 Namaste. Myself, Dr. Basaraj Emberadhar, Valchain Institute of Technology, Solopo. In this video, I explain solutions 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. Solution of simultaneous linear algebraic equation. Consider the system of equation a1x plus b1y plus c1z is equal to d1, a2x plus b2y plus c2z is equal to d2, a3x plus b3y plus c3z is equal to d3. The given system of equations can be put in the matrix form as ax is equal to b, where ax is the coefficient matrix, ax is the column matrix and b is the constant matrix that is the retinus and elements. The numerical solution of simultaneous linear algebraic equation can be obtained by the two different methods. One is the direct method which gives the exact solution and second one is the indirect method or iterative method which gives the approximation method. In the direct method, we are going to discuss the three different methods namely the Gauss-Elimination method, Gauss-Jordan method, LU decomposition or it is also called the Kraut's method whereas in the second type of the method that is indirect method, we are going to discussing the two different methods that is Jacobi's iterative method and Gauss-Seidel method. Gauss-Elimination method. This is an direct method means which gives the exact solution of the given equation. I explain this method by considering the three equations in the three unknowns. Consider the system of equations a1x plus b1y plus a1z is equal to d1, a2x plus b2y plus c2z is equal to d2, a3x plus b3y plus c3z is equal to d3. The given system of equations can be put in the matrix form as ax is equal to b, call it as equation 2 where a is the coefficient matrix, x is the column unknown matrix, b is the corner straight matrix that is the right hand side element that is a is equal to the first equation contains the coefficient of the unknowns of the first equation that is a1 b1c1, second equation contains the coefficient of the second equation e2b2c2 and third equation contains the coefficient of the third equation e3b3c3 and x is the unknown column matrix that is x, y, z and b is equal to column corner straight matrix that is d1, d2, d3. Now, from the argumentary matrix, now first I will know that what is the meaning of argumentary matrix inserting or adding a column which consists of the element of b to a coefficient matrix, a new matrix is so obtained it is called the argumentary matrix and it is denoted by ab which is equal to a1 b1 c1 d1 a2 b2 c2 d2 a3 b3 c3 d3. The method aims is reducing the coefficient matrix a to an upper triangular matrix means what is the meaning of upper triangular, upper triangular matrix means making the below that principle diagonals elements are to become as in 0. Now, in the first step here the use the element a1 means the first element of first true and the first column which is not equal to 0 make the element a2 and a3 0 by elementary row transformation. Applying the appropriate row transformation, argumentary matrix ab is equal to a1 b1 c1 d1 0 b2 c2 d2 0 b3 c3 d3 and step 2 use the element b2 dash which is not equal to 0 to make the element b3 0 by elementary row transformation that the argumentary matrix ab equal to a1 b1 c1 d1 0 b2 dash c2 dash d dash 0 0 c3 double dash d3 double dash called it is equation number 4. From equations for the equivalent system of equations are a1 x plus b1 y plus c1 z is equal to d1 b2 dash y plus c2 dash z is equal to d2 c3 double dash z is equal to d3 double dash. Now, from the equation 7 the z is equal to d3 double dash upon c3 double dash by cross multiplication. Now, step 3 by back substitution we get the value of z and substitute the value of z obtained from the 7 in equation 6 we get the value of y and then substitute both the value of y and z equation 5 you will use the exact solution of equation 1. Note in case a1 and b2 dash r0 we only need to rearrange the equations. First, you do reduce the following matrix into upper triangular matrix I hope you may get the result. Solution upper triangular means making the below the diagonal r to be 0 means these are the diagonal elements make these three elements r to become 0. It means using the row transformation only that is r2 is equal to r2 minus 2 times of r1 r3 is equal to r3 minus r1 that is changes applied in second row and third row write the first row as it is 1 1 0 minus 1 minus 3 0 minus 2 1 and r3 r3 means again these elements are to become 0 again you have to make it 0 for that you have to apply the transformation on the r3 that is r3 is equal to r3 minus 2 times of r2 that is 1 a equivalent to 1 1 1 0 minus 1 minus 3 0 0 7 which is an upper triangular matrix. Now, come to an example solve the system of equation by Gauss-Elmetsche method x plus y plus z is equal to 9 x minus 2 y plus 3 z is equal to 8 2 x plus y minus z is equal to 3 solution the given system of equation can be written in the matrix form as x is equal to b where a is the coefficient matrix that is a is equal to 1 1 1 1 minus 2 3 2 1 minus 1 and x is the column unknown matrix that is x y z and b is the column constant matrix that is retention element of the given system of equation that is 983 argumentary matrix form the argumentary matrix first of all you have to insert in the element of b to a column of the coefficient matrix a a new matrix. So, obtain it is called argumentary matrix a b that is a b is equal to matrix 1 1 1 9 1 minus 2 3 8 2 1 minus 1 3. Now, to make the coefficient matrix becomes the upper triangular means we have to make the these 3 elements are becomes 0. Now, first of all we make the second a first element of the second and the third row or becomes 0 by using the first element of the first row that is r 2 is equal to r 2 minus r 1 r 3 is equal to r 3 minus 2 times of r 1 that is argumentary matrix a b equal to 1 1 1 9 0 minus 3 2 minus 1 0 minus 1 minus 3 minus 15. Now, further now to make this is equal to the means here the second element of the third row by using the second element of the second row. Now, the transformation for the same is then the argumentary matrix a b equal to 1 1 1 9 0 minus 3 2 minus 1 0 0 11 44 from this matrix the equivalent system of equations are x plus y plus z is equal to that call it is equation minus 3 y plus 2 z is equal to minus 1 call it is equation 2 and from the third row that is 11 z is equal to 44 this implies the z is equal to 44 upon 11 that is by cross multiplication that is a z is equal to 4 now by back substitution. Now, the equation 2 is equal to implies minus 3 y plus 2 into 4 because z is equal to 4 is equal to minus 1 on simplifying this y is equal to 3 and substitute in the y is equal to 3 and z is equal to 4 in equation 1 it gives x plus 3 plus 4 is equal to 9 implies x is equal to 2 thus the x is equal to 2 y is equal to 3 and z is equal to 4 is the required solution of equation 1 references numerical methods by B. S. Graevaugh thank you.