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5.2 Inverse Variation

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5.2 Inverse Variation
If a quantity y  varies inversely as another quantity x, then
(a) y increases when x decreases,
(b) y decreases when x increases


5.2b Expressing an inverse variation in the form of an equation
An inverse variation can be written in the form of an equation, y = k x where k is a constant which can be determined.
 
Example 1:
Given y  varies inversely as x and y = 4 when x =10. Write an equation which relates x and y.

Solution:
y ∝ 1 x → y = k x 4 = k 10 → k = 40 ∴ y = 40 x


Example 2:
Given that y = 3 when x = 6, find the equation relates x and y if:
(a) y∝ 1 x 2   (b) y∝ 1 x 3   (c) y∝ 1 x

Solution:
( a ) y ∝ 1 x 2 → y = k x 2 3 = k 6 2 k = 3 × 36 = 108 ∴ y = 108 x 2 (b) y ∝ 1 x 3 → y = k x 3 3 = k 6 3 k = 3 × 216 = 648 ∴ y = 648 x 3 (c) y ∝ 1 x → y = k x 3 = k 6 k = 3 × 6 = 3 6 ∴ y = 3 6 x

5.4 SPM Practice (Short Questions)

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Question 1:
It is given that y varies directly as the cube of x and y = 192 when x = 4. Calculate the value of x when y = – 24.

Solution:
y α x³
y = kx³
192 = k (4)³
192 = 64 k
k = 3
 
y = 3 x³
when y = – 24 
– 24 = 3 x³
x³ = – 8
x = – 2


Question 2:
It is given that y varies directly as the square of x and y = 9 when x = 2. Calculate the value of x when y = 16.

Solution:
y α x²
y = kx²
9 = k (2)²
k= 9 4 y= 9 4 x 2 When y=16 16= 9 4 x 2 x 2 = 64 9 x= 8 3



Question 3:
Given that y varies inversely as w and x and y = 45 when w = 2 and x = 1 6 . Find the value of x when y = 15 and w = 1 3

Solution:
y α  1 wx y= k wx 45= k ( 2 )( 1 6 ) k=45× 1 3 =15 y= 15 wx

when y=15, w= 1 3 15= 15 ( 1 3 )x x 3 =1 x=3


Question 4:
Given that   p α 1 q r and p = 3 when q = 2 and r = 16, find the value of p when q = 3 and r = 4.

Solution:
p α  1 q r p= k q r 3= k ( 2 ) 16 k=24 p= 24 q r

when q=3, r=4 p= 24 3 4 p= 24 6 =4

4.8 Solving Simultaneous Linear Equations using Matrices

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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 f 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 y 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)

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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)

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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:
3u – v = 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



4.4 Multiplication of Matrix by a Number

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4.4 Multiplication of a Matrix by a Number
When matrix is multiplied by a number, every element in the matrix is multiplied by the number.

Example:
Given that  A = ( − 2 4 5 − 6 ) , find each of the following.
(a) 3A
(b) -2A 

Solution:
(a) 3 A = 3 ( − 2 4 5 − 6 ) = ( 3 × ( − 2 ) 3 × 4 3 × 5 3 × ( − 6 ) ) = ( − 6 12 15 − 18 )


(b) −2A=−2( −2 4 5 −6 ) =( −2×(−2) −2×4 −2×5 −2×(−6) ) =( 4 −8 −10 12 )




4.4 Multiplication of Matrix by a Number (Sample Questions)

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Example 1:
Express the following as single matrix.

(a) 3 ( 1 − 4 − 3 2 ) + 2 ( − 3 1 3 − 4 )

(b) 2 ( − 1 0 9 − 4 ) − 4 ( − 2 2 3 − 6 ) − 1 3 ( − 9 − 3 − 6 15 )

Solution:
(a) 3 ( 1 − 4 − 3 2 ) + 2 ( − 3 1 3 − 4 ) = ( 3 × 1 3 × ( − 4 ) 3 × ( − 3 ) 3 × 2 ) + ( 2 × ( − 3 ) 2 × 1 2 × 3 2 × ( − 4 ) ) = ( 3 − 12 − 9 6 ) + ( − 6 2 6 − 8 ) = ( 3 + ( − 6 ) − 12 + 2 − 9 + 6 6 − ( − 8 ) ) = ( − 3 − 10 − 3 14 )


(b) 2( −1 0 9 −4 )−4( −2 2 3 −6 )− 1 3 ( −9 −3 −6 15 ) =( −2 0 18 −8 )−( −8 8 12 −24 )−( 1 3 ×(−9) 1 3 ×(−3) 1 3 ×(−6) 1 3 ×15 ) =( −2 0 18 −8 )−( −8 8 12 −24 )−( −3 −1 −2 5 ) =( −2−(−8)−(−3) 0−8−(−1) 18−12−(−2) −8−(−24)−5 ) =( 9 −7 8 11 )


4.5 Multiplication of Two Matrices

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4.5 Multiplication of Two Matrices
 
1. Two matrices can be multiplied when the number of columns for the first matrix is the same as the number of rows for the second matrix.
 
For instance, if A is a m × n matrix and B is a n × t matrix, then the product of P = AB. The order of matrix P is m × t 


Examples: ( a ) ( a   b )   ( c d ) = ( a c  +  b d )   1 × 2  2 × 1     1 × 1 ( b ) ( a b c d ) ( e f ) = ( a e + b f c e + d f )    2 × 2  2 × 1  2 × 1 ( c ) ( a b c d ) ( e f g h ) = ( a e + b g a f + b h c e + d g c f + d h ) 2 × 2   2 × 2    2 × 2 ( d ) ( a b ) ( c   d ) = ( a c a d b c b d )  2 × 1 1 × 2    2 × 2 ( e ) ( a    b    c ) ( d e f ) = ( a d  +  b e + c f )  1 × 3    3 × 1   1 × 1 ( f ) ( a b c e d f ) ( g h ) = ( a g + b h c g + d h e g + f h )


Examples:
Determine whether the following pairs of matrices can be multiplied and state the order of the product of those that can be multiplied.

( a ) ( 3 − 5 1 2 ) ( 3      7 ) ( b ) ( 2 9 1 3 ) ( 8 6 ) ( c ) ( − 10       − 6 )   ( 7 2 ) ( d ) ( 8 6 ) ( 2 9 1 3 ) ( e ) ( 7 − 3 ) ( 2      10 )

Solution:
(a)( 3 −5 1 2 )( 3      7 )        2× 2         1 ×2    ← 2 ≠ 1 ∴ Cannot be multiplied (b)( 2 9 1 3 ) ( 8 6 )        2× 2     2 ×1    ← 2 = 2 ∴ Can be multiplied.                                             Order of product =2×1 (c)( −10      −6 ) ( 7 2 )          1× 2            2 ×1    ← 2 = 2 ∴ Can be multiplied.                                             Order of product =1×1 (d)( 8 6 )( 2 9 1 3 )        2× 1       2 ×2    ← 1 ≠ 2 ∴ Cannot be multiplied (e)( 7 −3 )( 2   10 )     2× 1      1 ×2    ← 1 = 1 ∴ Can be multiplied.                                             Order of product =2×2

4.5 Multiplication of Two Matrices (Sample Question 1)

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Example 1:
Find the product of the following pairs of matrices.
(a) ( 1      5      2 )( 2 4 3 ) (b) ( 2 8 −3 1 )( 1 0 4 −2 ) (c) ( −3 5 )( 2      6 ) (d) ( 0 4 −1 3 )( 7 −2 ) (e) ( 7      −4 )( −2 0 −1 3 )

Solution:
(a)  ( 1      5      2 )( 2 4 3 )← Matrices analysis 1×3 and 3×1  ↓          ↓ =1×1 matrix =( 1×2 ⊕ 5×4 ⊕ 2×3 ) =( 2+20+6 ) =( 28 )

(b)
( 2 8 −3 1 )( 1 0 4 −2 )← Matrices analysis 2×2 and 2×2  ↓            ↓ =2×2 matrix =( 2×1+8×4   2×0+8×−2 −3×1+1×4   −3×0+1×−2 ) =( 34 −16 1 −2 )

(c)
( − 3 5 ) ( 2      6 ) ← Matrices analysis 2 × 1  and 1 × 2   ↓             ↓ = 2 × 2  matrix = ( − 3 × 2    − 3 × 6 5 × 2      5 × 6 ) = ( − 6 − 18 10 30 )

(d)
( 0 4 −1 3 )( 7 −2 )← Matrices analysis 2×2 and 2×1  ↓           ↓ =2×1 matrix =( 0×7+4×−2 −1×7+3×−2 ) =( −8 −13 )

(e)
( 7      −4 )( −2 0 −1 3 )← Matrices analysis 1×2 and 2×2  ↓            ↓ =1×2 matrix =( 7×−2+( −4×−1 )         7×0+( −4×3 ) ) =( −14+4       0−12 ) =( −10     −12 )


4.7 Inverse Matrix

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4.7 Inverse Matrix
1. If A is a square matrix, B is another square matrix and A × B = B × A = I, then matrix A is the inverse matrix of matrix B and vice versa. Matrix A is called the inverse matrix of B for multiplication and vice versa.
 
2. The symbol A-1 denotes the inverse matrix of A.

3. Inverse matrices can only exist for square matrices but not all square matrices have inverse matrices.

4. If AB ≠ I or BA ≠ I, then A is not the inverse of B and B is not the inverse of A.

Example 1:
Determine whether matrix  A = ( 2 9 1 5 )  is an inverse matrix of matrix B = ( 5 − 9 − 1 2 ) .

Solution:
AB=( 2 9 1 5 )( 5 −9 −1 2 ) =( 2×5+9×−1 2×−9+9×2 1×5+5×−1 1×−9+5×2 ) =( 10+( −9 ) −18+18 5+( −5 ) −9+10 ) =( 1 0 0 1 )=I AB=( 5 −9 −1 2 )( 2 9 1 5 ) =( 5×2+( −9 )×1 5×9+( −9 )×5 −1×2+2×1 −1×9+2×5 ) =( 10+( −9 ) 18−18 −2+2 −9+10 ) =( 1 0 0 1 )=I AB=BA=I ∴A is the inverse matrix of B and vice versa.


5. The inverse of a matrix may also be found using a formula.
If A = ( a b c d ) , then the inverse matrix of A, A-1, is given by the formula below.
     A −1 = 1 ad−bc ( d −b −c a ), where ad−bc≠0    
6. ad – bc is known as the determinant of matrix A.

7. If the determinant, ad – bc = 0, then the inverse matrix of A does not exist.

Example 2:
Find the inverse matrix of A = ( 6 1 − 9 − 1 )  using the formula.

Solution:
A = ( 6 1 − 9 − 1 ) a = 6 , b = 1 , c = − 9 , d = − 1 A − 1 = 1 a d − b c ( d − b − c a ) A − 1 = 1 6 × − 1 − ( 1 × − 9 ) ( − 1 − 1 9 6 ) A − 1 = 1 − 6 + 9 ( − 1 − 1 9 6 ) A − 1 = 1 3 ( − 1 − 1 9 6 ) = ( − 1 3 − 1 3 3 2 )


Example 3:
The inverse matrix of ( − 7 2 − 9 2 ) is r ( 2 s 9 t ) .  Find the value of r, of s and of t.

Solution:
Let A = ( − 7 2 − 9 2 ) A − 1 = 1 − 7 × 2 − ( − 9 ) × 2 ( 2 − 2 9 − 7 ) A − 1 = 1 4 ( 2 − 2 9 − 7 ) ∴ r ( 2 s 9 t ) = 1 4 ( 2 − 2 9 − 7 ) By comparison, r = 1 4 , s = − 2 , t = − 7.