 In this video, I explain the solution of simultaneous linear algebraic equation by Gauss elimination method. Learning outcomes, at the end of this session, the student will be able to solve the simultaneous algebraic equation by Gauss elimination method. Now, first the video and answer this question. And Gauss elimination method, the original equations are transform 2. Number one is Gauss column transformation. Number two, row transformation. Number three, that is C option is both column and row transformation and D option is the subset transformation. I hope you get the answer. That is, answer is for the row transformation. In the Gauss elimination method, we cannot use the column transformation. Because you have to use the column transformation, then the coefficient of the unknowns are changes and which is not equal to the A and B. Therefore, we cannot use the column transformation while you have to solve the system of equation by the Gauss elimination method. Therefore, the required answer is the row transformation. Now, example, solve the system of equation by Gauss elimination method. Now, that method, that is 5x plus y plus z plus w is equal to 4. x plus 7y plus z plus w is equal to 12. x plus y plus 6z plus w is equal to minus 5. x plus y plus z plus 4w is equal to minus 6. The given system of equations can be written in the matrix form is the x is equal to b. Where is the coefficient matrix, which is defined as a equal to matrix. First row contains the coefficient of the unknowns of first equation, that is 1. Where a is the coefficient matrix, a is equal to the first row contains the coefficient of first equation, that is 5, 1, 1, 1. Second row contains the coefficient of the second equation, 1, 7, 1, 1. Third row contains the coefficient of the third equation 1, 1, 6, 1. And fourth row contains the coefficient of the fourth equation 1, 1, 1, 4. And x is the unknown column matrix, that is x, y, z, w. Where b is equal to, that is a retention element of the given system of equation, that is 4, 12, minus 5, minus 6. The augmented matrix means you are appending or adding or inserting a column to a coefficient matrix, which consists of the element of the b. That is a new matrix so obtained, it is called augmented matrix, that is a, b, which is equal to matrix 5, 1, 1, 1, 4, 1, 7, 1, 1, 12, 1, 1, 6, 1, minus 5, 1, 1, 1, 4, minus 6. Now, we are interchanging the first and the fourth row, that is r 1 interchanges to r 2, that is equal to 1, 1, 1, 4, minus 6, 1, 7, 1, 1, 1, 1, 1, 1, 6, 1, minus 5, 5, 1, 1, 1, 4. Now, the aim of, is in the Gauss-Elmetson. The aim of the Gauss-Elmetson method is reducing the coefficient matrix to upper triangular matrix. These are the principle diagonals. Make the below principle diagonals are to be the 0 by using the row transformation only. The row transformation for the same is the r 2 equal to r 2 minus r 1, r 3 equal to r 3 minus r 1, r 4 is equal to r 4 minus 5 times of r 1. That becomes the change in r 2, r 3 and r 4. Write r 1 as it is. That is 1, 1, 1, 4, minus 6, 0, 6, 0, minus 3, 18, 0, 0, 5, minus 3, 1, 0, minus 4, minus 4, minus 9, that is 34. Now, next one, the r 2 is equal to r 2 by 3 because all the elements of the second row are exactly divided by the 3. Therefore, for our convenience, we have to divide by 3. That is r 2 is equal to r 2 by 3. That becomes changes in only r 2. Write the remaining rows as it is. That is 1, 1, 1, 4, minus 6, 0, 2, 0, minus 1, 6, 0, 0, 5, minus 3, 1, 0, minus 4, minus 4, minus 19, 34. Next, you have to make that the second element of the fourth row will become 0 by using the second element of the second row. For that, the transformation is r 4 is equal to r 4 plus 2 times of r 2. That is, apply the transformation only the r 4. Write r 1, r 2, r 3 as it is. That is, augmented matrix A B is equal to 1, 1, 1, 4, minus 6, 0, 2, 0, minus 1, 6, 0, 0, 5, minus 3, 1, 0, 0, minus 4, minus 21, 46. Again, you have to make the third element of the fourth row to become 0 by using the third element of the third row. The transformation for the same is r 4 is equal to 5 times of r 4 plus 4 times of r 3. Then, augmented matrix A B is equal to 1, 1, 1, 4, minus 6, 0, 2, 0, minus 1, 6, 0, 0, 5, minus 3, 1, 0, 0, 0, minus 1, 1, 7, 2, 34. The equivalent system of equations is that is x plus y plus z plus 4 w is equal to minus 6, 2 y minus w is equal to 6. Call it as equation 2. 5 z minus 3 w is equal to 1. Call it as equation 3 and minus 117 w is equal to 234. And solving this one, the w is equal to minus 2. Now, by the back substitution, the substitute in the w is equal to minus 2 in equation 2 and 3. 2 and 3, we will get z is equal to y is equal to 2 and further, we have to substitute in the value of y which is obtained from the equation 2 and z is obtained from equation 3. And then, substitute in the w value in the equation 1. We get the x is equal to x is equal to 1. Thus, x is equal to 1, y is equal to 2, z is equal to minus 1 and w is equal to minus 2 is the required solution of the given equation. Now, come to another example. Solve the system of equations by Gauss-Elmetson method that is 2x plus 2y plus z is equal to 1, 3x plus 2y plus 2z is equal to 8, 5x plus 10y minus 8z is equal to 10. The solution, the given system of equation can be written in the matrix form is ax is equal to bm. Where a is the coefficient matrix which is given by 2, 2, 1, 3, 2, 2, 5, 10 minus 8, x is the column unknown matrix that is x, y, z. b is the column matrix which is the constant that is the retention element of the given system of equation that is 12, 8 and 10. The argumentary matrix is that means adding or appending the column matrix which consists of the element of b. A new matrix is obtained and that is called the argumentary matrix which is defined as ab is equal to 2, 2, 1, 12, 3, 2, 2, 8, 5, 10 minus 8, 10. Now, then again same reducing the coefficient matrix to a upper triangle element means these element you have to make it to be 0. Means the second element of sorry first element of the second row and third row are becomes 0, the first one by using the first element of the first row for that the using the row transformation that is the R 2 equal to 2 times of R 2 minus 3 times of R 1 and R 3 equal to 2 R 3 minus 5 R 1 that becomes the 2, 1, 1, 12, 0 minus 2, 1, minus 20, 0, 10, minus 40. Now, further now make the second element of the third row becomes 0 by using the second element of second row is equal by using second element of second row that transformation is the R 3 equal to R 3 plus 5 times of R 2 the argumentary matrix ab is equal to 2, 2, 1, 12, 0, minus 2, 1, minus 20, 0, 0, minus 16, minus 140. From this matrix the equivalent system of equations are 2x plus 2y plus z is equal to 12, quality is equational minus 2y plus z is equal to minus 20, quality is equation 2 that is and third one is minus 16 z is equal to minus 140 that is from this z is equal to 140 divided by 16 that is the z is equal to 35 by 4. Now, the by back substitution the 2y is equal to z plus 20 that is z is equal to 35 by 4 and simplify then the y is equal to 115 divided by 8 and substitute the both value of y substituting the value of y is equal to 115 by 8 and z is equal to 35 by 4 in equation 1 and the simplify that is x is equal to minus 51 by 4. Thus x is equal to minus 51 by 4 y is equal to 115 by 8 and z is equal to 35 by 4 is the required solution of the given system of equation. Reference numerical methods by B. S. Grivall. Thank you.