Section 4.1.3 – Multiplicative Inverse...

Notes 4.1.3

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


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


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


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


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


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


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


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


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


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


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


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


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


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


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


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


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


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


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

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