linear equations as a matrix equation. 3x – 2y = 7 5x – 4y = 9 Hence, calculate the values of x and y using matrices. [x = 2, y = Answer :

171

5 ] 2

SMK SEKSYEN 24(2)

5 1  M = 3 2

(a) State matrix M if 

5 1   3 2 1 0  ] 0 1

[ 

(b) Using matrices, calculate the value of u and w that satisfy the following matrix equations:

 5 1  u    6         3 2  w   2  [u = –2, w = 4]

Answer :

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HIKMAH MODULE FROM YAYASAN PELAJARAN JOHOR

 1 2   1 0    find matrix A.  3 5   0 1 

(a) Given that A 

5  2  ] 3 1

[A = 

(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 [r = –2, s = –3 ]

Answer :

173

SMK SEKSYEN 24(2)

 4 5 1 0  and matrix PQ =    6 8 0 1

Given matrix P = 

(a) Find the matrix Q. [Q =

1  8  5  ] 2   6 4 

(b) 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 [m = 3, n = –1] Answer :

174

HIKMAH MODULE FROM YAYASAN PELAJARAN JOHOR

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

It is given that matrix P =  (a) Find the value of k.

[k = 

4 ] 5

(b) 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 [x = –1, y = 3] Answer :

175

SMK SEKSYEN 24(2)

8(a)

 2 4  2 4 M =   1 3 1 3

Find matrix M such that 

1 0  ] 0 1

[M =  (b)

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

 2 4  x   6         1 3  y   5  [x = –1, y = 2] Answer :

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HIKMAH MODULE FROM YAYASAN PELAJARAN JOHOR

9(a)

 3 1    5 2 

Find the inverse of matrix 

 2  1  ]  5  3

[  (b)

Hence, using matrices, calculate the values of d and e that satisfy the following simultaneous equations : 2d – e = 7 5d – e = 16 [d = 5, e = 3]

Answer :

177

SMK SEKSYEN 24(2)

 1 2   , find 2 5 

Given matrix M = 

(a) the inverse matrix of M

1  5 2   9   2 1  (b) hence, using matrices, the values of u and v that satisfy the following simultaneous equations : u – 2v = 8 2u + 5v = 7 [u = 6, v = –1] Answer :

178

HIKMAH MODULE FROM YAYASAN PELAJARAN JOHOR

11(a)

 3 2   4 n   is m    5 4   5 3 

The inverse matrix of 

Find the value of m and of n. [m =  (b)

1 , n = 2] 2

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

Answer :

179

1 ] 2

SMK SEKSYEN 24(2)

12(a)

 m 3 1  4 3   ,  and the inverse matrix of G is 14  2 m   2 n

Given that G = 

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

[m = 5, n = 4] Hence, using matrices, calculate the value of p and of q that satisfies the following equation :

 p  1  G      q   8  [p = 2, q = 3] Answer :

180

HIKMAH MODULE FROM YAYASAN PELAJARAN JOHOR

MODULE 12 - ANSWERS TOPIC : MATRICES =

1  4  6   8  6   1 2 

1  4  6   2   1 2 

1. (a)

(b)

 k  1  4  6  0         m  2   1 2  1    3   1 

= 

k = –3, m = 1 2.

(a) (i) 4k – 3 = 0 k=

(ii) (a)

3 4

1  2  3   8  3   1 4  1  2  3  =  5   1 4  d 

1 2

(b)      e  5  1

 3  21   4  4 

6   1

= 

d = 6, e = –1 3. (a)

(b)

  4 2 1    12  10   5 3  1   4 2   =  2   5 3   x 1   4 2  7         y   2   5 3  9 

181

SMK SEKSYEN 24(2)

2

= 5

  2 5 x = 2, y = 2 1 0  0 1 

4. (a)

(b)

M= 

 u  1  2  1  6         w  7   3 5  2    2  =   4  u = –2, w = 4

5. (a)

(b)

5  2  3 1

A = 

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

6. (a)

(b)

1  8 5    32  30  6 4  1  8 5  =   2  6 4 

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

= 

m = 3, n = –1 7.

(a)

–4 – 5k = 0

k 

4 5

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HIKMAH MODULE FROM YAYASAN PELAJARAN JOHOR

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

(b)

 x 1  2 5  13       9  1 2  7   y x = –1, y = 3

8. (a)

(b)

1 0  0 1

M = 

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

9. (a)

(b)

1  2 1    6  5  5 3   2 1  =    5 3  =

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

= 

5  3

=   d = 5, e = 3 10. (a)

(b)

1  5 2   5  (4)  2 1  1  5 2 =   9  2 1 

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

183

SMK SEKSYEN 24(2)

1  54    9  9  6 =    1  =

u = 6, v = –1 11.

(a)

(b)

12. (a) b)

m= 

1 , n = 2 2

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

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

END OF MODULE 12

184