Post on 04-Jan-2020
Notes 4.1.3 Page 1
VCE Further Maths Mr Mark Judd Unit 4, Matrices Module Yr 12 Further Maths
Section 4.1.3 – Multiplicative Inverse Matrix VCAA “Dot Points” Matrices and their applications, including: • inverse of a matrix, its determinant, and the condition for a matrix to have an inverse • use of matrices to represent systems of linear equations and the solution of these equations as
an application of the inverse matrix; the concepts of dependent systems of equations and inconsistent systems of equations in the context of solving pairs of simultaneous equations in two variables; the formulation of practical problems in terms of a system of linear equations and their solution using the matrix inverse method.
Multiplicative Inverse A multiplicative inverse or reciprocal for the number x is given by 1
𝑥𝑥, or x-1 which when multiplied
by x gives the number 1. Example 1 The reciprocal for 5 is 1
5, as 5 × 1
5= 1
The reciprocal for 38 is 8
3, as 3
8× 8
3= 1
Similarly, if a matrix is multiplied by its inverse, the resultant matrix will be the identity matrix (I).
𝐴𝐴 × 𝐴𝐴−1 = 𝐼𝐼 Finding the Multiplicative Inverse of a matrix To find the multiplicative inverse for a matrix there are certain steps to follow. Consider Matrix A
𝐴𝐴 = �𝒂𝒂 𝒃𝒃𝒄𝒄 𝒅𝒅�
Step.1 Find the determinant of A, written as det A or |𝐴𝐴| det (A) = ad - cb Step.2 Swap the elements of the main diagonal
�𝒅𝒅 𝒃𝒃𝒄𝒄 𝒂𝒂�
Step.3 Multiply the elements on the other diagonal by -1
� 𝒅𝒅 −𝒃𝒃−𝒄𝒄 𝒂𝒂 �
Step.4 Multiply the result of step.3 by the reciprocal of the determinant, 1|𝐴𝐴|
∴A-1 = 1|𝐴𝐴| �𝑑𝑑 −𝑏𝑏−𝑐𝑐 𝑎𝑎 �
Notes 4.1.3 Page 2
VCE Further Maths Mr Mark Judd Unit 4, Matrices Module Yr 12 Further Maths
Example 2
If 𝐴𝐴 = �2 31 4�, calculate A-1
Step.1 Find the determinant of A, written as det A or |𝐴𝐴| det(𝐴𝐴) = 𝑎𝑎𝑑𝑑 − 𝑐𝑐𝑏𝑏 = (2 × 4) − (1 × 3) = 8 − 3 = 5 Step.2 Swap the elements of the main diagonal
�4 31 2�
Step.3 Multiply the elements on the other diagonal by -1
� 4 −𝟑𝟑−𝟏𝟏 2 �
Step.4 Multiply the result of step.3 by 1|𝐴𝐴|
A-1 = 𝟏𝟏𝟓𝟓� 4 −3−1 2 � or �
45
−35
−15
25
�
So the multiplicative inverse of �2 31 4� is 𝟏𝟏
𝟓𝟓� 4 −3−1 2 �
We can check our answer to show that A x A-1 = I
𝐴𝐴 × 𝐴𝐴−1 = �2 31 4� ×
15� 4 −3−1 2 �
= 15
× �2 31 4� �
4 −3−1 2 �
= 15
× �5 00 5�
= �1 00 1�
= 𝐼𝐼
Notes 4.1.3 Page 3
VCE Further Maths Mr Mark Judd Unit 4, Matrices Module Yr 12 Further Maths
Applications using an inverse matrix Consider the following algebraic equation: 5x = 10, and solve for x 5x = 10 [Divide b.s. by 5] 5x𝟓𝟓
=10𝟓𝟓
∴x = 2 Or alternatively 5x = 10 [Multiply b.s. by 𝟏𝟏
𝟓𝟓] NB: 1
5 = 5-1(the inverse of 5)
𝟏𝟏𝟓𝟓5 x = 𝟏𝟏
𝟓𝟓10
∴x = 2 Matrices equations can also be solved by using an inverse matrix. Example 3 Solve the following matrix equation:
� 3 4−1 2� �
𝑥𝑥𝑦𝑦� = � 2
−4�
You cannot simply divide both sides by � 3 4−1 2� to leave �
𝑥𝑥𝑦𝑦�. However you can multiply both
sides by the inverse of � 3 4−1 2�
Step.1 Calculate the inverse
� 3 4−1 2�
−1= 110
�2 −41 3 �
Step.2 Multiply both sides by the inverse 𝟏𝟏𝟏𝟏𝟏𝟏
�𝟐𝟐 −𝟒𝟒𝟏𝟏 𝟑𝟑 � � 3 4
−1 2� �𝑥𝑥𝑦𝑦� = 𝟏𝟏
𝟏𝟏𝟏𝟏 �𝟐𝟐 −𝟒𝟒𝟏𝟏 𝟑𝟑 � � 2
−4�
�1 00 1� �
𝑥𝑥𝑦𝑦� = 1
10 �2 −4
1 3 � � 2−4�
∴�𝑥𝑥𝑦𝑦� = 1
10 �2 −4
1 3 � � 2−4�
= 110
� 20−10�
= � 2−1�
So for this example x = 2 and y = -1.
Notes 4.1.3 Page 4
VCE Further Maths Mr Mark Judd Unit 4, Matrices Module Yr 12 Further Maths
The general rules for multiplying by the inverse of a matrix are:
If AX = B, then X = A-1B If XA = B, then X = BA-1
Where A is a known matrix, X is the unknown matrix to be solved and B is a second known matrix, resulting from the product of the previous two matrices. Example 4
If A = �2 12 −4� and B = �50�, solve for the unknown matrix X if AX = B
As, �2 12 −4� �
𝑥𝑥𝑦𝑦� = �50� is of the format AX = B,
Then the unknown matrix X can be found by using the above rules: ∴ X = A-1B
= �2 12 −4�
−1�50�
= �21� Example 5
If A = �3 24 1� and B = �3 2
2 3�, solve for the unknown matrix X if XA = B
As,�𝑥𝑥𝑦𝑦� �
3 24 1� = �3 2
2 3� is of the format XA = B,
Then the unknown matrix X can be found by using the above rules:
∴X = BA-1
= �3 22 3� �
3 24 1�
−1
= �1 02 −1�
Notes 4.1.3 Page 5
VCE Further Maths Mr Mark Judd Unit 4, Matrices Module Yr 12 Further Maths
Using inverse matrices to solve simultaneous equations Most year 12 Further Maths students have solved a simultaneous equation somewhere during their junior mathematics classes.
Technique.1 Elimination method
Solve the following simultaneous equations and find the values for x and y.
Equation 1: 2x + y = 7 Equation 2: 3x – y = 8 Add the two equations to eliminate the y’s. 2x +y = 7 3x – y = 8 5x = 15
∴ x =3 Now substitute x=3 back into either of the original equations to solve for y. Substitute x=3 into the equation 2x + y = 7
2(3) + y = 7 6 + y = 7
∴7 = 1
So the answers are x=3 and y=1 Technique.2 Substitution method
Solve the following simultaneous equations and find the values for x and y.
Equation 1: y – 2x = 1 Equation 2: 2y – 3x = 5 Rearranging equation 1, you get y = 1 + 2x Now substitute this “y” equation into equation 2 Equation 2 now becomes: 2(1 + 2x) – 3x = 5 2 + 4x – 3x = 5 2 + x = 5 ∴x = 3 Substituting x = 3 into Equation 1 gives us y - 6 = 1, ∴y = 7 So the answers are x=3 and y=7 Both of these two examples can also be solved using the matrices method.
+
Notes 4.1.3 Page 6
VCE Further Maths Mr Mark Judd Unit 4, Matrices Module Yr 12 Further Maths
Solving Simultaneous Equations Using Matrices Simultaneous equations are of the format: ax + by = e cx + dy = f This can be also expressed using a matrices equation: A × X = B
�𝑎𝑎 𝑏𝑏𝑐𝑐 𝑑𝑑� �
𝑥𝑥𝑦𝑦� = �
𝑒𝑒𝑓𝑓�
(And can be solved using: X = A-1B)
�𝑥𝑥𝑦𝑦� = �𝑎𝑎 𝑏𝑏
𝑐𝑐 𝑑𝑑�−1�𝑒𝑒𝑓𝑓�
Example 6
Use matrices to solve the following simultaneous equations: 4x + y = 20 x – y = 0 Step.1 Set up the matrix equation
�4 11 −1� �
𝑥𝑥𝑦𝑦� = �20
0 � Where:
�4 11 −1� is the coefficient matrix
�𝑥𝑥𝑦𝑦� is the matrix of the pronumerals of the equations
�200 � is the matrix numbers on the right hand side of the equations
NB: this matrix equation is of the format AX=B,
∴the solution matrix can be found by following: X=A-1B
Step.2 Calculate determinant of the coefficient matrix det(𝐴𝐴) = 𝑎𝑎𝑑𝑑 − 𝑐𝑐𝑏𝑏 = (4 × −1) − (1 × 1) = −4 − 1 = −5
Notes 4.1.3 Page 7
VCE Further Maths Mr Mark Judd Unit 4, Matrices Module Yr 12 Further Maths
Step.3 Calculate the coefficient inverse
The inverse is −15�−1 −1−1 4 �
Step.4 Multiply b.s. by the inverse matrix
−15�−1 −1−1 4 � �4 1
1 −1� �𝑥𝑥𝑦𝑦� =−
15�−1 −1−1 4 � �20
0 �
�1 00 1� �
𝑥𝑥𝑦𝑦� =−
15�−1 −1−1 4 � �20
0 �
�𝑥𝑥𝑦𝑦� = −1
5�−1 −1−1 4 � �20
0 �
�𝑥𝑥𝑦𝑦� = −1
5 �−20−20�
∴ �𝑥𝑥𝑦𝑦� − �44�
∴ x = 4 and y = 4 are to solutions to the simultaneous equations
Notes 4.1.3 Page 8
VCE Further Maths Mr Mark Judd Unit 4, Matrices Module Yr 12 Further Maths
Let’s now solve the previous two simultaneous equations solutions using the matrices method. Example 7 Equation 1: 2x + y = 7 Equation 2: 3x – y = 8 Set up the matrix equation
�2 13 −1� �
𝑥𝑥𝑦𝑦� = �78�
This is of the form: AX=B, ∴ X=A-1B Solve using the TI-Nspire CAS calculator
So the answers are x=3 and y=1 Example 8 Equation 1: y – 2x = 1 Equation 2: 2y – 3x = 5 Set up the matrix equation
�1 −22 −3� �
𝑦𝑦𝑥𝑥� = �15�
This is of the form: AX=B, ∴ X=A-1B Solve using the TI-Nspire CAS calculator
So the answers are y=7 and x=3
NB: The reverse order of the y and x in both equations. This is reflected in the order of the y and x in the pronumeral matrix �𝑦𝑦𝑥𝑥�
Notes 4.1.3 Page 9
VCE Further Maths Mr Mark Judd Unit 4, Matrices Module Yr 12 Further Maths
A Matrix with “No Unique Solutions” There are some pairs of simultaneous equations that are classified as having “no unique solutions”. When such simultaneous equations are constructed using a matrix format, it is very easy to establish and identify which pairs of equations have no unique solutions. Consider the following pairs of simultaneous equations:
2𝑥𝑥 + 6𝑦𝑦 = 4 𝑥𝑥 + 3𝑦𝑦 = 5
Now construct a matrix to represents this pair of simultaneous equations.
�2 61 3� �
𝑥𝑥𝑦𝑦� = �45�
This matrix equation is of the format AX=B, ∴ X=A-1B So, to find the solution for the unknown matrix, the following expression is used:
�𝑥𝑥𝑦𝑦� = �2 6
1 3�−1�45�
However, when this solution is entered into the Ti-Nspire CAS calculator the following response is generated:
A matrix such as �2 61 3� does not have an inverse. It is described as a “singular matrix”. A
singular matrix is easy to identify as it has a determinant of zero (0). Examine the determinant for
�2 61 3�.
det = (2 × 3) − (1 × 6) = 0
In summary:
• A singular matrix has not inverse • A singular matrix has a determinant of zero (0) • A singular matrix has “no unique solutions”
NB: A matrix with a determinant other than zero does have an inverse matrix and does have a
unique solution.
Notes 4.1.3 Page 10
VCE Further Maths Mr Mark Judd Unit 4, Matrices Module Yr 12 Further Maths
Exam Styled Questions – Multiple Choice Question 1 (2017 Exam 1, Module 1, Qn 3) Which one of the following matrix equations has a unique solution?
To have a unique solutions, the determinant of the coefficient matrix must not equal 0. Option A 𝑑𝑑𝑒𝑒𝑑𝑑(𝐴𝐴) = 1 × 1 − 1 × 1 = 1 − 1 = 0, ∴No unique solution Option B 𝑑𝑑𝑒𝑒𝑑𝑑(𝐵𝐵) = 6 × 4 − (−4 × −6) = 24 − 24 = 0, ∴No unique solution Option C 𝑑𝑑𝑒𝑒𝑑𝑑(𝐶𝐶) = 8 × 2 − (4 × −4) = 16 − (−16) = 32, ∴Has a unique solution Option D 𝑑𝑑𝑒𝑒𝑑𝑑(𝐷𝐷) = 7 × 0 − 5 × 0 = 0 − 0 = 0, ∴No unique solution Option E 𝑑𝑑𝑒𝑒𝑑𝑑(𝐸𝐸) = (4 × −3) − (6 × −2) = −12 − (−12) = 0, ∴No unique solution ∴ Option C
C
Notes 4.1.3 Page 11
VCE Further Maths Mr Mark Judd Unit 4, Matrices Module Yr 12 Further Maths
Question 2 (2016 Exam 1, Module 1, Qn 3) The matrix equation below represents a pair of simultaneous linear equations
These simultaneous linear equations have no unique solution when m is equal to A. –4 B. –3 C. 0 D. 3 E. 4
To have no unique solutions, the determinant of the coefficient matrix must equal 0. ∴ 12 × 3 −𝑚𝑚 × 9 = 0 36 − 9𝑚𝑚 = 0 36 = 9𝑚𝑚
𝑚𝑚 =369
𝑚𝑚 = 4 ∴ Option E
E
Notes 4.1.3 Page 12
VCE Further Maths Mr Mark Judd Unit 4, Matrices Module Yr 12 Further Maths
Question 3 (2015 Exam 1, Module 1, Qn 3) Four systems of simultaneous linear equations are shown below.
12x + 8y = 26 3x − 2y = 14 −4x − 2y = 17 x + 0.5y = 8 3x + 2y = 15 −7x + 5y = 9 −6x + 3y = 10 0.5x + y = 8
How many of these systems of simultaneous linear equations do not have a unique solution? A. 0 B. 1 C. 2 D. 3 E. 4
Rewrite simultaneous equations as matrices �12 8
3 2� �𝑥𝑥𝑦𝑦� = �26
15� �13 −2−7 5 � �
𝑥𝑥𝑦𝑦� = �14
9 � �−4 −2−6 3 � �
𝑥𝑥𝑦𝑦� = �17
10� � 1 0.50.5 1 � �
𝑥𝑥𝑦𝑦� = �88�
To have a no unique solutions, the determinant of the coefficient matrix must equal 0. Matrix A 𝑑𝑑𝑒𝑒𝑑𝑑(𝐴𝐴) = 12 × 1 − 3 × 8 = 12 − 24 = −12, ∴ Has a unique solution Matrix B 𝑑𝑑𝑒𝑒𝑑𝑑(𝐵𝐵) = 13 × 5 − (−7 × −2) = 65 − 14 = 51, ∴ Has a unique solution Matrix C 𝑑𝑑𝑒𝑒𝑑𝑑(𝐶𝐶) = −4 × 3 − (−6 × −2) = −12 − (−12) = 0, ∴ Has no unique solution Matrix D 𝑑𝑑𝑒𝑒𝑑𝑑(𝐷𝐷) = 1 × 1 − 0.5 × 0.5 = 1 − 0.25 = 0.75 ∴Has a unique solution ∴ Only 1 of the above simultaneous equations do not have a unique solution? ∴ Option B
B
Notes 4.1.3 Page 13
VCE Further Maths Mr Mark Judd Unit 4, Matrices Module Yr 12 Further Maths
Question 4 (2015 Exam 1, Module 1, Qn 6) A carpenter can make four coffee tables and seven stools in a total of 33 hours. The carpenter can make two coffee tables and three pencil boxes in a total of 12 hours. The carpenter can make five stools and one pencil box in a total of 10 hours. The time, in hours, that it takes to make one coffee table is closest to A. 2 B. 3 C. 4 D. 5 E. 6
Let c represent the time to make 1 Coffee table Let s represent the time to make 1 Stool Let p represent the time to make 1 Pencil box Set up the worded description into 3 separate simultaneous equations.
4c +7s= 33 2c + 3p = 12 5s + p = 10
As a matrix equation:
�4 7 02 0 30 5 1
� �𝑐𝑐𝑠𝑠𝑝𝑝� = �
331210�
This matrix equation is on the format AX=B, ∴ X=A-1B So to find a solution, use the following equation:
�𝑐𝑐𝑠𝑠𝑝𝑝� = �
4 7 02 0 30 5 1
�−1
�331210�
�𝑐𝑐𝑠𝑠𝑝𝑝� = �
4.991.860.68
�
∴ The time it takes to make 1 coffee table is 4.99, or approximately 5 hours. ∴ Option D
D
Notes 4.1.3 Page 14
VCE Further Maths Mr Mark Judd Unit 4, Matrices Module Yr 12 Further Maths
Question 5 (2015 Exam 1, Module 1, Qn 7) Matrix P has inverse matrix P–1. Matrix P is multiplied by the scalar w (w ≠ 0) to form matrix Q. Matrix Q–1 is equal to A. 1
𝑤𝑤𝑃𝑃−1
B. 1𝑤𝑤2 𝑃𝑃−1
C. 𝑤𝑤𝑃𝑃−1 D. 𝑤𝑤2𝑃𝑃−1 E. 𝑃𝑃−1
𝑃𝑃 × 𝑃𝑃−1 = 𝐼𝐼,
∴ (𝑤𝑤𝑃𝑃) × �1𝑤𝑤𝑃𝑃−1� = 𝐼𝐼,
∴ 𝑄𝑄 �1𝑤𝑤𝑃𝑃−1� = 𝐼𝐼,
∴ 𝑄𝑄−1 =1𝑤𝑤𝑃𝑃−1
∴ Option A
A
Notes 4.1.3 Page 15
VCE Further Maths Mr Mark Judd Unit 4, Matrices Module Yr 12 Further Maths
Question 6 (2014 Exam 1, Module 1, Qn 2)
y – z = 8 5x – y = 0 x + z = 4
The system of three simultaneous linear equations above can be written in matrix form as
In each of the above options the 3×1 “letter” matrix includes an x, y and z term. Accordingly, the coefficient matrix needs to be a 3×3 matrix including all 3 coefficients, even those which are zeros (0). The three equations entered into a matrix format are as follows:
�0 1 −15 −1 01 0 1
� �𝑥𝑥𝑦𝑦𝑧𝑧� = �
804�
∴ Option B
B
Notes 4.1.3 Page 16
VCE Further Maths Mr Mark Judd Unit 4, Matrices Module Yr 12 Further Maths
Question 7 (2014 Exam 1, Module 1, Qn 9) A and B are square matrices such that AB = BA = I, where I is an identity matrix. Which one of the following statements is not true? A. ABA = A B. AB2A = I C. B must equal A D. B is the inverse of A E. both A and B have inverses
Option A. ABA=A, substitute in AB = I, ∴IA=A, ∴true Option B. AB2A=I, expand to form ABBA=I, substitute AB=I and BA=I ∴II=I, ∴true Option C. B must equal A. There is no proof to support this statement, ∴ not true Option D. B is the inverse of A (B=A-1), sub into AB=I from above ∴AA-1=I, ∴true Option E. Both A and B have inverses, Ab=BA=I proves they both have inverse, ∴true ∴ Option C
C
Notes 4.1.3 Page 17
VCE Further Maths Mr Mark Judd Unit 4, Matrices Module Yr 12 Further Maths
Question 8 (2013 Exam 1, Module 1, Qn 4)
2.8x + 0.7y = 10 1.4x + ky = 6
The set of simultaneous linear equations above does not have a solution if k equals A. –0.35 B. –0.250 C. 0 D. 0.25 E. 0.35
Rewrite as a matrix: �2.8 . 71.4 𝑘𝑘 � �
𝑥𝑥𝑦𝑦� = �10
6 � To not have a solution, the determinant of the coefficient matrix must equal 0. ∴ 2.8 × 𝑘𝑘 − 1.4 × .7 = 0 2.8𝑘𝑘 − 0.98 = 0 2.8𝑘𝑘 = 0.98
𝑘𝑘 =0.982.8
𝑘𝑘 = 0.35 ∴ Option E
E
Notes 4.1.3 Page 18
VCE Further Maths Mr Mark Judd Unit 4, Matrices Module Yr 12 Further Maths
Question 9 (2013 Exam 1, Module 1, Qn 6) A worker can assemble 10 bookcases and four desks in 360 minutes, and eight bookcases and three desks in 280 minutes. If each bookcase takes b minutes to assemble and each desk takes d minutes to assemble, the matrix �𝑏𝑏𝑑𝑑�will be given by
Set up the worded description into a pair of simultaneous equations. 10b + 4d = 360 8b + 3d = 280 As a matrix equation: �10 4
8 3� �𝑏𝑏𝑑𝑑� = �360
280� This matrix equation is on the format AX=B, ∴ X=A-1B So to find a solution, use the following equation:
�𝑏𝑏𝑑𝑑� = �10 48 3�
−1�360280�
�𝑏𝑏𝑑𝑑� = −12� 3 −4−8 10� �
360280�
�𝑏𝑏𝑑𝑑� = �−
32
42
82
−102
� �360280�
∴ Option A
A
Notes 4.1.3 Page 19
VCE Further Maths Mr Mark Judd Unit 4, Matrices Module Yr 12 Further Maths
Question 10 (2013 Exam 1, Module 1, Qn 9) P, Q, R and S are matrices such that the matrix product P = QRS is defined. Matrix Q and matrix S are square, non-zero matrices for which Q + S is not defined. Which one of the following matrix expressions is defined? A. R – S B. Q + R C. P2 D. R–1 E. P × S
Consider Matrix Q has the following order: m×m (as it is a square matrix) Consider Matrix S has the following order: n×n (as it is a square matrix) Therefore Matrix R has to be of the order: m×n (as matrix R must have the same number of rows as matrix Q has columns and the same number of columns as Matrix S has rows) Matrix Q and Matrix S are both square matrices but have different orders, as Q + S is not defined (so m ≠n) This can be represented as follows:
(m × m) × (m × n) × (n × n) R – S is not defined as they are different orders Q+R is not defined as they are different orders P is the product matrix of QRS, which has the order m×n, it is not squared so cannot be squared R-1is not possible as R is not a square matrix P×S is possible and therefore defined P is the product of QRS and has the order m×n, and S has the order n×n. The multiplication of Matrix P and Matrix S is possible, as shown below:
(m × n) × (n × n) ∴ Option E
Matrix Q Matrix R Matrix S
Matrix P Matrix S
E