In-Class Demonstration for MATH 544, Section 1.5Douglas B. MeadeUniversity of South CarolinaJanuary 24, 2011restart;with( LinearAlgebra ):libname := "C://Documents and Settings\134\134meade\134\134Desktop\134\134laylinalg", libname;with( laylinalg );c1s5();Example #1: Exercise 12 (p. 55)c1s5( 12 );This matrix is already in echelon form. From this we can see that the homogeneous equation has 3 free variables (3 columns without a pivot). To find these solutions, it's best to do the row operations to put this matrix in reduced echelon form.Unlike what I did in class, to be consistent, let's add the right-hand side of zeros:zero := < 0, 0, 0, 0 >;M := augment( A, zero );M1 := replace( M, 2, 8, 3 );M2 := replace( M1, 1, -2, 2 );Now it is very simple to write down the general solution to this (homogeneous) system: x1 = -5x2 -8x4 - x5 x2 = x2 x3 = 7x4 - 4x5 x4 = x4 x5 = x5 x6 = 0or, as a linear combination of vectors with the free variables as coefficients: [ x1 ] [ -5 ] [ -8 ] [ -1 ] [ x2 ] [ 1 ] [ 0 ] [ 0 ] [ x3 ] = [ 0 ] x2 + [ 7 ] x4 + [ -4 ] x5 [ x4 ] [ 0 ] [ 1 ] [ 0 ] [ x5 ] [ 0 ] [ 0 ] [ 1 ] [ x6 ] [ 0 ] [ 0 ] [ 0 ]The importance of this fact is that you now know that if you have a solution to a consistent nonhomogeneous problem with this same coefficient matrix, you can add any linear combination of the three vectors found above and still have a solution to the same nonhomogeneous problem. Another way to say this is that you and I get different solutions to the same problem with this coefficient matrix, we would expect that our solutions differ by a vector that is a solution to the homogeneous problem (and so can be written as found above).TTdSMApJNlJUQUJMRV9TQVZFLzEwNzQwNzA0NFgqJSlhbnl0aGluZ0c2IkYlW2dsISMlISEhIiUiJSIiIUYmRiZGJkYl