Department of MathematicsMathematics For Computing-19MT1101

I/IV-B.Tech-(I st Sem), Academic Year: 2019-2020

MATRIX ALGEBRA & Game Theory (CO-II)Session-11 &12

Introduction of Matrix Algebra:Cayley a French mathematician, discovered matrices in the year 1860. But it was not until the twentieth century was well – advanced that engineers heard of them. These days, however, matrices have been found to be of great utility in many branches of applied mathematics, such as algebraic and differential equations, mechanics, theory of electrical circuits, nuclear physics, aerodynamics and astronomy. It is, therefore, necessary for the young engineer to learn the elements of matrix algebra in order to keep up with the fast developments of physics and engineering. It is with this aim, that the present chapter has been written.Definition: A system of elements arranged in rows and columns within the square brackets is called a matrix. Matrices are denoted by a single capital letter.

Thus A = ¿

It has m-rows and n- columns. Each of the mn numbers is called an element of the matrix.Elementary Row Operations1. Interchange any two rows. 2. Multiply a row with a nonzero number. 3. Add a row to another one multiplied by a number. Definition: Two matrices are row equivalent if and only if one may be obtained from the other one via elementary row operations. Elementary TransformationsAny one of the following operations on a matrix is called an elementary transformation.

(a) Interchanging of any two rows (/columns). This transformation is indicated by Ri ↔ R j, if the i-th and j-th rows are interchanged.

(b) Multiplication of the elements of any row (/column) by a non–zero scalar quantity, k. It is denoted by Ri → kR j, if the i – th row is replaced by ‘k’ times the j – th row.

(c) Addition of constant multiplication of the elements of any row to the corresponding elements of any other row.

It is denoted by ¿¿ + kR j ¿, if the i – th row is replaced by sum of the i – th row and ‘k’ times of j-th row.

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Definition: Two matrices are said to be equivalent if one is obtained from the other by elementary transformations.

Session : 12

Types of Matrices

Upper triangular matrix: If in a square matrix, all the elements below the principal diagonal are zero, the matrix is called an upper triangular matrix.

Example: Reduce the matrix A= [1 2 32 5 73 1 2] to upper triangular matrix form.

Sol: -Let A = [1 2 32 5 73 1 2]R2→ R2−2 R1

≃ [1 2 30 1 10 −5 −7] R3 → R3−3 R1

≃ [1 2 30 1 10 0 −2]R3 → R3 +5 R2

≃ [1 2 30 1 10 0 −2]

∴ Required upper triangular matrix is [1 2 30 1 10 0 −2]

Diagonal matrix: Diagonal Matrix is a symmetric matrix with all of its entries equal to zero except may be the ones on the diagonal. So a diagonal matrix has at most n different numbers. For example, the matrices

(a 00 b) and (a 0 0

0 0 00 0 b)

are diagonal matrices. Identity matrices are examples of diagonal matrices. Diagonal matrices play a crucial role in matrix theory. We will see this later on.

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Symmetric matrix: A Square Matrix A is said to be symmetric if aij=aji for all i and j. In other words, we can say that  matrix A is said to be symmetric if  transpose of matrix A is equal to Matrix A i.e (AT=A).Skew-Symmytric matrix:Square Matrix A is said to be skew-symmetric if aij=−aji for all i and j. In other words, we can say that matrix A is said to be skew-symmetric if transpose of matrix A is equal to negative of Matrix A i.e (AT=−A).Note that all the principle diagonal elements in skew-symmetric matrix are zero.Orthogonal Matrix: A square matrix A is said to be orthogonal if AAT = ATA = I (If A is orthogonal then, AT = A -1 )

Complex MatricesWe have considered matrices whose elements were real numbers. The elements of a matrix can, however, be complex numbers also.

(1) Conjugate of a matrix: If the elements of a matrix A = [ars] are complex numbers ∝rs+iβ rs and β rsbeing real, then the matrix A=[ars ]=[∝rs−iβrs ]is called the conjugate matrix of A.

(2) Hermitian matrix: A square matrix A such that AT = A is said to be a Hermitian Matrix. The elements of the leading diagonal of a Hermitian matrix are evidently real , while every other element is the complex conjugate of the element is the transposed position. For instance

A = [ 2 3+4 i3−4 i −5 ] is a Herimitian matrix, since AT = [ 2 3+4 i

3+4 i −5 ] A

(3) Skew – Hermitian Matrix : A square matrix A such thatAT = -A is said to be a Skew – Hermitian matrix. This implies that the leading diagonal elements of skew – hermitian matrix are either all zeros or all purely imaginary.

Observation: A hermitian matrix is a generalization of a real symmetric matrix as every real symmetric matrix is Hermitian. Similarly, a skew – Hermitian matrix is a generalization of a real skew – symmetric matrix.

(4) Unitary matrix: A square matrix, A is said to Unitary if AA* = A*A = I (where A* = AT

). Note: If A is Unitary then A* = A-1

Note: (a) The inverse of a unitary matrix is Unitary (b) The product of two n – rowed unitary matrices is Unitary.

(5) Sparse matrix or Sparse array is a matrix in which most of the elements are zero. By contrast, if most of the elements are nonzero, then the matrix is considered dense. The number of zero-valued elements divided by the total number of elements (e.g., m × n for an m × n matrix) is called the sparsity of the matrix (which is equal to 1 minus the density of the matrix). Using those definitions, a matrix will be sparse when its sparsity is greater than 0.5.

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(6)  Block matrix or a partitioned matrix is a matrix that is interpreted as having been broken into sections called blocks or submatrices. Intuitively, a matrix interpreted as a block matrix can be visualized as the original matrix with a collection of horizontal and vertical lines, which break it up, or partition it, into a collection of smaller matrices.

(7) Rotation matrix is a matrix that is used to perform a rotation in Euclidean space. For example, using the convention below, the matrix

R = cosθ −sinθsinθ cosθ

Rotates points in the xy-plane counterclockwise through an angle about the origin of a two-dimensional Cartesian coordinate system.

(8) Logical Matrix: A matrix with all entries either 0 or 1.(9) Sign Matrix: A matrix whose entries are either +1, 0, or −1.(10) Signature Matrix: A diagonal matrix where the diagonal elements are either +1 or −1.Exercise

1. Given that A = [ 0 1+2 i−1+2i 0 ], show that (1-A) (1+ A)-1 is a Unitary matrix.

2. Prove that : (a) Every Hermitian matrix can be written as (A + iB), where A si real and symmetric

and B is real and skew – sysmmetric.(b) If A is a Hermitian matrix, (iA) is a skew – Hermitian matrix.

3. Show that [ 3 7−4 i −2+5 i7+4 i −2 3+i

−2−5 i 3−i 4 ] is a Hermitian matrix.

4. Verify is Hermitian matrix or not

5. Verify is Skew-Hermitian matrix or not.

6. If

A=[2+i 3 −1+3 i−5 i 4−2i ]

, verify A*A is Hermitian or notZero–One Matrices

A matrix whose entries are either 0 or 1 is called a zero–one matrix. Zero–one matricesare often used to represent discrete structures. These structures are based on Boolean arithmetic with zero–one matrices. This arithmetic is based on the Boolean operations ∧ and ∨, which operate on pairs of bits, defined by

b1 ∧ b2 = 1 if b1 = b2 = 1

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= 0 otherwise,b1 ∨ b2 = 1 if b1 = 1 or b2 = 1

= 0 otherwise.

DEFINITION Let A = [aij ] and B = [bij ] be m × n zero–one matrices. Then the join of A and B is the Zero–one matrix with (i, j )th entry aij ∨ bij . The join of A and B is denoted by A ∨ B. The meet of A and B is the zero–one matrix with (i, j)th entry aij ∧ bij . The meet of A and B is denoted by A ∧ B.

DEFINITION Let A = [aij ] be an m × k zero–one matrix and B = [bij ] be a k × n zero–one matrix. Then the Boolean product of A and B, denoted by , is the m × n matrix with (i, j ) th entry cij where cij

= (ai1 ∧ b1j ) ∨ (ai2 ∧ b2j ) ∨---- ∨ (aik ∧ bkj ).Note that the Boolean product of A and B is obtained in an analogous way to the ordinary Product of these matrices, but with addition replaced with the operation and with multiplication replaced with the operation ∧.We give an example of the Boolean products of matrices.

DEFINITION Let A be a square zero–one matrix and let r be a positive integer. The rth Boolean power of A is the Boolean product of r factors of A. The rth Boolean product of A is denoted by A[r].

Session - 13

Rank of a Matrix : The rank of a matrix is said to be ‘r ’ If (a) It has at least one non – zero minor of order ‘ r’.

(b) Every minor of matrix of order higher than ‘r’ is zero.

∴ Rank of a matrix = Number of non – zero rows in an upper triangular matrix.

Note: Non – zero row is that row which does not contain all the elements zero.

Examples: Determine the rank of the following matrices:

(a) [1 2 31 4 22 6 5 ] (b) [0 1 −3−1

0 0 113 1 02

11−20 ]Sol: - (a) Operate (R2 - R1) and (R3 - 2 R1) so that the given matrix

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≃ [1 2 30 2 −10 2 −1] = A (say)

Obviously, the 3rd order minor of the matrix A vanishes. Also its 2nd order minor formed by its

2nd and 3rd rows are all zero. But another 2nd order minor is [1 30 −1] = -1≠0.

Hence the rank of matrix A is 2.

Rank of a Matrix by using Echelon Form

1. Any row consisting of zeros is below any row that contains at least one nonzero number; 2. The first (from left to right) nonzero entry of any row is to the left of the first nonzero entry of any lower row. Now if we make sure that the first nonzero entry of every row is 1, we get a matrix in row echelon form.

For example [1 2 1−10 0 130 0 0 0 ]

This matrix is in echelon form. Example1:

Consider the matrix [0 0 1 32 4 0−81 2 1−1]

First we will transform the first column via elementary row operations into one with the top number equal to 1 and the bottom ones equal 0. Indeed, if we interchange the first row with the last one, we get

[1 2 1−12 4 0−80 0 1 3 ]

Next, we keep the first and last rows. And we subtract the first one multiplied by 2 from the second one. We get

[1 2 1−10 0 −2−60 0 13 ]

We are almost there. Looking at this matrix, we see that we can still take care of the 1 (from the last row) under the -2. Indeed, if we keep the first two rows and add the second one to the last one multiplied by 2, we get

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[1 2 1−10 0 −2−60 0 00 ]

Hence Rank of the matrix is 2.Example2: Consider the matrix

[0 1 −3−11 0 003 1 −3−1

11−3−1 ]Operating (R3−R1¿, (R4−R1 ¿

[0 1 −3−11 0 0 03 0 0 0

1 0 0 0 ] Operating (R3−3 R2¿, (R4−R2 ¿

[0 1 −3−11 0 0 00 0 0 0

0 0 0 0 ] [0 1 0 01 0 0 00 0 0 0

0 00 0 ] = A (say)

Obviously, the 4th order minor of the matrix A is zero. Also every 3rd order minor of A is

zero. But, of all the 2nd order minor, only |0 11 0| = -1 ≠0.

Hence the rank of matrix A is 2

Exercise

(1) Determine the rank of the following matrices:

(a) [ 3 −1 2−6 2 4−3 1 2] (b) [2 −1 31

1 4 −215 2 43 ]

Session -14 & 15

Consistency of a system of linear equationsConsider the system of ‘m’ linear equations

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a11 x1+a12 x2+−−−+a1n xn=K1

a21 x1+a22 x2+−−−−+a2 n xn=K2

--------------------------------------------------------- ---------------------------------------------------------

am1 x1+am2 x2+−−−−+amn xn=K m

------------------------------------(1)Containing the ‘n’ unknowns x1 , x2 ,−−−− , xn . To determine whether the equations (1) are consistent (i.e., possess a solution) or not, we consider the ranks of the matrices.

A = ¿ and K = ¿

A is the coefficient matrix and K is called the augmented matrix of the equations (1).

Find the ranks of the coefficient matrix A and the augmented matrix, K by reducing A to the triangular form by elementary row operations. Let the rank of A be r and that of K be r .

(i) If r ≠ r , the equations are in – consistent, ie., there is no solution.(ii) If r = r =n, the equations are consistent and there is a unique solutions.(iii) If r = r ¿n, the equations are consistent and there are infinite number of solutions.

Solutions of linear simultaneous equations: There are two methods to solve the linear simultaneous equations:1. Direct method: Gauss elimination method2. Iterative method: Jacobi and Gauss Seidal method.

Simultaneous linear equations occur in various engineering problems. These equations can be solved by Cramer’s rule and matrix method. But these methods become tedious for large systems. However, there exist other numerical methods of solutions which are well suited for computing machines. We now discuss some iterative methods of solutions. Note: Gauss seidal method converges if in each equation, the absolute value of the largest cooefficient is greater than the sum of the absolute values of the remaining coefficients.Gaussian Elimination method:The purpose of this article is to describe how the solutions to a linear system are actually found. The fundamental idea is to add multiples of one equation to the others in order to eliminate a variable and to continue this process until only one variable is left. Once this final variable is determined, its value is substituted back into the other equations in order to evaluate the remaining unknowns. This method, characterized by step‐by‐step elimination of the variables, is called Gaussian elimination.

Example consider the following system of algebraic equations:

x+y+z = 62x+y-z = 1

-x+2y+2z = 9If the first equation is multiplied by -2 and is added to the second equation and if the first equation is added to the third equation, the system of linear equation can be written as follows:

x+y+z = 68

0-y-3z = -110+3y+3z =15

The x’s coefficient of the last two equations has be eliminated. This technique can be used for the last equation again. If the second equation is multiplied by 3 and is added to the third one, the system of linear equation is transformed to the following one:

x+y+z = 60-y-3z = -110+0-6z = -18

Using the third equation, z can be evaluated easily and then y can be determined by the second equation and x can be determined by using the first one. 

x =1; y = 2; z = 3Jacobi’s iterative method: Consider the linear system with three equations and three unknowns;

a11 x + a12 y + a13 z = b1

a21 x + a22 y + a23 z = b2

a31 x + a32 y + a33 z = b3. If a11, a22, a33 are absolutely large as compared to other coefficients, then the equations can be written in the following form

x = (b1 – a12 y – a13 z)/ a11

y = (b2 – a21 x – a23 z)/ a22

z = (b3. – a31 x – a32 y)/ a33. The Jacobi’ iterative formula is

xn+1 = (b1 – a12 yn – a13 zn)/ a11

yn+1 = (b2 – a21 xn – a23 zn)/ a22

zn+1 = (b3 – a31 xn – a32 yn)/ a33, n = 0, 1, 2, … Let us start with the initial approximations x0 = y0 = z0 = 0 for the values of x, y, z. Substituting in iterative formula, we get first approximations. This process is repeated till the difference between two consecutive approximations is negligible.Example:1. Determine the solution of the following system of linear equations

20x + y – 2z = 173x + 20y – z = –182x – 3y + 20z = 25

using Jacobi method.Sol. The diagonal coefficients are larger than the other coefficients, the equations expressed in the following form;

3y).2x(25201z

z)3x18(201y

2z)y(17201x

The Jacobi iterative formula is

9

2, 1, 0,n ),3y2x(25201z

)z3x18(201y

)2zy(17201x

nn1n

nn1n

nn1n

Taking the initial approximation x0 = y0 = z0 = 0. Therefore, the first approximate to the solution is

.25.12025)0025(

201)3y2x(25

201z

9.02018)0018(

201)z3x18(

201y

85.02017)0017(

201)2zy(17

201x

001

001

001

The second approximate to the solution is

.1515.1)]9.0(3)85.0(225[201)3y2x(25

201z

965.0)25.1)85.0(318[201)z3x18(

201y

02.1)]25.1(2)9.0(17[201)2zy(17

201x

112

112

112

The third approximate to the solution is

.003.1)]965.0(3)02.1(225[201)3y2x(25

201z

9954.0)1515.1)02.1(318[201)z3x18(

201y

0134.1)]1515.1(2)965.0(17[201)2zy(17

201x

223

223

223

The fourth approximate to the solution is

.9993.0)]9954.0(3)0134.1(225[201)3y2x(25

201z

0018.1)003.1)0134.1(318[201)z3x18(

201y

0009.1)]003.1(2)9954.0(17[201)2zy(17

201x

334

334

334

The fifth approximate to the solution is

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.9996.0)]0018.1(3)0009.1(225[201)3y2x(25

201z

0002.1)9993.0)0009.1(318[201)z3x18(

201y

0000.1)]9993.0(2)0018.1(17[201)2zy(17

201x

445

445

445

The fifth approximate to the solution is

.0000.1)]0002.1(3)0000.1(225[201)3y2x(25

201z

0000.1)9996.0)0000.1(318[201)z3x18(

201y

0000.1)]9996.0(2)0002.1(17[201)2zy(17

201x

556

556

556

The difference between fifth and sixth approximations is negligible. Therefore the solution of the given system of equations are x = 1, y = –1, z = 1.

Gauss-Seidel iterative method: This is a modification of the Jacobi’s iterative method. In this method the most recent approximation of the unknowns are used while proceeding to the next step. Solution procedure is same as in Jacobi method. Here we use the followingGauss-Seidel iterative formula

xn+1 = (b1 – a12 yn – a13 zn)/ a11

yn+1 = (b2 – a21 xn+1 – a23 zn)/ a22

zn+1 = (b3 – a31 xn+1 – a32 yn+1)/ a33, n = 0, 1, 2, … for the given system of three equation

a11 x + a12 y + a13 z = b1

a21 x + a22 y + a23 z = b2

a31 x + a32 y + a33 z = b3.Example:. Determine the solution of the following system of linear equations

20x + y – 2z = 173x + 20y – z = –182x – 3y + 20z = 25

using Gauss-Seidel iterative method.

Sol. The diagonal coefficients are larger than the other coefficients, the equations expressed in the following form;

3y).2x(25201z

z)3x18(201y

2z)y(17201x

The Gauss-Seidel iterative formula is

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2, 1, 0,n ),3y2x(25201z

)z3x18(201y

)2zy(17201x

1n1n1n

n1n1n

nn1n

Taking the initial approximation x0 = y0 = z0 = 0. Therefore, the first approximate to the solution is

.0109.1)]0275.1(3)85.0(225[201)3y2x(25

201z

0275.1]0)85.0(318[201)z3x18(

201y

85.02017)0017(

201)2zy(17

201x

111

011

001

The second approximate to the solution is

.9998.0)]9998.0(3)0025.1(225[201)3y2x(25

201z

9998.0]0109.1)0025.1(318[201)z3x18(

201y

0025.1)]0109.1(2)0275.1(17[201)2zy(17

201x

222

122

112

The third approximate to the solution is

.0000.1)]0000.1(3)0000.1(225[201)3y2x(25

201z

0000.1]9998.0)0000.1(318[201)z3x18(

201y

0000.1)]9998.0(2)9998.0(17[201)2zy(17

201x

333

233

223

The difference between second and third approximations is negligible. Therefore, the solution of the given system of equations are x = 1, y = –1, z = 1. Example: A firm can produce three types of cloths A, B and C. Three kinds of wool are required for it, say red, green and blue wool. One unit of type ‘A’ cloth needs 2 yards of red wool, 8 yards of green and one yard of blue wool; one unit length of type ‘B’ cloth needs one yard of red, 3 yards of green and 5 yards of blue wool; one unit length of type ‘C’ cloth needs 6 yards red, 2 yards of green and one yard of blue wool. The firm has only a stock of 9 yards red, 13 yards green and 7 yards of blue wool. If total stock is used, then determine the number of units of cloth A, B and C. Sol. Let x, y and z be the number of units of cloth types ‘A’, ‘B’ and ‘C’ produced by the firm.Write the given data in the following form Types of the cloth stock A B C availableRed 2 1 6 9Green 8 3 2 13

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Blue 1 5 1 7The total quantity of red wool required to prepare x, y and z yards of cloths A, B and C is

2x + y + 6zSince the stock of red wool available is 9 and we used whole. Therefore

2x + y + 6z = 9.Similarly total quantity of green and blue wool required

8x + 3y + 2z = 13,x + 5y + z = 7.

Hence the problem of firm formulated as to find x, y and z which satisfies the following equations

2x + y + 6z = 98x + 3y + 2z = 13,

x + 5y + z = 7.Solving these equations using Jacobi or Gauss-Seidel method, we get x = y = z = 1.Example. Determine the current flowing in each branch of the following circuit using Jacobi method.

Exercise1. Determine an approximate solution of the following system of simultaneous linear equations, using Jacobi iterative method starting with (0.5, 1.5, 2.5); x + 2y + 5z = 20 5x + 2y + z = 12 x + 4y + 2z = 15.2. Determine an approximate solution of the following system of simultaneous linear equations, using Gauss-Seidel iterative method; 2x + 17y + 4z = 35 x + 3y + 10z = 24 28x + 4y - z = 32. 3. Solve by Jacob’s iteration method, the equations 20x+y-2z=17; 3x+20y-z=-18; 2x-3y+20z=254. Solve the equations 54x+y+z=110; 2x+15y+6z=72; -x+6y+27z=85 by Gauss-Seidel method.5. Apply Gauss-Seidal iteration method to solve the equations 10x1-2x2-x3-x4=3; -2x1+10x2-x3-x4=15; -x1-x2+10x3-2x4=27 and –x1-x2-2x3+10x4= -9.

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6. A civil engineer involved in construction requires 4800, 5800 and 5700 m3 of sand, fine gravel, and coarse gravel respectively for a building project. There are three pits from which these materials can be obtained. The composition of these pits is;

Sand Fine gravel Coarse gravelPit 1 55 30 15Pit 2 20 50 30Pit 3 25 20 55

How many cubic meters must be hauled from each pit in order to meet the engineer’s need?

7. The following system of equations is designed to determine concentrations (the c’s in g/m3) in a series of coupled reactors as a function of the amount of mass input to each reactor(the right-hand sides in g/day);15c1 – 3c2 – c3 = 3800-3c1 + 18c2 – 6c3 = 1200-4c1 – c2 + 12c3 =2350.

Determine the concentrations using Gauss-Seidel method, the approximate relative error fall below 5%.

8. Solve the following system of non-homogenous linear equation by Gaussian elimination method.

9. Evaluate Boolean product of A and B, Where A=

10. The management of Hartman Rent-A-Car has allocated $1.5 million to buy a fleet of new automobiles consisting of compact, intermediate-size, and full-size cars. Compacts cost $12,000 each, intermediate size cars cost $18,000 each, and full-size cars cost $24,000 each. If Hartman purchases twice as many compacts as intermediate-size cars and the total number of cars to be purchased is 100, determine how many cars of each type will be purchased. (Assume that the entire budget will be used.)Session -16 & 17

Characteristic Equation

If A is any matrix of order n, we can form the matrix (A- λI ), where I is the n the order unit matrix. The determinant of this matrix equated to zero i.e,

(A- λI ) = ¿ = 0 is called the characteristic equation of A. on expanding the determinant, the characteristic equation taken the form

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(-1) λn+k1 λn−1+k2 λn−2+−−−+kn=0 where k’ s are expressible in terms of the elements aij. The roots of this equation are called the characteristic roots or latent roots or eigen values of the matrix A.

1. Calculate Eigen Values and Eigen Vectors of the matrix,

A=[2 0 10 2 01 0 2 ]

2. Calculate Eigen Values and Eigen Vectors of the matrix,

A=[ 2 1 −11 1 −2

−1 −2 1 ]

3. Using Cayley Hami;ton theorem evaluate A-1, here

A=[ 6 −2 2−2 3 −12 −1 3 ]

4. Using

Cayley Hamilton theorem evaluate A3, where

Eigen Vector

If X = [ x1

x2

−−¿−−¿ x3] and A = ¿then the linear transformation Y = AX ----- (i)

carries the column vector, X into the column vector, Y by means of the square matrix A. In practice, it is often required to find such vectors which transform into themselves or to a scalar multiple of themselves. Let X be such a vector which transforms into λX by means of the transform (i) then λX = AX (or) AX – λIX = 0 (or)(A- λI ¿X = 0 ------- (ii) .This matrix equation represents n homogeneous linear equations

¿ ------ ( iii)which will have a non – trivial solution only if the coefficient matrix is singular, i.e., if |A−λI|=0. This is called the characteristic equation of the transformation and is same as the characteristic equation of the matrix, A. It has n roots and corresponding to each root the equation (ii) will have a non – zero solution X = ¿¿, -- xn].which is known as the Eigen vector or latent vector. Example:

Find the eigen values and eigen vectors of the matrix A = [ 8 −6 2−6 7 −42 −4 3 ] .

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Sol: - The characteristic equation is

|A−λI|=|8− λ −6 2−6 (7− λ) −42 −4 (3−λ)|=−λ3+18 λ2−45 λ=0

(or) λ (λ−3¿ ( λ−15 )=0

λ=0 , 3,15.

If x, y, z be the components of an eigen vectors corresponding to the eigen value, λ we have

[ A− λI ] X=|8− λ −6 2−6 (7−λ) −42 −4 (3−λ)|[ x

yz ]=0−−( i)

Putting λ =0, we have 8x- 6y+ 2z = 0, -6x + 7y – 4z = 0, 2x – 4y + 3z = 0.

These equations determine a single linearly independent solution which may be taken as (1,2,2) so that every non – zero multiple of this vector is an eigen vector corresponding to λ = 0.

Similarly, the eigen vectors corresponding to λ = 3 and λ = 15 are the arbitrary non – zero multiples of the vectors (2,1,-2) and (2, -2, 1) which are obtained from (i)

Hence the three eigen vectors may be taken as (1,2,2), (2, 1, -2), (2, -2, 1).

Properties of Eigen values

(1) The sum of the eigen values of a matrix is the sum of the elements of the principal diagonal.

(2) If λ is an eigen value of a matrix, A then 1/λ is the eigen value of A−1.(3) If λ is an eigen value of an orthogonal matrix, then 1/λ is also its eigen value.(4) If λ1 , λ2 ,−−−−, λn are the eigen values of a matrix A then Am has the eigen values

λ1m, λ2

m, λ3m, ………………λn

m

GAME THEORY

SESSION -18

Introduction, Basic Concept & Related Methods

What Is a Game?

A game is defined as an activity between two or more persons involving activities by each person with some rules, and at the end of which each person receives some benefit or satisfaction or suffers loss (Negative Benefit).

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Game Theory: The mathematical analysis of competitive problems

Game theory is the study of mathematical models of strategic interaction between two or more opponents. Game Theory helps us to understand the situations in which decision-makers interact. A game in the everyday sense—“a competitive activity . . . in which players contend with each other according to a set of rules. Few examples of game theory utilization are as follows:

Firms competing for business Arms race between the countries Political candidates competing for votes Bidders competing in an auction

And the list goes on….. Like other sciences, game theory consists of a collection of models.

Limitation of Game Theory 1. Infinite Number of Strategy. In a game theory we assume that there is finite number of possible courses of action available to each player. But in practice a player may have infinite number of strategies or courses of action. 2. Knowledge about Strategy. Game theory assumes that each player has the knowledge of strategies available to his opponent. But sometimes knowledge about strategy about the opponent is not available to players. This leads to the wrong conclusions.

Some Basic Terms in Game Theory: Pay-off:

A quantitative measure of satisfaction a person gets at the end of the play is called pay-off (gain or loss).

Mini-Max or (Maxi-Min) Criterion: Game theory is fundamentally based upon the Mini-Max or (Maxi-Min) criterion. This criterion implies the assumption of rationality from which it is argued that each player will act so as to maximize his minimum gain or to minimize his maximum loss.

Strategy: A strategy for player is defined as set of alternative courses of action available to him in advance, by which player decides the course of action that he should adopt.

The strategy may be two types:(i) Pure Strategy: Pure strategy is when a player chooses one of his two preferences but not

mix them.(ii) Mixed Strategy: Mixed strategy is when individual chooses either of his actions with some

probability.

For example, choosing to either confess or deny in Prisoners Dilemma is pure strategy. While choosing to confess half of the times and to deny other half of the times is a mixed strategy by the player.

Optimal Strategy: 17

The strategy that optimizes a player’s pay-off is called optimal strategy. It relates to a course of action when the player remains in the most preferred position regardless of the action of his competitor(s).

Value of the Game: When each player plays his optimal strategy, the resulting pay-off is called the value of the game.

Fair Game: If value of the game is zero i.e. there is no loss or gain for any player, the game is called fair game.

Type of GamesGame theory models can be classified in the following manner:

Two-Person Zero Sum Game:-A game with only two persons (player A and player B) is called two-person zero sum game, if the losses of one player are equal into gains of the other, so that the sum of their net gains is zero. Two-person zero sum games are called rectangular games as these are usually presented by payoff Matrix in rectangular form.

Two-Person Non-Zero Sum Game:- A game in which, the gain and loss are not equal, and hence the outcome is not obvious.

N−¿Persons Game:- A game in which there are a large number of players (competitors), all trying for the best trade-off.

Pay-off MatrixA matrix of order (m× n), which gives the possible outcome of a two-person zero-sum game when player A has m possible strategies and player B has n strategies. The analysis of the matrix in order to determine optimal strategies is the aim of game theory. A pay-off matrix can be formed by adopting the following rules.

1. Row designations for each strategy of player A.2. Column designations for each strategy of player B.3. When A chooses the activity ‘i’, and B chooses the activity ‘ j’, the cell entry a ij is the payment to player A in the pay-off matrix.

The so-called pay-off matrix is defined as follows:

Player A

Player BStrategies(pure)

B1 B2 Bn

A1 a11 a12 … a1n

A2 a21 a22 … a2n

… … … … …

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Am am1 am 2 … amn

Saddle Point: A saddle point of a payoff matrix is the position of such an element in the pay-off matrix which is minimum in its row and maximum in its column. That is, if a pay-off matrix is such that

Max ¿ then the matrix is said to have a saddle point (r, s).

Remark: For a pay-off matrix of a game problem, 1. There may be no saddle point. 2. There may be one saddle point, two saddle points or more. 3. If saddle point exists, then the value of the game is equal to the saddle point.

Strictly Determinable Game:A game is said to be strictly determinable if saddle point exists.

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Figure: Flow Chart of Game TheoryGames with Saddle Point (Pure Strategy)

Method of Maximin (or Minimax) Criteria: To solve the games with saddle point, the method consists of the following steps:

Step-1: Write down the minimum pay-off in each row.

Step-2: Select the maximum among there. It is maximin value.

Step-3: Write down the maximum pay-off in each column.

Step-4: Select the minimum among these. It is minimax value.

Step-5: If maximin value=minimax value ,

then mark the position of that pay-off in the matrix. This element is the saddle

point, and represents the value of the game.

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Rule of Dominance In a two-person zero sum game, if there are large number of strategies available for each player,

then the pay-off matrix becomes very large. By the rule of dominance, this can be reduced to

2 ×2 order pay-off matrix, and then can be easily solved. This concept is very useful when the

saddle point does not exist.

The rules of dominance are as follows:

(i) Rule of Rows: When all the elements in a row are less than or equal to the corresponding elements in another row, this row is said to be dominated by the other row, and hence can be deleted from the pay-off matrix.

(ii) Rule of Columns:When all the elements in a column are greater than or equal to the corresponding elements in another column, this column is said to be dominated by the other column, and hence can be deleted from the pay-off matrix.

(iii) Rule of Averages:If the average of the elements of two rows or two columns is observed to be dominated as per rule of rows or columns, then these rows or columns can be deleted from the pay-off matrix.

Games without Saddle Point (Mixed Strategy)

Algebraic Method: In situations where a saddle point does not exist, the method of maximin (or minimax) criteria

for solving a game problem breaks down. Such games can be solved by Algebraic Method.

The optimal strategy mixture for each player may be determined by assigning to each strategy its

probability of being chosen. Let 2×2 matrix so obtained from the rule of dominance can be

written as:

Player B

Player A

Probability q1 q2=1−q1

Probability Pure Strategy

B1 B2

p1 A1 a11 a12

p2=1−p1 A2 a21 a22

Let p1and p2 be the probability for player A.Let q1 and q2 be the probability for player B.Let the optimal strategy be SA for player A and SB for player B.

Then the optimal strategies are given by:

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(a) SA=[A1 A2

p1 p2 ] , and (b) SB=[B1 B2

q1 q2] (1)

Values of p1 and p2 are determined by using the formulae,

p1=a22−a21

(a11+a22)−(a12+a21) , (2)

¿ p2=1−p1 , (3)

q1=a22−a12

(a11+a22)−(a12+a21) (4)

¿q2=1−q1 (5)

The value of the game w . r . t . player A is given by,

V = a11a22−a12 a21

(a11+a22)−(a12+a21) (6)

Solved ExamplesExample 1: Find the saddle point of the game:

Player A

Player BB1 B2

A1 1 3A2 −¿1 6

Solution: The game is worked out using minimax procedure. Find the smallest value in each row and select the largest value of these values. Next, find the largest value in each column and select the smallest of these numbers. The procedure is shown in the following table.

Player B Row

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Playe

r A

minB1 B2

A1 1 3 1A2 −¿1 6 −¿1

Column max 1 6

If the maximum value in row is equal to the minimum value in column, then saddle point exists.Max Min = Min Max

1 = 1Therefore, there exists a saddle point. Value of game = 1. The strategies are: Player A plays Strategy a1, and Player B plays Strategy b1.

Example 2: Solve the following matrix game: T = (−2 33 −4 ).

Solution: In this game, the saddle point does not exist. Therefore, this game problem cannot be

solved by the maximin (or minimax) criteria. Such a game is solved by Algebraic Method.

T = (−2 33 −4 )

p1=a22−a21

(a11+a22)−(a12+a21) ¿

−4−3(−2−4)−(3+3)

= 712 , ¿ p2=1− 7

12= 5

12,

q1=a22−a12

(a11+a22)−(a12+a21)= −4−3

(−2−4 )−(3+3)= 7

12 , ¿q2=1− 712

= 512

Optimal Strategies

(a) SA=[ A1 A2

712

512 ] , and (b) SB=[ B1 B2

712

512 ]

Value of the Game V = a11 a22−a12 a21

(a11+a22)−(a12+a21) = 8−9

(−2−4)−(3+3)= 1

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Example 3: Find out the saddle point, game value, and optimal strategies for the given pay-off matrix:

X

YY 1 Y 2 Y 3

X1 2 18 4X2 16 10 8

Solution: First, we find the minimum pay-off value for each row. Then, we find the maximum pay-off value for each column. The Saddle point will be where; Largest of row minima (Maximin) equals Smallest of column maxima (Minimax).

Y Row

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X

minY 1 Y 2 Y 3

X1 2 18 4 2

X2 16 10 8 8

Col. Max.

16 18 8

Max (Row Min) = 8 = Min (Column Max), therefore the saddle point is 8. The optimal strategies are, Player X plays Strategy X2, and Player Y plays Strategy Y3.Value of game = 8.

Example 4: Determine the best strategy for the firm A at different gains as maintained in the pay off matrix given below against B’s strategies using maximin criteria.

A

BB1 B2 B3 B4

A1 15 14 23 11A2 11 10 12 12A3 13 13 13 13

Solution: Stage one: - Find the minimum among each row.Stage Two: - Find the maximum value of calculated minimum as MAXIMIN.

Example 5: Apply dominance rules to reduce the size of the following pay-off matrix to 2 × 2 order, and hence determine the optimal strategies and value of the game. B

A (1 7 26 2 75 1 6)

Solution: The given pay-off matrix is

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A

B MiniRowB1 B2 B3 B4

A1 15 14 23 11 11A2 11 10 12 12 10A3 13 13 13 13 13

No saddle point exists. Row-II dominates over Row-III, and therefore Row- III can be deleted.

Column-I dominates over Column-III, and hence Column-III may be deleted. We obtain

Applying Algebraic method, we obtain

p1= 25 , p2=

35 , q1=

12 , q2=

12 , and value of the game V = 4.

Optimal Strategies

(a) SA=[ A1 A2 A3

2/5 3 /5 0 ] , and (b) SB=[ B1 B2 B3

1/2 1/2 0 ] Self Learning:

Decision Making Under Uncertainty

A decision under uncertainty is when there are many unknowns and no possibility of knowing what could occur in the future to alter the outcome of a decision. We feel uncertainty about a situation when we can't predict with complete confidence what the outcomes of our actions will be.Launching a new product, a major change in marketing strategy or opening your first branch could be influenced by such factors as the reaction of competitors, new competitors, technological changes, changes in customer demand, economic shifts, government legislation and a host of conditions beyond your control. These are the type of decisions facing the senior executives of large corporations who must commit huge resources.The small business manager faces the same type of conditions which could cause decisions that result in a disaster from which he or she may not be able to recover. A situation of uncertainty arises when there can be more than one possible consequences of selecting any course of action.

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A

B MiniRowB1 B2 B3

A1 1 7 2 1A2 6 2 7 2A3 5 1 6 1

Column Max. 6 7 7

A

BB1 B2 B3

A1 1 7 2A2 6 2 7

A

BB1 B2

A1 1 7A2 6 2

In terms of the pay-off matrix, if the DM selects A1, his payoff can be X11, X12, X13, etc., depending upon which state of nature S1, S2, S3, etc., is going to occur.

Methods of Decision Making Under UncertaintyThere are a variety of criteria that have been proposed for the selection of an optimal course of action under the environment of uncertainty. Each of these criteria makes an assumption about the attitude of the decision-maker. The methods of decision making under certainty are:

1. Maximin Criterion: This criterio is used when the decision-maker is pessimistic about future. Maximin implies the maximization of minimum payoff. The pessimistic decision-maker (DM) locates the minimum payoff for each possible course of action. The maximum of these minimum payoffs is identified and the corresponding course of action is selected. This is explained in the following example.

Example 6: Let there be a situation in which a decision-maker has three possible alternatives A1, A2 and A3, where the outcome of each of them can be affected by the occurrence of any one of the four possible events S1, S2, S3 and S4. Monetary payoffs of each combination of Ai and Sj are given in table:

Solution: Since 17 is maximum out of the minimum payoffs, the optimal action is A2.

2. Maximax Criterion: This criterion is used when the decision-maker is optimistic about future. Maximax implies the maximisation of maximum payoff. The optimistic decision-maker (DM) locates the maximum payoff for each possible course of action. The maximum of these payoffs is identified and the corresponding course of action is selected. The optimal course of action in the above example, based on this criterion, is A3.

SESSION - 19

Class Delivery Questions- Level - 1

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1. [JK Sharma, Page-475] Two players A & B, without showing each other put a coin on a table with head or tail up. If the coins show the same side (both head or tail), the player A takes both the coins, otherwise B gets them. Write the pay-off matrix of the game.

(Ans. ( 2 −2−2 2 )

2. [SD Sharma, Page-384] Find out whether there is any saddle point in the following problem: Player B

Player A (−3 13 −1)

(Ans. Saddle point does not exit).

3. [SD Sharma, Page-386] Find saddle point (or points) and hence solve the game : (6 8 64 12 2)

(Ans. (I, I), (I, III), v=6).4. Determine the saddle point for the matrix game:

(−2 0 0 5 34 2 1 3 2

−4 −3 0 −2 65 3 −4 2 −6

) (Ans. Saddle point = 1).

5. [SD Sharma, Page-383] Find the range of the values of p and q which will render the entry (2,

2) a saddle point for the game: ( 2 4 510 7 q4 p 6)

(Ans. p≤7, q≥7).6. [SD Sharma, Page-384] Consider the game G with the following pay-off matrix:

G= ( 2 6−2 µ)

(i) Verify that G is strictly determinable whatever µ may be.(ii) Determine the value of G.(Ans. (1,1), v=2)

7. [SD Sharma, Page-386] For what value of k, the game with the following pay-off matrix is strictly determinable.

( k 6 2−1 k −7−2 4 k )

8. [SD Sharma, Page-384] For the game with pay-off matrix: (−1 2 −26 4 −6), determine the best

strategies for players A and B, and also the values of the game. Is this game (i) fair, (ii) strictly determinable?(Ans. (I, II), v¿−¿2, Game is strictly determinable, and not fair).

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SESSION -20

Class Delivery Questions- Level-2

1. [JK Sharma, Page-496] Solve the following game by using maximin (minimax) criteria whose pay-off matrix:

(−2 15 −2−5 −6 −4−5 20 −8 )

2. [SD Sharma, Page-394] Two companies A and B are competing for the same product. Their different strategies are given in the following pay-off matrix:

Company B

Company A ( 2 −2 3−3 5 −1)

Determine the best strategies and find the value of the game.3. [SD Sharma, Page-386] For the following pay-off matrix for firm A, determine

the optimal strategies for both the firms and the value of the game (using maximin-minimax criteria) :

Firm B

Firm A (3 −1 4 6 7

−1 8 2 4 1216 8 6 14 121 11 −4 2 1

)(Ans. (III, III), v = 6 for A)

4. Reduce by dominance to a 2×2 matrix game and solve:

(5 4 1 04 3 2 −10 −1 4 31 −2 1 2

)5. Determine the game-value and an optimal (mixed) strategy for both players.

(5 −10 9 06 7 8 18 7 15 13 4 −1 4

)6. Two companies promote two competing products. Currently, each product controls 50 % of

the market. Because of recent improvements in the two products, each company is preparing to launch an advertising campaign. If neither company advertises, equal market shares will

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continue. If either company launches a stronger campaign, the other is certain to lose a proportional percentage of its customers. A survey of the market shows that 50 % of the potential customers can be reached through television, 30 % through newspapers, and 20 % through radio.

(a) Formulate the problem as a two person zero sum game, and select the appropriate advertising media for each company.

(b) Determine a range for the value of the game.

7. On a picnic outing, 2 two-person teams are playing hide-and-seek. There are four hiding locations (A, B, C, D), and two members of the hiding team can hide separately in any two of the four locations. The other team will then have the chance to search any two locations. The searching team gets a bonus point if they find both members of the hiding team. If they miss both, they lose a point. Otherwise, the outcome is a draw.

(a) Set up the problem as a two-person zero-sum game.(b) Determine the optimal strategy and the value of the game.

Tutorial -429

1. A firm can produce three types of cloths A, B and C. Three kinds of wool are required for it, say red, green and blue wool. One unit of type ‘A’ cloth needs 2 yards of red wool, 1 yards of green and one yard of blue wool; one unit length of type ‘B’ cloth needs one yard of red, 2 yards of green and 1 yards of blue wool; one unit length of type ‘C’ cloth needs 1 yards red, 1 yards of green and one yard of blue wool. The firm has only a stock of 4 yards red, green and blue wool each. If total stock is used, then determine the number of units of cloth A, B and C by applying Gauss Seidal iteration method

1. Ram, Raj, and Ravi go to a restaurant for lunch and order three different items. Ram orders 2 plates of fried rice, 3 plates of chicken pieces and 1-plate of curd rice. Raj orders 1 plate of fried rice, 2 plates of chicken pieces and 3 plates of curd rice. Ravi orders 3 plates of fried rice, 1 plate of chicken pieces and 2 plates of curd rice. Ram’s bill costs $9, Raj’s costs $6, and Ravi’s costs $8. Determine plate cost of each item, by using suitable method.

3. Cantwell Associates, a real estate developer, is planning to build a new apartment complex consisting of one-bedroom units and two- and three-bedroom townhouses. A total of 192 units is planned, and the number of family units (two- and three-bedroom townhouses) will equal the number of one-bedroom units. If the number of one-bedroom units will be 3 times the number of three bedroom units, find how many units of each type will be in the complex.

4. Your aunt receives an inheritance of $20,000. She wants to put some of the money into a savings account that earns 2% interest annually and invest the rest in certificates of deposit (CDs) and bonds. A broker tells her that CDs pay 5% interest annually and bonds pay 6% interest annually. She wants to earn $1000 interest per year, and she wants to put twice as much money in CDs as in bonds. How much should she put in each type of investment?

5. The Johnson Farm has 500 acres of land allotted for cultivating corn and wheat. The cost of cultivating corn and wheat (including seeds and labor) is $42 and $30 per acre, respectively. Jacob Johnson has $18,600 available for cultivating these crops. If he wishes to use all the allotted land and his entire budget for cultivating these two crops, how many acres of each crop should he plant?

Tutorial -530

1. Calculate Eigen Values and Eigen Vectors of the matrix,

A=[−2 2 −32 1 −6

−1 −2 0 ]

2. Calculate Eigen Values and Eigen Vectors of the matrix A and A

2, here

A=[ 8 −6 2−6 7 −42 −4 3 ]

3. Calculate Eigen Values and Eigen Vectors of the matrix,

4. Using

Cayley Hamilton theorem, evaluate A-1, where A=

[ 3 −1 1−1 5 −11 −1 3 ]

5. Using Cayley Hamilton theorem, evaluate A3 , where A= .

6. Find the Eigen values and Eigen vectors of A, A2, A-1 and A+ 4 I, where A= .

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Tutorial - 61) Solve the game with the following pay-off matrices:

1. (1 3 −10 −4 −31 5 −1)

2. (−2 15 −2−5 −6 −4−5 20 −8 )

3. (−3 4 2 97 8 6 106 2 4 −1)

4. Apply Algebraic method to solve the matrix game: ( 15 2 36 5 7

−7 4 0)5. Reduce by dominance to a 2×2 matrix game and solve: (30 40 −80

0 15 −2090 20 50 )

6. Reduce by dominance to a 2×2 matrix game and solve: (1 3 2 7 43 4 1 5 66 5 7 6 52 0 6 3 1

)7. Reduce each of the following games by using the rule of dominance and then solve the

reduced game by any of the method you have studied:

(3 8 56 2 74 5 6)

8. Reduce each of the following games by using the rule of dominance and then solve the reduced game by any of the method you have studied:

(3 8 56 2 74 5 6)32

9. Solve the following matrix game by using dominance property: (−1 −2 87 5 −16 0 12 )

10. What is the optimal plan for both the players?

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