 In this video, I explain the solution of simultaneous algebraic equation by LU-decomposition method. Learning outcomes. At the end of this session, the student will be able to solve the simultaneous algebraic equation by LU-decomposition method. A-decomposition is also called Kraut's method. This is a direct method. Consider the system of equations 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. Call it equation 1. given system of equation can be put in the matrix form as x is equal to b, where a is the coefficient matrix, x is the column unknown matrix, b is the constant column matrix here. That is a is equal to a 1, b 1, c 1, a 2, b 2, c 2, a 3, b 3, c 3 and x is the column unknown matrix that is x, y, z and b is equal to the constant column matrix that is d 1, d 2, d 3. In this method the coefficient matrix A is decomposed or factorized into the product of the two matrix L, U, which are the lower and upper triangular matrix respectively. The diagonal element of either L or U can be unity for a convenience. Let the coefficient matrix is A is equal to a 1, b 1, c 1, a 2, b 2, c 2, a 3, b 3, c 3. We express this in the form A is equal to L U call it is equation 3 here, where A is the coefficient matrix a 1, b 1, c 1, a 2, b 2, c 2, a 3, b 3, c 3 which is equal to L lower triangular matrix here means all the above diagonals are become 0 and defined as L 1 1, 0 0, L 2 1, L 2 2, 0, L 3 1, L 3 2, L 3 3 into upper triangular matrix means the all the below diagonals are become 0 and here all the diagonal elements are to be considered as a unity that is 1, U 1 2, U 1 3, 0 1, U 2 3, 0 0 1. Now, taking the product of the lower and upper triangular matrix and the coefficient matrix A become the first true becomes L 1 1, L 1 1, U 1 2, L 1 1, U 1 3. The second rule becomes L 2 1, L 2 1, U 1 2 plus L 2 2, L 2 1 into U 1 3 plus L 2 2 into U 2 3 and third rule becomes L 3 1, L 3 1 into U 1 2 plus L 3 2, L 3 1 into U 1 3 plus L 3 2 into U 2 3 plus L 3 3. Equating the corresponding elements on the both side we obtain the value of L and U using equation 3 into we get that is L U into x is equal to be L into U x is equal to be call it is equation 4. Now, you have to put U x is equal to y that is call it is equation 5 y is the column unknown matrix that is y 1 y 2 y 3. Now, equation 4 becomes L into y is equal to be call it is equation 6 here. Now, using equation 6 y can be found substituting y in equation 5 we get U x is equal to y we obtain the required x here being the exact the solution. First the video find the product of the matrix A x is equal to 0 where A is equal to 2 3 0 6 x is equal to column matrix that is x y. I hope all of you will get the result that solution A x is equal to 0 that is A is the matrix of order 2 2 3 0 6 x is the column matrix x y is equal to 0 0. Now, taking the product of the matrix 2 x plus 3 y and 6 y is equal to 0 0 equating the corresponding elements we get 2 x plus 3 y is equal to 0 6 y is equal to 0 implies y is equal to 0 by back substituting the x is also is equal to 0. Now, come to an example use the L U decomposition method solve the system of equation 2 x plus y plus 4 z is equal to 12 4 x plus 11 y minus z is equal to 33 8 x minus 3 y plus 2 z is equal to 20 solution. The given system of equation can be written in the matrix form as A x is equal to b where A is the coefficient matrix it differed as the A is equal to 1 4 4 11 minus 1 8 minus 3 2 and x is the column matrix that is x y z and b is the column matrix that is 12 33 and 20. In this method the coefficient matrix A is decomposed or factorized into the product of the 2 matrices L U which are the lower and upper triangular matrices respectively that L U is equal to A L is lower triangular matrix it differed as L 11 0 0 L 2 1 L 2 2 0 L 3 1 L 3 2 L 3 3 in 2 U is the upper triangular matrix that is 1 U 1 2 U 1 3 0 1 U 2 3 0 0 1 is equal to the matrix A there is 2 1 4 1 2 2 2 2 2 2 2 2 2 2 4 11 minus 1 8 minus 3 2. Now, I am taking the product of the lower and upper triangular matrix we get the matrix as L 11 L 11 into U 1 2 L 11 U 1 3 the second 2 becomes L 2 1 L 2 1 into U 1 2 plus L 2 2 L 2 1 into U 1 3 plus L 2 2 into U 2 3 the third row becomes L 3 1 L 3 1 into U 1 2 plus L 3 2 L 3 1 into U 1 3 plus L 3 2 into U 2 3 plus L 3 3 is equal to 2 1 4 4 11 minus 1 8 minus 3 2 equities the corresponding coefficient on both side that is L 11 is equal to 2 that and L 11 into U 1 2 is equal to 1 means substitute the value of L 11 and simplify the U 1 2 is equal to 1 by 2 that is L 11 into U 1 3 is equal to 4 implies U 1 3 is equal to 2 L 2 1 is equal to 4 L 2 1 into U 1 3 is equal to 4 L 2 1 into U 1 2 plus L 2 2 that is substitute in the value of L 2 1 and U 1 2 and solving that is L 2 2 is equal to 9 L L 2 1 into U 1 3 plus L 2 2 into U 2 3 is equal to minus 1 on solving this that is implies U 2 3 is equal to minus 1 L 3 1 is equal to 8 L 3 1 into U 1 2 plus L 3 2 simplify it becomes the L 3 2 is equal to minus 7 L 3 1 into U 1 3 plus L 3 2 into U 2 3 plus L 3 3 is equal to 2 this implies L 3 3 is equal to minus 21 consider A x is equal to B from equation 1 but L is equal to U x from equation 1 to that is L U into x is equal to B that can be written as L into U x is equal to B call it is equation 3 put U x is equal to y call it is equation 4 where y is equal to the column unknown matrix y 1 y 2 y 3 now from equation 3 L y is equal to B L is the lower triangular matrix that is 2 0 0 4 9 0 8 minus 7 minus 21 into y y 1 is equal to minus 7 minus 7 y 2 y 3 and B 12 33 20 and taking the product on the left hand side it becomes the 2 y 1 4 y 1 plus 9 y 2 8 y 1 minus 7 y 2 minus 21 y 3 is equal to 12 33 20 equating the corresponding elements on both sides we get 2 y 1 is equal to 12 this implies y 1 is equal to 6 and 4 y 1 plus 9 y 2 is equal to 33 the implies y 2 is equal to 1 8 y 1 minus 7 y 2 minus 21 y 3 is equal to 20 implies y is equal to 1 substituting the values of y then equation 4 becomes U x is equal to y U is the upper triangular matrix and x is the column unknown matrix that is 1 1 by 2 2 0 1 minus 1 0 0 1 into x y z is equal to 6 1 1 now taking the product on the L U and x that is x plus 1 by 2 y plus 2 z y minus z z is equal to 6 1 1 equating the corresponding elements on both sides we get x plus 1 by 2 into y plus 2 z is equal to 6 y minus z is equal to 1 and z is equal to 1 by the back substituting we get z is equal to 1 y is equal to 2 and x is equal to 3 thus x is equal to 2 and z is equal to 3 y is equal to 2 z is equal to 1 reference numerical methods by B. S. Grebel thank you.