Section Gaussian Elimination


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1 Section.  Gaussian Elimination A matrix is said to be in row echelon form (REF) if it has the following properties:. The first nonzero entry in any row is a. We call this a leading one or pivot one.. The leading one in any successive row must be to the right of the leading one in the previous row. This gives the matrix an "echelon" form.. Any rows consisting entirely of zeros must be grouped together at the bottom of the matrix. A matrix is said to be in reduced row echelon form (RREF) if it is in REF and each column that contains a leading one has zeros everywhere else in that column. Determine whether matrices are in REF only, RREF, or not REF The location of leading ones in any REF for a given matrix will always be located in the same position. For augmented matrices, variables that correspond to the leading ones in any REF are called leading variables or fixed variables. The remaining variables are called free variables. When we have free variables for a consistent system of linear equations, then that system will have infinitely many solutions. In this case, we will express the solution using a set of parametric equations. Here is a REF for the augmented matrix of a system of linear equations. State which variables are fixed and which are free
2 In each case, a REF for the augmented matrix of a system of linear equations is given. Give the number of equations, the number of unknowns, and the solution (if any). Parameterize solution as needed. # of equations # unknowns Solution # of equations # unknowns Solution # of equations # unknowns Solution # of equations # unknowns Solution
3 In each case, write out a possible RREF of the augmented matrix for the system described: 5. Three equations and unknowns with solution x 4, x, x Three equations and unknowns with solution x t, x 5t, x t. 7. Three equations and unknowns with no solution. 8. Four equations and unknowns with solution x, x. 9. A homogeneous system of equations in unknowns that has infinitely many solutions.
4 We will now show how to get any augmented matrix into a REF using a stepbystep elimination method that uses only elementary row operations. Recall the elementary row operations:. swap any two rows.. multiply row by a nonzero constant.. Add a multiple of one row to another. The method we will show that reduces an augmented matrix to a REF is called Gaussian elimination (in honor of the great German mathematician Carl Friedrich Gauss). This procedure (or algorithm) consists of what is referred to as the forward phase in which zeros are introduced below the leading ones. If we continue this algorithm and add the backward phase (in which zeros are introduced above the leading ones), we can obtain the RREF of the augmented matrix. The procedure that uses both forward and backward phases to put the augmented matrix into its RREF is called GaussJordan elimination (to also honor the German engineer Wilhelm Jordan). When solving with Gaussian elimination, the solution is obtained by writing out the system of equations obtained from the REF of augmented matrix and then using back substitution. Here is the idea of the algorithm: Step : Get a leading one in the upper right corner by either interchanging rows if this entry is zero and/or multiplying the row by the reciprocal of this entry. Step : Get zeros in the rows beneath this leading (or pivot) one by adding appropriate multiples of first row to the row with the entry you want to zero out. The multiple is the opposite of the entry you want to zero out. 0 0 Step : Shift down to the second row second column. If there is a zero here, look to see if any entry beneath it is nonzero. If not move to the next column in the second row and repeat this step. If there is a nonzero entry, interchange rows to get the nonzero entry in the second row. Multiply the second row by the reciprocal of this nonzero entry to get another leading one. 0 0 Step 4: Get zeros in the rows beneath this leading (or pivot) one by adding appropriate multiples of first row to the row with the entry you want to zero out. The multiple is the opposite of the entry you want to zero out Continue this process until you have it in a REF. By adding multiples of rows and using the pivot ones, you can now continue to the RREF. 4
5 Solve the system shown to the left below using Gaussian Elimination with back substitution & then with GaussJordan Elimination. x x x 5 5x x x 0 x x x 6x 4x x The system of equations yields the augmented matrix shown above to the right of the system. We first will show Gaussian Elimination with back substitution. We proceed by using the elementary row operations to get a REF. R R R R 5R R R R R 4 R 4 6R R R R R R R R 7 The resulting system from this is and we have x x x 5 x x 75 x 7 This is now in REF form x 5 x x x 75 x so that by back substitution we have x 7 x 75 7 x
6 We could have also obtained solution directly had we solved using the GaussJordan elimination. This requires transforming the REF form above into RREF R R R 0 75 R R 7R R R R This is now in RREF form The corresponding system to the RREF gives the same solution directly. x 7 x x 4 6
7 Solve the following system using GaussJordan elimination 5x 0x x x 4 5x 5 0 x 4x x 4 7x 5 7 x x x 5 7
8 Recall that a homogeneous system is one in which all the constant terms are zero. That is, if the system is of the form a x a x a x a n x n 0 a x a x a x a n x n 0 a x a x a x a n x n 0 a m x a m x a m x a mn x n 0 Every homogeneous system is consistent because all such systems have x 0, x 0,, x n 0 as a solution. This solution is called the trivial solution. If there are other solutions to the system than they are called nontrivial solutions. Examples of homogeneous systems: x 4x x 0 5x 0x 4x x 4 0 x 5x x 0 6x 9x x 4 0 6x x 7x 0 7x x 8x 0 Since every homogeneous system is consistent, there are only two possibilities for its solution:. The system has only the trivial solution.. The system has infinitely many solutions in addition to the trivial solution. When performing row operations on matrices, columns of zeros are not altered. Thus, any REF of a homogeneous system will have a last column of all zeros. Theorem: If a homogeneous linear system has n unknowns (variables), and if the RREF of its augmented matrix has r nonzero rows (i.e. rows with a leading one), then the system has n r free variables. Theorem: A homogeneous linear system with fewer equations than unknowns has infinitely many solutions. Note that this last theorem does NOT apply to nonhomogeneous linear systems as such a system may not be consistent. However, it can be proved that a nonhomogeneous linear systems with fewer equations than unknowns that is consistent has infinitely many solutions. 8
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