Solving Sstems of Linear Equations in Matri Form with rref Learning Goals Determine the solution of a sstem of equations from the augmented matri Determine the reduced row echelon form of the augmented matri, rref Kewords: sstem of linear equations, matri form or sstem of linear equations, augmented matri, rref Eample : Solve the sstem of linear equations in the two variables and 4 4 4 The matri form of this equation, AX B, the matri of coefficients times the vector of variables set equal to the right hand side vector of constants, is written as 4 2 4 4 The matri form, the vector form, and the augmented matri form, respectivel, for the sstem of equations are 4 4 4 2 4 4 2 4 4 2 4 4 Matri form Vector form of equations Augmented form The augmented matri will be entered into the MATLAB and the rref command will be used to help find the solution. The rref command computes the reduced row echelon form of a matri. We will store our augmented matri into the 2 3 (two rows and three columns) matri A. The MATLAB commands to complete the problem with the associated MATLAB results follow. A=[ - 4;2 4 4] A = - 4 2 4 4 rref(a) ans = 0 5 0 The reduced row echelon form of the augmented matri means that the original problem has been transformed to a sstem of equations where the solution is eas to determine. The rref
version of the augmented matri has 2 pivots and no free variables so the sstem of equations will have a unique solution. The equation form of the rref result is: Solution tpe: Unique solution; solution: 5,. The solution of the sstem of equations is the ordered pair 5,. The equations will both be true at the same time if 5 and. Graphicall, this means that the two equations represent straight lines that intersect at eactl one point. The solution, in vector form, is 5 Eample 2: Solve the sstem of linear equations in the two variables and 5 3 3 2 To put the sstem of equations into matri form and then the augmented matri form we need to rewrite the sstem so that the s and the s are lined up on one side of the equation and the constant part is on the other side. 3 0 3 2 The matri, vector, and augmented matri forms, respectivel, of the equations are 3 0 3 0 3 0 2 3 2 2 3 2 2 3 2 Matri form Vector form of equations Augmented form Enter the augmented matri into the MATLAB and use the rref command to help find the solution of the sstem of equations. MATLAB displas 0.0909 0 3.2727 Again: 2 pivots and no free variables so the sstem of equations has a unique solution.
Solution tpe: Unique solution; solution:.0909, 3.2727. 2 36 The eact solutions are and. To displa the answer in fraction form (as rational numbers) enter format rat, then redo the rref work. The solution is the ordered pair, 2 36,. Graphicall, the equations represent two straight lines that intersect at eactl one point, a unique solution. Eample 3: Solve the sstem of linear equations in the two variables and 4 6 3 6 The matri and augmented matri forms of the equations are 2 4 2 4 6 3 6 6 3 6 Matri form Augmented form Enter the augmented matri into the MATLAB and use the rref command to help find the solution of the sstem of equations. MATLAB displas -.5 0 0 0 The rref version of the augmented matri has pivot and free variable so the sstem of equations will not have a unique solution. From the reduced row echelon form of the augmented matri, our original problem has been transformed to one of the form 0.5 0 0 Solution tpe: Contradiction; solution: No Solution or inconsistent. There are no pairs of, that satisf both equations at the same time. Graphicall, the equations represent parallel lines with no points in common. Eample 4: Solve the sstem of linear equations in the two variables and 4 2 2 8 4 The matri and augmented matri forms of the equations are
4 2 4 2 8 2 4 8 2 4 Enter the augmented matri into the MATLAB and use the rref command to help find the solution of the sstem of equations. MATLAB displas -0.25 0.5 0 0 0 The transformed equations are: 0.25 0.5 0 0 The rref result indicates pivot and free variable so no unique solution. Solution tpe: Consistent and dependent or Identit; solution: 0.250.5, ( is an real number). Alternatel, one ma write: 0.25t 0.5, t or, 0.25 tt,. Some sample solutions:, 2, 2, 0. Vector form of solution. The vector form of the solutions is determined from the transformed equations that are written using the rref version of the augmented matri. The regular variables are solved in terms of the free variables from the transformed equations and the free variables set equal to themselves. The vector form of the solution is: 0.25 0.5 Parametric form of solution. The general solution of AX B is X vh p, where v h is the solution of the homogeneous problem, AX 0, and p is the particular part of the solution due to a nonzero right hand side vector, B. The parametric form of the solution to the linear sstem of equations AX B with the parameter t is: 0.25 0.5 0.25 0.5 t, t ; then : vh t, p 0 0 Eample 5: Three equations with three unknowns. Solve the sstem of linear equations in the three variables,, and z. 4z 4 4 z 4 4 3z 8 The matri and vector forms of the equations and the augmented matri of the equations are
4 4 4 4 4 4 2 4 4 2 4 z 4 2 4 4 4 0 3 z 8 4 0 3 8 4 0 3 8 Matri form Vector form of equations Augmented form The augmented matri is entered into MATLAB and the rref command applied to determine the solution of the sstem of equations. Store the augmented matri into the 3 4 (three rows and four columns) matri A using the EDIT option of the MATLAB MATRIX menus. MATLAB commands and results are A=[ - 4 4;2 4-4;4 0 3 8] A = - 4 4 2 4-4 4 0 3 8 rref(a) ans =.0000 0 0 0.743 0.0000 0 3.574 0 0.0000.743 The rref version of the augmented matri has 3 pivots with no free variables so the sstem of equations will have a unique solution. Solution tpe: Unique solution. Solution: 0.74285, 3.5742, z.7428. Vector form of the solution:.74285... 3.5742... z.7428... The solution of the sstem of equations is the ordered triple 0.74285, 3.5742,.7428, that is, the equations will all be true at the same time if 0.74285, 3.5742, and z.7428. Geometricall the three equations represent three planes that intersect at eactl one point. The general solution of AX B is X vh p, so the parametric form of the solution to the linear sstem above is: 0 0.74285 0.74285 0 0.74285 t 0 3.5742 3.5742 ; with : v h 0, p 3.5742 z 0.7428.7428 0.7428 Note: Here the homogeneous part of the solution is the zero vector since AX B has a unique solution for an right hand side. This also means that AX 0 has onl the zero solution X 0. Thus the particular solution is the unique solution of the problem for right hand side B.
Solve the following sstems of equations using augmented matrices and the rref command in MATLAB. Determine the augmented matri, the equivalent form as transformed b rref, write down the new transformed sstem of equations from the rref result, and the solution, if an. Also write out the vector form of the equations and the solution in parametric form. Identif the particular, p, and homogeneous, v h, parts of the vector form. Indicate the number of pivots and circle the pivot columns of the augmented matri.. Determine: augmented matri, rref equivalent form, and the solution, if an. 3 3 Number of pivots Free variables Solution tpe: Solution: 2. Determine: augmented matri, rref equivalent form, and the solution, if an. 0520 34 Number of pivots Free variables Solution tpe: Solution:
3. Determine: augmented matri, rref equivalent form, and the solution, if an. 463 33 Number of pivots Free variables Solution tpe: Solution: 4. Determine: augmented matri, rref equivalent form, and the solution, if an. z 4 46z 6 32z 2 Number of pivots Free variables Solution tpe: Solution: z
5. Determine: augmented matri, rref equivalent form, and the solution, if an. z 4 4z 5 4z 5 Number of pivots Free variables Solution tpe: Solution: z 6. Determine: augmented matri, rref equivalent form, and the solution, if an. 5 2 3 2 8 3 2 3 52 3 4 Number of pivots Free variables Solution tpe: Vector form of solution: p =, v,
7. Determine: augmented matri, rref equivalent form, and the solution, if an. 3 2 2 3 4 2 3 4 32 344 6 Number of pivots Free variables Solution tpe: Solution: