PowerPoint Presentation : Matrix
PowerPoint Presentation : For Math1
PowerPoint Presentation : Definition:- Matrix is the arrangement of numbers in rows and columns. If a matrix has m rows and n columns it is said to be a matrix of (m x n) order matrix. For example Think what is the order of this matrix ? Give an example of (4 x 5) order matrix by your self.
PowerPoint Presentation : Matrix algebra Addition and Subtraction Matrices of same order can only be added or subtracted and the result can be obtained by adding or subtracting the respective elements. for example:
PowerPoint Presentation : 2. Scalar multiplication: If a scalar is multiplied to a matrix the scalar will get multiplied to each elements of the matrix. For example:
PowerPoint Presentation : 3 .Matrix Multiplication: Suppose A and B are matrices such that the number of column of the first matrix that is A is same as the number of rows of the second matrix that is B, then we can find the product of those matrices by multiplying and adding the elements of each rows to the corresponding elements of the each columns of the second matrix. For example:
PowerPoint Presentation : Special matrices: Null matrix If each elements of a matrix are zero then it is called null matrix or zero matrix. For example: Echelon Matrix: Suppose A is a square matrix where all the pricipal diagonal elements are 1 and bellow to it are zero, then A is a echelon matrix. (Example give by your self) Identity matrix: Suppose A is a square matrix (no. of rows is equal to the no. of columns) and all the elements in it are zero except principal diagonal which are equal to 1. For example: Write the (5 x 5) order Identity matrix by your self.
PowerPoint Presentation : Speciality of Identity matrix If A is a (n x n) order matrix and I is a (n x n) order Identity matrix then I x A = A x I= A Take an example to verify this. By (2 x 2) order By (3x 3) order.
PowerPoint Presentation : TUTORIAL SHEET 1
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PowerPoint Presentation : Where a 11 , a 12 , …. a 21 ,a 22 ,….are the coefficient of x 1 ,x 2, in different equations Let us define three different matrices from the above system of equation. A= X= B= Now we can translate the above system of equation in short notation by using matrices, that is A X = B
PowerPoint Presentation : Example 1: Let us consider the system of equations 2x – 3y =8 x + 2y = 1 Express in matrix notation. Ans: 2 -3 x 8 A= X= B= 1 2 y 1 Short notation is AX = B Give Some examples in the class.
PowerPoint Presentation : Elementary Row Operation. Let us consider the system of equation 3x + 2y = 2--------------(1) 5x + y = 2--------------(2) system 1 Let us try to solve by elementary method that is by eliminating one variable, Let us interchange the position of the equations. 5x + y = 2 3x + 2y = 2 system 2 Multiply 2 in equation 1, and write both the equations together, 10x + 2y = 4 --------------(3) 3x + 2y = 2 --------------(4) system 3 ------------------- 7x = 2 ----------(5) let us write Eq 3 an 5 together 10x + 2y = 4 7x = 2
PowerPoint Presentation : Elementary Row Operation. Let us consider the system of equation Which is called Augmented matrix
PowerPoint Presentation : Which is said to be in echelon form .
PowerPoint Presentation : To solve the above equation we started with system 1 and we made system 2, 3, 4, 5, and 6 which are mathematically equivalent systems. Hence the corresponding matrices are also equivalent. From the above observation we can make certain comments that the set of solutions of a system of linear equations is unaltered by the following three operations, 1- Multiply both side of an equation by a non zero constant. 2- Add a multiple of one equation to another. 3- Interchange the equations. These observations form the motivation behind the row operations method to solve linear equations.
PowerPoint Presentation : ROW OPERATION
PowerPoint Presentation : Steps to solve a system of linear equations.
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PowerPoint Presentation : Solve the system of equation. x + 2y +z = 1 2x + 2y = 2 3x + 5y + 4z = 1 Let us write the corresponding augment matrix.
PowerPoint Presentation : 1 2 1 1 2 2 0 2 3 5 4 1 Let us try to make to the format below by using the row operation. 1 * * * 0 1 * * 0 0 1 *
PowerPoint Presentation : 1 2 1 1 2 2 0 2 3 5 4 1 R2=R2-2R1 Let us try to make the 1 st element of the Principle diagonal element to 1. It is already 1. Now let try to make its below elements 0 by using R1.
PowerPoint Presentation : 1 2 1 1 0 -2 -2 0 3 5 4 1 Let us try to make This element 0 by using R1. R3=R3 - 3*R1
PowerPoint Presentation : 1 2 1 1 0 -2 -2 0 0 -1 1 -2 Let us try to make 2 nd element of the principle diagonal element 1. R2=R2*(-1/2)
PowerPoint Presentation : 1 2 1 1 0 1 1 0 0 -1 1 -2 Let us try to make its below element 0 by using R2. R3=R3+R2
PowerPoint Presentation : 1 2 1 1 0 1 1 0 0 0 2 -2 Let us try to make last element of the Principle diagonal element 1. R3=1/2 *R3
PowerPoint Presentation : 1 2 1 1 0 1 1 0 0 0 1 -1 Target achieved. Now let us transform the matrix to system of equation.
PowerPoint Presentation : Z= -1 Y+Z= 0 X+2Y+Z= 1 Let us start from 3 rd equation, Z = -1 From 2 nd equation Y= - z = -(-1) = 1 F rom 1 st equation X= 1-2y-z = 1- 2(1) –(-1) = 0
PowerPoint Presentation : 2. Solve the system of equation. 2x +3y +3z = 1 x – y + z = 3 x – 2y + 3z = 5 Let us transform the system to a matrix form.
PowerPoint Presentation : 2 3 3 1 1 -1 1 3 1 -2 3 5 R1=1/2 R1 1 3/2 3/2 1/2 1 -1 1 3 1 -2 3 5 Let us try to make the 1 st element of the principle diagonal element to 1 by row operation.
PowerPoint Presentation : 1 3/2 3/2 1/2 1 -1 1 3 1 -2 3 5 R2=R2-R1 R3=R3-R1 1 3/2 3/2 1/2 0 -5/2 - 1/2 5/2 0 -7/2 3/2 9/2 Let us try to make these element 0 by using R1
PowerPoint Presentation : 1 3/2 3/2 1/2 0 -5/2 - 1/2 5/2 0 -7/2 3/2 9/2 R3=-2/5 * R2 1 3/2 3/2 1/2 0 1 1/5 -1 0 -7/2 3/2 9/2
PowerPoint Presentation : 1 3/2 3/2 1/2 0 1 1/5 -1 0 -7/2 3/2 9/2 R3=R3 + 7/2*R2 1 3/2 3/2 1/2 0 1 1/5 -1 0 0 11/5 1
PowerPoint Presentation : 1 3/2 3/2 1/2 0 1 1/5 -1 0 0 11/5 1 R3=R3* 5/11 1 3/2 3/2 1/2 0 1 1/5 -1 0 0 1 5/11 Target achieved. Let us transform this matrix to system of equation.
PowerPoint Presentation : x + 3/2 y +3/2z = ½ y +1/5 z = -1 z = 5/11 y = -1 – 1/5 z = -1 – 1/5(5/11) = -12/11 x = ½ - 3/2y – 3/2z = ½ -3/2(-12/11) -3/2(5/11) = 16/11
PowerPoint Presentation : APPLICATION PROBLEMS
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PowerPoint Presentation : TUTIRIAL SHEET
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PowerPoint Presentation : MATRIX ASSINMENT(MATH1) 1. 2. (8 MARKS) (10MARKS)
PowerPoint Presentation : End for Math1
PowerPoint Presentation : For Math2
PowerPoint Presentation : Inverse of a matrix Let A is a square matrix, then B is said to be the inverse matrix of A if A x B = B x A = I The inverse matrix of a is denoted by A -1 If the inverse of a matrix is existing then it is called a invertible or non-singular. If does not have then A is said to non-invertible or singular.
PowerPoint Presentation : Inverse of a 2x2 order matrix Solve some examples in the class.
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PowerPoint Presentation : Evaluating the value of Determinant By using Row operation Row operation Row operation in matrix in determinant
PowerPoint Presentation : Important property of determinant For example 4 4 5 0 3 4 8 = 2 x 3 x 6 x 4 = 144 0 0 6 2 0 0 0 4
PowerPoint Presentation : NOTE: Given a determinant, if we could able to convert to a triangular determinant (numbers below the Principal diagonal are zero) by row operation it is easy to determine the value of it.
PowerPoint Presentation : Example:
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PowerPoint Presentation : The value = 6 x 1 x 1 x -10 = - 60
PowerPoint Presentation : NOTE : There are different ways to convert a given determinant to a triangular matrix form. But the answer is unique.
PowerPoint Presentation : CRAMER’S RULE In Math1 you studied to solve a system of linear equation AX = B by using Row operation. An alternate method to solve the system of linear equation AX = B is Cramer’s Rule. Where A is a (n x n) order invertible matrix. Note: A is said to be invertible matrix if the determinant value is not equal to zero.
PowerPoint Presentation : Cramer’s Formula X = D1/D D1 is formed by replacing 1 st column by B. Y = D2/D D2 is formed by replacing 2 nd column by B. Z = D3/D D3 is formed by replacing 3 rd column by B. The example in the next slide will certainly help you to understand the formula. Let us consider a 3 variable simultaneous equations.
PowerPoint Presentation : D D1 D2 D D3=
PowerPoint Presentation : X1 = D1/D = 0/-4 = 0 X2 = D2/D = -4/4 = -1 X3 = D3/D = 4/4 = 1. NOTE : It is always a good idea to verify your answer by applying your values in the given equations.
PowerPoint Presentation : A square linear system when A is not invertible ( | A | = 0) Case 1 : When D=0 as well as D1= D2=….=0 then the system has infinitely many solution, and you need to give the a general formula. (Examples are given in the next slide) Case 2 : When only D = 0 and D1,D2,…are non zero values, then the system has no solution at all.
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PowerPoint Presentation : Solving The system AX = B by Row operation and appropriate comments. Example 1
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PowerPoint Presentation : Let s=1 x 3 =1 x 2 = -1, and x 1 = 1+1=2
PowerPoint Presentation : Note: If there are at least one set of solution for a system of equation, then it is called consistent.
PowerPoint Presentation : Example2 Ans:
PowerPoint Presentation : The third equation Will never satisfies for any real values for x 1 , x 2 , and x 3 . Hence there is no solution for the given system of equation.
PowerPoint Presentation : Note: If there is no solution for a system of equation, then it is called inconsistent.
PowerPoint Presentation : Calculating inverse of a matrix by Determinant method.
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PowerPoint Presentation : (Changed the rows to column of the cofactor matrix) Multiplied the position value + - + - + - + - + correspondingly.
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PowerPoint Presentation : A1= A2= A3= B1= B2= B3= C1= C2= C3=
PowerPoint Presentation : Now minor of A = -2 -1 1 -6 -3 -5 4 -2 2 Cofactor matrix of A= Adjoint matrix of A=
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PowerPoint Presentation : Finding Inverse of a square matrix by ROW operation. We can find the inverse of a matrix by elementary raw operation. Let A is a square matrix and we have to find A -1 . Then we form a larger matrix by placing the identity matrix to the right of A, obtaining the matrix denoted by (A I). We then use row operations to reduce to (I B). Then B is the inverse of A. i.e. A -1 =B If it is not possible to reduce to (I B) form then the inverse of A is does not exist.
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PowerPoint Presentation : Show one more example of 3x3 order matrix.
PowerPoint Presentation : TUTORIAL SHEET
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PowerPoint Presentation : FINDING EIGENVALUES Let A is a square matrix and I is the corresponding identity matrix, Then the eigen values of A can be found by solving the equation Further the corresponding eigen vectors can be found by using the eigen values.
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PowerPoint Presentation : Two Industry model Let us start a with a hypothetical economy with only two industries. Electric company E and Water company W. Output for both companies is measured in dollars. The electricity company uses both electricity and water (input) in the production of electricity (output). The water company uses both electricity and water (input) in the production of water. Let for the production of 1dollar worth of electricity requires $0.30 worth of electricity and $0.10 worth of water. And Let for the production of 1dollar worth of water requires $0.20 worth of electricity and $0.40 worth of water. Let the final demand for the outside sector of the economy is d 1 = $12 million for electricity d 2 = $8 million for water.
PowerPoint Presentation : Basic Input-output problem Given the internal demand for each industry’s output, determine output level for the various industries that will meet a given final (outside) level of demand as well as the internal demand. If x 1 = total output from electricity company x 2 = total output from water company Then the internal demands are 0.3x 1 + 0.2x 2 internal demands for electricity. 0.1x 1 + 0.4x 2 internal demand for water.
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PowerPoint Presentation : SOLVE EXAMPLE 6.1, PAGE 89
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