Difference between revisions of "Scilab/C4/Linear-equations-Gaussian-Methods/English-timed"
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| − | |Dear Friends, | + | |Dear Friends, Welcome to the spoken tutorial on '''Solving System of Linear Equations using Gauss Elimination and Gauss-Jordan Methods'''. |
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| At the end of this tutorial, you will learn how to: | | At the end of this tutorial, you will learn how to: | ||
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| − | ||Press '''Enter'''. | + | ||Press '''Enter'''. The next prompt is for '''matrix b.''' |
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| − | |So we type | + | |So we type '''open square bracket four point seven two semi colon three point one semi colon two point nine one close square bracket.''' |
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| − | |So we type | + | |So we type '''open square bracket zero point seven comma one seven two five semi colon''' |
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Revision as of 11:24, 10 March 2017
| Time | Narration |
| 00:01 | Dear Friends, Welcome to the spoken tutorial on Solving System of Linear Equations using Gauss Elimination and Gauss-Jordan Methods. |
| 00:12 | At the end of this tutorial, you will learn how to: |
| 00:15 | Solve system of linear equations using Scilab |
| 00:20 | Develop Scilab code to solve linear equations. |
| 00:25 | To record this tutorial, I am using |
| 00:27 | Ubuntu 12.04 as the operating system |
| 00:31 | with Scilab 5.3.3 version. |
| 00:36 | To practice this tutorial, a learner should have basic knowledge of Scilab |
| 00:40 | and should know how to solve Linear Equations. |
| 00:45 | To learn Scilab, please refer to the relevant tutorials available on the Spoken Tutorial website. |
| 00:52 | A system of linear equations is a |
| 00:55 | finite collection of linear equations of the same set of variables. |
| 01:00 | Let us study Gauss elimination method. |
| 01:04 | Given a system of equations |
| 01:06 | A x equal to b |
| 01:08 | with m equations and |
| 01:10 | n unknowns. |
| 01:12 | We write the coefficients of the variables a one to a n |
| 01:16 | along with the constants b one to b n of the system of equations |
| 01:22 | in one matrix called the augmented matrix. |
| 01:27 | How do we convert the augmented matrix to an upper triangular form matrix? |
| 01:33 | We do so by performing row wise manipulation of the matrix. |
| 01:40 | Let us solve this system of equations using Gaussian elimination method. |
| 01:45 | Before we solve the system, let us go through the code for Gaussian elimination method. |
| 01:52 | The first line of the code is format e comma twenty. |
| 01:58 | This defines how many digits should be displayed in the answer. |
| 02:04 | The letter 'e' within single quotes denotes that the answer should be displayed in scientific notation. |
| 02:12 | The number twenty is the number of digits that should be displayed. |
| 02:17 | The command funcprot is used to let Scilab know what to do when variables are redefined. |
| 02:26 | The argument zero specifies that Scilab need not do anything if the variables are redefined. |
| 02:33 | Other arguments are used to issue warnings or errors if the variables are redefined. |
| 02:40 | Next we use the input function. |
| 02:43 | It will display a message to the user and get the values of A and b matrices. |
| 02:51 | The message should be placed within double quotes. |
| 02:55 | The matrices that the user enters, will be stored in the variables A and b. |
| 03:02 | Here A is the coefficient matrix and b is the right-hand-side matrix or the constants matrix. |
| 03:11 | Then we define the function naive gaussian elimination. |
| 03:15 | And we state that A and b are the arguments of the function naive gaussian elimination. |
| 03:22 | We store the output in variable x. |
| 03:27 | Then we find the size of matrices A and b using the size command. |
| 03:34 | Since they are two dimensional matrices, we use n and n one to store the size of matrix A. |
| 03:42 | Similarly we can use m one and p for matrix b. |
| 03:48 | Then we have to determine if the matrices are compatible with each other and |
| 03:53 | if A is a square matrix. |
| 03:57 | If n and n one are not equal then we display a message that Matrix A must be square. |
| 04:05 | If n and m one are not equal, we display a message |
| 04:10 | incompatible dimension of A and b. |
| 04:15 | If the matrices are compatible, we place matrices A and b in one matrix, C. |
| 04:23 | This matrix C is called augmented matrix. |
| 04:28 | The next block of code performs forward elimination. |
| 04:32 | This code converts the augmented matrix to upper triangular matrix form. |
| 04:39 | Finally, we perform back substitution. |
| 04:42 | Once the upper triangular matrix is obtained, we take the last row and find the value of the variable in that row. |
| 04:52 | Then once one variable is solved, we take this variable to solve the other variables. |
| 04:59 | Thus the system of linear equations is solved. |
| 05:03 | Let us save and execute the file. |
| 05:06 | Switch to Scilab console to solve the example. |
| 05:10 | On the console, we have a prompt to enter the value of the coefficient matrix. |
| 05:17 | So we enter the values of matrix A. |
| 05:20 | Type: square bracket three point four one space one point two three space minus one point zero nine semi colon |
| 05:33 | two point seven one space two point one four space one point two nine semi colon |
| 05:41 | one point eight nine space minus one point nine one space minus one point eight nine close square bracket. |
| 05:53 | Press Enter. The next prompt is for matrix b. |
| 05:57 | So we type open square bracket four point seven two semi colon three point one semi colon two point nine one close square bracket. |
| 06:10 | Press Enter. |
| 06:13 | Then we call the function by typing |
| 06:16 | naive gaussian elimination open parenthesis A comma b close parenthesis |
| 06:24 | Press Enter. |
| 06:26 | The solution to the system of linear equations is shown on Scilab console. |
| 06:32 | Next we shall study the Gauss-Jordan method. |
| 06:36 | In Gauss–Jordan Method, |
| 06:38 | the first step is to form the augmented matrix. |
| 06:42 | To do this, place the coefficient matrix A and the right hand side matrix b together in one matrix. |
| 06:50 | Then we perform row operations to convert matrix A to diagonal form. |
| 06:56 | In diagonal form, only the elements a i i are non-zero. Rest of the elements are zero. |
| 07:05 | Then we divide the diagonal element and corresponding element of right hand side element, by the diagonal element. |
| 07:14 | We do this to get diagonal element equal to one. |
| 07:19 | The resulting value of the elements of each row of the right hand side matrix gives the value of each variable. |
| 07:27 | Let us solve this example using Gauss-Jordan Method. |
| 07:33 | Let us look at the code first. |
| 07:36 | The first line of the code uses format function to specify the format of the displayed answers. |
| 07:44 | The parameter e specifies the answer should be in scientific notation. |
| 07:49 | Twenty (20) denotes that only twenty digits should be displayed. |
| 07:55 | Then we get the A and b matrix using the input function. |
| 08:00 | We define the function Gauss Jordan Elimination with input arguments A and b and output argument x. |
| 08:11 | We get the size of matrix A and store it in m and n. |
| 08:17 | Similarly, we get the size of matrix b and store it in r and s. |
| 08:23 | If the sizes of A and b are not compatible, we display an error on the console using error function. |
| 08:33 | Then we perform row operations to get diagonal form of the matrix. |
| 08:38 | Here pivot refers to the first non-zero element of a column. |
| 08:45 | Then we create a matrix of zeros called x with m rows and s columns. |
| 08:52 | Once we have the diagonal form, |
| 08:54 | we divide the right hand side part of augmented matrix by the corresponding diagonal element to get the value of each variable. |
| 09:04 | We store the value of each variable in x. |
| 09:08 | Then we return the value of x. |
| 09:11 | Finally, we end the function. |
| 09:13 | Now let us save and execute the function. |
| 09:18 | The prompt requires us to enter the value of matrix A. |
| 09:22 | So we type open square bracket zero point seven comma one seven two five semi colon |
| 09:31 | zero point four three five two comma minus five point four three three close square bracket. |
| 09:41 | Press Enter. |
| 09:43 | The next prompt is for vector b. |
| 09:45 | So we type open square bracket one seven three nine semi colon |
| 09:51 | three point two seven one close square bracket. |
| 09:55 | Press Enter. |
| 09:58 | Then we call the function by typing |
| 10:01 | Gauss Jordan Elimination open parenthesis A comma b close parenthesis |
| 10:08 | Press Enter. |
| 10:10 | The values of x one and x two are shown on the console. |
| 10:15 | Let us summarize this tutorial. |
| 10:18 | In this tutorial, we have learnt to: |
| 10:21 | Develop Scilab code for solving system of linear equations. |
| 10:25 | Find the value of the unknown variables of a system of linear equations. |
| 10:32 | Watch the video available at the link shown below. |
| 10:35 | It summarizes the Spoken Tutorial project. |
| 10:38 | If you do not have good bandwidth, you can download and watch it. |
| 10:43 | The spoken tutorial project Team: |
| 10:45 | Conducts workshops using spoken tutorials. |
| 10:48 | Gives certificates to those who pass an online test. |
| 10:52 | For more details, please write to conatct@spoken-tutorial.org. |
| 10:59 | Spoken Tutorial Project is a part of the Talk to a Teacher project. |
| 11:03 | It is supported by the National Mission on Eduction through ICT, MHRD, Government of India. |
| 11:10 | More information on this mission is available at http://spoken-tutorial.org/NMEICT-Intro. |
| 11:21 | This is Ashwini Patil, signing off. |
| 11:23 | Thank you for joining. |
