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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.

4.6 Identity Matrix

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4.6 Identity Matrix
1. Identity matrix is a square matrix, usually denoted by the letter I and is also known as unit matrix.
 
2. All the diagonal elements (from top left to bottom right) of an identity matrix are 1 and the rest of the elements are 0.
For example,
( 1 0 0 1 )  and  ( 1 0       0 0 0 1       0 0      1 )  are identity matrices .

3. If I is the identity matrix of order n × n and A is a matrix of the same order, then IA = A and AI = A. 


Example 1:
Determine whether each of the following is an identity matrix of ( − 2 4 3 7 ) .
( a ) ( 1 0 0 1 ) ( b ) ( 0 1 1 0 )
 
Solution:
( a ) ( − 2 4 3 7 ) ( 1 0 0 1 ) = ( − 2 × 1 + 4 × 0 − 2 × 0 + 4 × 1 3 × 1 + 7 × 0 3 × 0 + 7 × 1 ) = ( − 2 4 3 7 ) Therefore, ( 1 0 0 1 ) is an identity matrix . ( b ) ( − 2 4 3 7 ) ( 0 1 1 0 ) = ( − 2 × 0 + 4 × 1 − 2 × 1 + 4 × 0 3 × 0 + 7 × 1 3 × 1 + 7 × 0 ) = ( 4 − 2 7 3 ) ≠ ( − 2 4 3 7 ) Therefore, ( 0 1 1 0 ) is not an identity matrix .  


Example 2:
Find the product of the following pairs of matrices and determine whether the given matrix is an identity matrix.

( a ) ( − 3 2 5 7 ) ( 1 0 0 1 ) and ( 1 0 0 1 ) ( − 3 2 5 7 ) ( b ) ( 0 0 1 1 ) ( 1 8 5 3 ) and ( 1 8 5 3 ) ( 0 0 1 1 )  
 
Solution:
( a ) ( − 3 2 5 7 ) ( 1 0 0 1 ) = ( − 3 × 1 + 2 × 0 − 3 × 0 + 2 × 1 5 × 1 + 7 × 0 5 × 0 + 7 × 1 ) = ( − 3 2 5 7 ) ( 1 0 0 1 ) ( − 3 2 5 7 ) = ( 1 × − 3 + 0 × 5 1 × 2 + 0 × 7 0 × − 3 + 1 × 5 0 × 2 + 1 × 7 ) = ( − 3 2 5 7 ) ∴ ( 1 0 0 1 ) is an identity matrix for ( − 3 2 5 7 ) . ( b ) ( 0 0 1 1 ) ( 1 8 5 3 ) = ( 0 × 1 + 0 × 5 0 × 8 + 0 × 3 1 × 1 + 1 × 5 1 × 8 + 1 × 3 ) = ( 0 0 6 11 ) ( 1 8 5 3 ) ( 0 0 1 1 ) = ( 1 × 0 + 8 × 1 1 × 0 + 8 × 1 5 × 0 + 3 × 1 5 × 0 + 3 × 1 ) = ( 8 8 3 3 ) ∴ ( 0 0 1 1 ) is NOT an identity matrix for ( 1 8 5 3 ) .

4.1 Matrix

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4.1 Matrices
 
A matrix is a rectangular array of numbers enclosed in large brackets.
For example ( 2 0 3 1 ) is a matrix .
  
(A)   Rows, Columns and Order of Matrices
1. A matrix which has m rows and n columns is known as a matrix of order m × n.
 


2. A row matrix is a matrix which has only one row.
Example:
( 4 ),          ( 2     6 ),      ( 3     8    5 )  1 ×1         1 ×2          1 ×3           Only 1 row     
 
3. A column matrix is a matrix which has only one column.
Example:
( 3 ),           ( 2 6 ),         ( −5 7 −9 ) 1× 1        2× 1            3× 1              Only 1 column  
4. A square matrix is a matrix which has equal number of rows and columns.
Example:
( 3 ),         ( 7 0 2 5 ),       ( 1 3    9 0 6 4    1 3    5 ) 1×1           2×2                 3×3 Number of rows = Number of columns

4.2 Equal Matrices (Sample Questions)

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Example 1:
State the values of the unknowns in the following pairs of equal matrix.
( 1 x + 2 4 − y − 1 ) = ( 1 3 2 − 1 )

Solution:
( 1 x + 2 4 − y − 1 ) = ( 1 3 2 − 1 )

x + 2 = 3
x = 1

4 – y = 2
–y = –2
y = 2


Example 2:
Calculate the values of p and q in each of the following matrix equations.
(a) ( 3 2 p + q p − 3 ) = ( 3 1 8 − 2 q − 3 ) (b) ( 10 0 5 p − 8 1 ) = ( p − 2 q 0 − 4 q 1 )

Solution:
(a) ( 3 2 p + q p − 3 ) = ( 3 1 8 − 2 q − 3 )

2p + q = 1
q = 1 – 2p ----(1)
p = 8 – 2q ----(2)

Substitute (1) into (2),
p = 8 – 2(1 – 2p)
p = 8 – 2 + 4p
p – 4p = 6
–3p = 6
p = –2

Substitute p = –2 into (1),
q = 1 – 2(–2)
q = 5


 (b) ( 10 0 5 p − 8 1 ) = ( p − 2 q 0 − 4 q 1 )

10 = p – 2q
p = 10 + 2q ----(1)
5p – 8 = –4q ----(2)

Substitute (1) into (2),
5 (10 + 2q) – 8 = –4q
50 + 10q – 8 = –4q
14q = –42
q = –3

Substitute q = –3 into (1),
p = 10 + 2(–3)
p = 4

4.3 Addition and Subtraction of Matrices

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4.3 Addition and Subtraction of Matrices
 
(A) Determining whether addition or subtraction can be performed on two given matrices
1. Two matrices can be added or subtracted if both matrices have the same order.
2. The addition or subtraction of two matrices is performed by adding or subtracting the corresponding elements of the matrices.



Example 1:
State whether the following matrices can be added or subtracted. Give reason to your answer.
  (a)  ( 2 3 )  and  ( 1    8 ) (b)  ( 1 2 7 1 )  and  ( 10 0 − 3 − 1 )  
(c) (p   2   4) and (2   6   q)

Solution:
(a) Cannot because the matrices are not of the same order.
(b) Can because the matrices are of the same order.
(c) Can because the matrices are of the same order.


Example 2:
Express each of the following matrices as a single matrix.
If A=( 3 2 ) and B=( 5 1 ), then (a) A+B=( 3 2 )+( 5 1 ) =( 3+5 2+1 ) =( 8 3 )


( b ) B − A = ( 3 2 ) − ( 5 1 ) = ( 3 − 5 2 − 1 ) = ( − 2 1 )



4.2 Equal Matrices

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4.2 Equal Matrices

(A) Determining whether two matrices are equal
1. Two matrices are equal if they have the same order and their corresponding elements are equal.
For example,  ( a b c d ) = ( e f g h )

Therefore, a = e, b = f, c = g and d = h.


Example 1:
Determine whether the following pairs of matrices are equal.
(a) A=( 10 8 −3 1 ) and B=( 10 8 −3 1 ) (b) P=( 2 −4 10 ) and Q=( 2 −3 10 ) (c) M=( 3 5 ) and N=( 4   7 )

Solution:
(a) Equal
(b) Not equal, because -4 ≠ -3.
(c) Not equal, because the orders of the matrices are not equal.




(B) Solving problem involving equal matrices
1. When matrices are equal, elements whose values are unknown can be determined.

Example 2:
State the value of the unknowns in the following pairs of equal matrix. 
( 2 x x + 2 y ) = ( − 8 10 )

Solution:
( 2 x x + 2 y ) = ( − 8 10 )

2x = -8
x = -4

x + 2y = 10
(-4) + 2y = 10
2y = 10 + 4
2y = 14
y = 7
 

4.3 Addition and Subtraction of Matrices (Sample Questions)

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Example 1:
Find the addition of the following matrices.
(a)( 18   −7 )+( 3        6 ) (b) ( −13 0 7 −1 )+( −1 3 5 6 )

Solution:
(a) ( 18   −7 )+( 3       6 ) =( 18+3   −7+6 )=( 21   −1 ) (b) ( −13 0 7 −1 )+( −1 3 5 6 ) =( −13+( −1 ) 0+3 7+5 −1+6 )=( −14 3 12 5 )



Example 2:
Find the subtraction of the following matrices.
( a ) ( 9 6 ) − ( 7 − 2 ) ( b ) ( 3 − 4 0 5 ) − ( − 7 − 3 − 6 1 )

Solution:
( a ) ( 9 6 ) − ( 7 − 2 ) = ( 9 − 7 6 − ( − 2 ) ) = ( 2 8 ) ( b ) ( 3 − 4 0 5 ) − ( − 7 − 3 − 6 1 ) = ( 3 − ( − 7 ) − 4 − ( − 3 ) 0 − ( − 6 ) 5 − 1 ) = ( 3 + 7 − 4 + 3 0 + 6 5 − 1 ) = ( 10 − 1 6 4 )


Example 3:
Given that ( p q ) + ( 5 p 9 ) = ( − 12 3 q + 1 )  , find the values of p and q.

Solution:
( p q ) + ( 5 p 9 ) = ( − 12 3 q + 1 ) ( p + 5 p q + 9 ) = ( − 12 3 q + 1 )
 
p + 5p = –12
6p = –12
p = –2

q + 9 = 3q + 1
q – 3q = 1 – 9
–2q = –8
q = 4



Example 4:
Find the values of m and n in the following matrix equation.
( 7 6 n 1 ) − ( − 1 m − 4 − 2 ) = ( 5 12 6 11 )

Solution:
( 7 6 n 1 ) − ( − 1 m − 4 − 2 ) = ( 5 12 6 11 )

6 – m = 12
 –m = 6
 m = –6
n – (–4) = 6
n + 4 = 6
n = 2 

SPM Practice (Long Questions)

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Further Practice:
Transformation III, Long Questions (Question 1)
 
Question 1:
(a) Transformation T is a translation ( − 4 2 ) and transformation P is an anticlockwise rotation of 90oabout the centre (1, 0).

State the coordinates of the image of point (5, 1) under each of the following transformation:
(i) Translation T,
(ii) Rotation P,
(iii) Combined transformation T2.

(b) Diagram below shows three quadrilaterals, ABCD, EFGH and JKLM, drawn on a Cartesian plane.
 
(i) JKLM is the image of ABCD under the combined transformation VW.
Describe in full the transformation:
(a) W   (b) V

(ii) It is given that quadrilateral ABCD represents a region of area 18 m2.
Calculate the area, in m2, of the region represented by the shaded region.
 
Solution:
(a)


(b)
 
(i)(a)
W: A reflection in the line x = –2
 
(i)(b)
V: An enlargement of scale factor 3 with centre (0, 4).
 
(b)(ii)
Area of EFGH = area of ABCD = 18 m2
Area of JKLM = (Scale factor)2 × Area of object
= 32 × area of EFGH
= 32 × 18
= 162 m2
Therefore,
Area of the shaded region
= Area of JKLM – area of EFGH
= 162 – 18
= 144 m2