4.10 SPM Practice (Long Questions)


Question 11 (5 marks):
Table 8 shows the information of books purchased by Maslinda.
 
Table 8

Maslinda purchased x history books and y Science books. The total number of books purchased is 5. The total price of the books purchased is RM17.

(a)
 Write two linear equations in terms of x and y to represent the above information.

(b)
 Hence, by using matrix method, calculate the value of x and of y.


Solution:
(a)
x + y = 5
4x + 3y = 17

(b)
( 1    1 4    3 )( x y )=(  5 17 )             ( x y )= 1 1( 3 )4( 1 ) (   3    1 4       1 )(  5 17 )             ( x y )=1( 3( 5 )+( 1 )( 17 ) 4( 5 )+( 1 )( 17 ) )             ( x y )=1( 1517 20+17 )             ( x y )=1( 2 3 )             ( x y )=( 2 3 ) x=2 and y=3



Question 12 (6 marks):
Given A=( 4    2 3    1 ), B=m( 1    n 3    4 ) and I=( 1    0 0    1 ).
(a) Find the value of m and of n if AB = I.
(b) Write the following simultaneous linear equation as matrix equation:
4x – 2y = 3
3xy = 2
Hence, by using matrix method, calculate the value of x and of y.

Solution:
(a)
If AB=I, then B= A 1 A 1 = 1 4( 1 )( 2 )( 3 ) ( 1    2 3    4 ) A 1 = 1 2 ( 1    2 3    4 ) By comparison: B= A 1 m( 1    n 3    4 )= 1 2 ( 1    2 3    4 ) m= 1 2  ; n=2


(b)

4x2y=3 3xy=2 ( 4    2 3    1 )( x y )=( 3 2 )                ( x y )= 1 2 ( 1    2 3    4 )( 3 2 )                ( x y )= 1 2 ( ( 1 )( 3 )+( 2 )( 2 ) ( 3 )( 3 )+( 4 )( 2 ) )                ( x y )= 1 2 (   1 1 )                ( x y )=(    1 2 1 2 ) x= 1 2  and y= 1 2


4.10 SPM Practice (Long Questions)


Question 9:
(a) Given  1 s ( 4 2 5 3 )( t 2 5 4 )=( 1 0 0 1 ), find the value of s and of t.

(b) Using matrices, calculate the value of x and of y that satisfy the following matrix equation:
( 4 2 5 3 )( x y )=( 1 2 )


Solution:
(a) 1 s ( t 2 5 4 )= ( 4 2 5 3 ) 1 = 1 ( 4 )( 3 )( 2 )( 5 ) ( 3 2 5 4 ) = 1 2 ( 3 2 5 4 ) s=2, t=3

(b) ( 4 2 5 3 )( x y )=( 1 2 )   ( x y )= 1 2 ( 3 2 5 4 )( 1 2 )   ( x y )= 1 2 ( ( 3 )( 1 )+( 2 )( 2 ) ( 5 )( 1 )+( 4 )( 2 ) )   ( x y )= 1 2 ( 1 3 )   ( x y )=( 1 2 3 2 ) x= 1 2 ,  y= 3 2



Question 10 (5 marks):
During the sport day, students used coupons to buy food and drink. Ali and Larry spent RM31 and RM27 respectively. Ali bought 2 food coupons and 5 drinks coupons while Larry bought 3 coupons and 1 drinks coupon.
Using the matrix method, calculate the price, in RM, of a food coupon and of a drinks coupon.

Solution:
Ali spent RM31. He bought 2 food coupons and 5 drinks coupons.
Larry spent RM27. He bought 3 food coupons and 1 drinks coupons.
x = price of one food coupon
y = price of one drinks coupon

( 2    5 3    1 )( x y )=( 31 27 )             ( x y )= 1 2( 1 )5( 3 ) ( 1    5 3     2 )( 31 27 )             ( x y )= 1 215 ( 1( 31 )+( 5 )( 27 ) 3( 31 )+2( 27 ) )             ( x y )= 1 13 ( 104 39 )             ( x y )=( 8 3 ) x=8 and y=3 Thus, the price for a food coupon is RM8 and the price for drink coupon is RM3.

4.10 SPM Practice (Long Questions)


Question 7:
(a) Find the inverse matrix of ( 3 2 5 4 ).
(b) Ethan and Rahman went to the supermarket to buy cucumbers and carrots. Ethan bought 3 cucumbers and 2 carrots for RM9. Rahman bought 5 cucumbers and 4 carrots for RM16.
By using matrix method, find the price, in RM, of a cucumber and the price of a carrot. 


Solution:
(a) Inverse matrix of ( 3 2 5 4 ) = 1 1210 ( 4 2 5 3 ) = 1 2 ( 4 2 5 3 ) =( 2 1 5 2 3 2 )

(b) 3x+2y=9.................( 1 ) 5x+4y=16...............( 2 ) ( 3 2 5 4 )( x y )=( 9 16 )               ( x y )=( 2 1 5 2 3 2 )( 9 16 )               ( x y )=( ( 2 )( 9 )+( 1 )( 16 ) ( 5 2 )( 9 )+( 3 2 )( 16 ) )               ( x y )=( 1816 45 2 +24 )               ( x y )=( 2 3 2 ) x=2,  y= 3 2 Price of a cucumber=RM2    Price of a carrot=RM1.50



Question 8:
The inverse matrix of   ( 4 1 2 5 ) is t( 5 1 2 n ).
(a) Find the value of n and of t.
(b) Write the following simultaneous linear equations as matrix equation:
4xy = 7
2x + 5y = –2
Hence, using matrix method, calculate the value of x and of y.


Solution:
(a) t( 5 1 2 n )= ( 4 1 2 5 ) 1 = 1 ( 4 )( 5 )( 1 )( 2 ) ( 5 1 2 4 ) = 1 22 ( 5 1 2 4 ) t= 1 22 , n=4

(b) ( 4 1 2 5 )( x y )=( 7 2 )   ( x y )= 1 22 ( 5 1 2 4 )( 7 2 )   ( x y )= 1 22 ( ( 5 )( 7 )+1( 2 ) ( 2 )( 7 )+( 4 )( 2 ) )   ( x y )= 1 22 ( 352 148 )   ( x y )= 1 22 (  33 22 )   ( x y )=(   3 2 1 ) x= 3 2 ,  y=1

4.10 SPM Practice (Long Questions)


Question 5:
(a) Given  1 14 ( 2 s 4 t )( t 1 4 2 )=( 1 0 0 1 ), find the value of s and of t.
(b) Write the following simultaneous linear equations as matrix form:
3x – 2y = 5
9x + y = 1
Hence, using matrix method, calculate the value of x and y.

Solution:
(a) 1 14 ( 2 s 4 t )( t 1 4 2 )=( 1 0 0 1 ) 1 14 ( 2t+4s 2+2s 4t+4t 4+2t )=( 1 0 0 1 ) 2+2s 14 =0   2s=2      s=1 4+2t 14 =1 4+2t=14 2t=10 t=5

(b) ( 3 2 9 1 )( x y )=( 5 1 )   ( x y )= 1 21 ( 1 2 9 3 )( 5 1 )   ( x y )= 1 21 ( ( 1 )( 5 )+( 2 )( 1 ) ( 9 )( 5 )+( 3 )( 1 ) )   ( x y )= 1 21 ( 7 42 )   ( x y )=( 1 3 2 ) x= 1 3 ,  y=2



Question 6:
It is given that matrix P=( 6 3 5 2 ) and matrix Q= 1 m ( 2 3 5 n )  such that PQ=( 1 0 0 1 ).
(a) Find the value of m and of n.
(b) Write the following simultaneous linear equations as matrix form:
6x – 3y = –24
–5x + 2y = 18
Hence, using matrix method, calculate the value of x and y.

Solution:
(a) m=6( 2 )( 3 )( 5 )   =1215 m=3 n=6

(b) ( 6 3 5 2 )( x y )=( 24 18 )   ( x y )= 1 1215 ( 2 3 5 6 )( 24 18 )   ( x y )= 1 3 ( ( 2 )( 24 )+( 3 )( 18 ) ( 5 )( 24 )+( 6 )( 18 ) )   ( x y )= 1 3 ( 6 12 )   ( x y )=( 2 4 ) x=2,  y=4

4.10 SPM Practice (Long Questions)


Question 3:
It is given that Q  ( 3 2 6 5 ) = ( 1 0 0 1 ) , where Q is a 2 x 2 matrix.
(a) Find Q.
(b) Write the following simultaneous linear equations as matrix equation:
3u + 2v = 5
6u + 5v = 2
Hence, using matrix method, calculate the value of u and v.

Solution
:

(a) 
Q = ( 3 2 6 5 ) 1 Q = 1 3 ( 5 ) 2 ( 6 ) ( 5 2 6 3 ) Q = 1 3 ( 5 2 6 3 ) Q = ( 5 3 2 3 2 1 )

(b)

( 3 2 6 5 ) ( u v ) = ( 5 2 ) ( u v ) = 1 3 ( 5 2 6 3 ) ( 5 2 ) ( u v ) = 1 3 ( ( 5 ) ( 5 ) + ( 2 ) ( 2 ) ( 6 ) ( 5 ) + ( 3 ) ( 2 ) ) ( u v ) = 1 3 ( 21 24 ) ( u v ) = ( 7 8 ) u = 7 , v = 8


Question 4:
It is given that Q ( 3 2 5 4 ) = ( 1 0 0 1 )  , where Q is a 2 × 2 matrix.
(a) Find the matrix Q.
(b) Write the following simultaneous linear equations as matrix equation:
3x – 2y = 7
5x – 4y = 9
Hence, using matrix method, calculate the value of x and y.

Solution:

(a)
Q = 1 3 ( 4 ) ( 5 ) ( 2 ) ( 4 2 5 3 ) = 1 2 ( 4 2 5 3 ) = ( 2 1 5 2 3 2 )

(b)

( 3 2 5 4 ) ( x y ) = ( 7 9 ) ( x y ) = 1 2 ( 4 2 5 3 ) ( 7 9 ) ( x y ) = 1 2 ( ( 4 ) ( 7 ) + ( 2 ) ( 9 ) ( 5 ) ( 7 ) + ( 3 ) ( 9 ) ) ( x y ) = 1 2 ( 10 8 ) ( x y ) = ( 5 4 ) x = 5 , y = 4

4.9 SPM Practice (Short Questions)


4.9.2 Matrices, SPM Practice (Short Questions)
Question 5:
Given that ( 3   x ) ( x 1 ) = ( 18 ) ,  find the value of x.

Solution:
( 3   x ) ( x 1 ) = ( 18 )
[3 × x + x (–1)] = (18)
3xx = 18
2x = 18
x = 9



Question 6:
( 3 4 2 3 ) ( 5 2 ) =

Solution:
( 3 4 2 3 ) ( 5 2 ) = ( ( 3 ) ( 5 ) + ( 4 ) ( 2 ) ( 2 ) ( 5 ) + ( 3 ) ( 2 ) ) = ( 15 8 10 6 ) = ( 7 16 )



Question 7:
( 2 4 3 4 0 1 ) ( 1 3 ) =

Solution:
Order of the product of the two matrices
= ( 3 × 2 ) ( 2 × 1 ) = ( 3 × 1 )

( 2 4 3 4 0 1 ) ( 1 3 ) = ( 2 ( 1 ) + 4 ( 3 ) ( 3 ) ( 1 ) + 0 ( 3 ) ( 4 ) ( 1 ) + 1 ( 3 ) ) = ( 2 12 3 + 0 4 3 ) = ( 10 3 1 )



Question 8:
( 1  1   2 )( 5 1 3 2 0 4 )=

Solution:
Order of the product of the two matrices
= ( 1 × 3 ) ( 3 × 2 ) = ( 1 × 2 )

( 1  1   2 )( 5 1 3 2 0 4 )=
= (1×5 + (–1)(–3) + (2)(2)  1×1 + (–1)(0) + (2)(4))
= (5 + 3 + 4   1 + 0 + 8)
= (12   9)

4.5 Multiplication of Two Matrices (Sample Question 2)


Example 2:
Find the values of m and n in each of the following matrix equations.
( a ) ( 3 m ) ( 1       n ) = ( 3 12 2 8 ) ( b ) ( m 2 3 1 ) ( 2 n ) = ( 12 4 + 2 n ) ( c ) ( m 3 1 1 ) ( 1 2 4 n ) = ( 14 11 5 3 )
 
Solution:
(a) ( 3 m ) ( 1   n ) = ( 3 12 2 8 ) ( 3 3 n m m n ) = ( 3 12 2 8 ) m = 2 ,   3 n = 12 n = 4

(b) ( m 2 3 1 ) ( 2 n ) = ( 12 4 + 2 n ) ( 2 m + 2 n 6 + n ) = ( 12 4 + 2 n ) 6 + n = 4 + 2 n n = 10 2 m + 2 n = 12 2 m + 2 ( 10 ) = 12 2 m 20 = 12 2 m = 32 m = 16

(c) ( m 3 1 1 ) ( 1 2 4 n ) = ( 14 11 5 3 ) ( m + ( 12 ) 2 m + ( 3 n ) 1 + 4 2 + n ) = ( 14 11 5 3 ) m 12 = 14 m = 2 m = 2 2 + n = 3 n = 5


4.8 Solving Simultaneous Linear Equations using Matrices


4.8 Solving Simultaneous Linear Equations using Matrices
1. Two simultaneous linear equations can be written in the matrix equation form.
For example, in the simultaneous equations:
ax + by = e
cx + dy = f
can be written in the matrix form as follows:

( a b c d ) ( x y ) = ( e f ) ,

Where a, b, c, d, e and are constant while x and y are unknowns.

Example 1:
Write the following simultaneous linear equations in the matrix form.
y– 6x – 19 = 0
2y + 3x + 22 = 0

Solution:
– 6x + y = 19
3x + 2y = – 22
The matrix form is:
( 6 1 3 2 ) ( x y ) = ( 19 22 )



2. Matrix equations in the form ( a b c d ) ( x y ) = ( e f )
can be solved for the unknowns x and as follows.

(a) Let A = ( a b c d ) and find A-1.

(b) Multiply both sides of the equation by A-1.

A 1 ( a b c d ) ( x y ) = A 1 ( e f )

(c)  A 1 A ( x y ) = A 1 ( e f )   I ( x y ) = A 1 ( e f )   A 1 A = I = ( 1 0 0 1 )   ( x y ) = A 1 ( e f )   ( x y ) = 1 a d b c ( d b c a ) ( e f )



Example 2:
Solve the following simultaneous equations by using the matrix method.
2x = 5 – 3y
7x = 1 – 5y

Solution:
2x + 3y = 5
7x + 5y = 1 
( 2 3 7 5 ) ( x y ) = ( 5 1 ) write the simultaneous equations in matrix form . Let A = ( 2 3 7 5 ) A 1 = 1 a d b c ( d b c a ) A 1 = 1 10 21 ( 5 3 7 2 ) A 1 = 1 11 ( 5 3 7 2 ) ( x y ) = 1 11 ( 5 3 7 2 ) ( 5 1 ) ( x y ) = A 1 ( e f ) ( x y ) = 1 11 ( 5 × 5 + ( 3 ) × 1 7 × 5 + 2 × 1 ) ( x y ) = 1 11 ( 22 33 ) ( x y ) = ( 22 11 33 11 ) = ( 2 3 ) x = 2 , y = 3.



4.9 SPM Practice (Short Questions)


Question 1:
( 1 4 6 2 ) + 3 ( 2 0 4 3 ) ( 3 0 2 5 )

Solution:

( 1 4 6 2 ) + 3 ( 2 0 4 3 ) ( 3 0 2 5 ) = ( 1 4 6 2 ) + ( 6 0 12 9 ) ( 3 0 2 5 ) = ( 7 4 18 7 ) ( 3 0 2 5 ) = ( 7 ( 3 ) 4 0 18 ( 2 ) 7 ( 5 ) ) = ( 10 4 20 2 )



Question 2:
Find the value of m in the following matrix equation:
( 9 4 5 0 ) + 1 2 ( 8 m 6 10 ) = ( 13 7 2 1 )

Solution:

( 9 4 5 0 ) + 1 2 ( 8 m 6 10 ) = ( 13 7 2 1 ) 4 + 1 2 m = 7 1 2 m = 3 m = 6



Question 3:
Given ( 2x 3y )4( 2 3 )=( 2 6 ) Find the values of x and y.

Solution:

2 x + 8 = 2 2 x = 10 x = 5 3 y 12 = 6 3 y = 18 y = 6




Question 4:
Given that matrix equation 3 ( 6   m ) + n ( 3   4 ) = ( 12   7 ) ,   find the value of  m   +   n

Solution:

3 ( 6   m ) + n ( 3   4 ) = ( 12   7 ) 18 + 3 n = 12 3 n = 6 n = 2 3 m + 4 n = 7 3 m + 4 ( 2 ) = 7 3 m = 15 m = 5 m + n = 5 + ( 2 ) = 3

4.10 SPM Practice (Long Questions)


Question 1:
It is given that matrix A = ( 3 1 5 2 )
(a) Find the inverse matrix of A.
(b) Write the following simultaneous linear equations as matrix equation:
3uv = 9
5u – 2v = 13
Hence, using matrix method, calculate the value of u and v.

Solution
:

(a)
A 1 = 1 3 ( 2 ) ( 5 ) ( 1 ) ( 2 1 5 3 ) = 1 ( 2 1 5 3 ) = ( 2 1 5 3 )

(b)

( 3 1 5 2 ) ( u v ) = ( 9 13 ) ( u v ) = 1 ( 2 1 5 3 ) ( 9 13 ) ( u v ) = 1 ( ( 2 ) ( 9 ) + ( 1 ) ( 13 ) ( 5 ) ( 9 ) + ( 3 ) ( 13 ) ) ( u v ) = 1 ( 5 6 ) ( u v ) = ( 5 6 ) u = 5 , v = 6


Question 2:
It is given that matrix A = ( 2 5 1 3 ) and matrix B = m ( 3 k 1 2 ) such that AB = ( 1 0 0 1 )
(a) Find the value of m and of k.
(b) Write the following simultaneous linear equations as matrix equation:
2u – 5v = –15
u + 3v = –2
Hence, using matrix method, calculate the value of u and v.

Solution:

(a) Since AB = ( 1 0 0 1 ) , B is the inverse of A.

m = 1 ( 2 ) ( 3 ) ( 5 ) ( 1 ) = 1 11
k = 5

(b)
( 2 5 1 3 ) ( u v ) = ( 15 2 ) ( u v ) = 1 11 ( 3 5 1 2 ) ( 15 2 ) ( u v ) = 1 11 ( ( 3 ) ( 15 ) + ( 5 ) ( 2 ) ( 1 ) ( 15 ) + ( 2 ) ( 2 ) ) ( u v ) = 1 11 ( 55 11 ) ( u v ) = ( 5 1 ) u = 5 , v = 1