A Poorly Conditioned System. Matrix Form

Transcription

1 Possibilities for Linear Systems of Equations A Poorly Conditioned System A Poorly Conditioned System Results No solution (inconsistent) Unique solution (consistent) Infinite number of solutions (consistent) Graphical interpretation Equations determine lines, planes, hyperplanes Parallel inconsistent Single point intersection unique solution Congruence infinite number of solutions y solution y x solution y = m 2 x + b 2 y = m 1 x + b 1 Slopes m 1 m 2 x Previous solution y Large change in the solution for a small change of m 2 y = m 2 x + b 2 New solution y = m 1 x + b 1 x Example of Solving a System of Equations Matrix Form Strategy: Use Row Operations to Convert This: x x 2 = x x x 2 = x 3-9

2 To This: Which is the Same as: Row Operations x 1 x sol x 2 = x sol x 3 x sol3 Identity matrix x 1 = x sol1 x 2 = x sol2 x 3 = x sol3 Interchange two rows. Multiply all entries in a row by a nonzero constant. Replace a row by the sum of itself and a multiple of another row. Start With the Augmented Matrix: Original System and Augmented Matrix Side by Side, Ready for Row Operations Step 1: Add 4 Times Equation 1 to Equation 3, And Write This in Place of Equation 3 4 4x 1-8x 2 + 4x 3 =

3 This Yields: Step 2: Multiply Equation 2 by 1/2, Yielding: Step 3: Add 3 Times Equation 2 to Equation 3, And Write This in Place of Equation 3 x 2-4x 3 = x 2-4x 3 = Triangular form At this point, we have successfully solved for x 3. Step 4: Add 4 Times Equation 3 to Equation 2, And Write This in Place of Equation 2 Step 5: Subtract Equation 3 from Equation 1, And Write This in Place of Equation 1 Step 6: Add 2 Times Equation 2 to Equation 1 And Write This in Place of Equation 1 x 1-2x 2 = x 1 = At this point, we have successfully solved for x 2 and x 3. At this point, we have successfully solved for x 1, x 2, and x 3.

4 This Procedure Can be Formalized with the Following Definitions General Echelon Form of a Matrix apple * * * * * * * General Reduced Echelon Form of a Matrix 1 * 0 0 * * 0 * Matrix echelon form Matrix reduced echelon form Pivots 0 0 apple * * * * * apple * * * * apple * Here, the leading entries apple may have any nonzero value, and the * entries may have any values (including zero) * * 0 * * * 0 * * Here, the leading entries are 1 s, and the * entries may have any values (including zero). Note that there are 0 s below and above each leading 1. Uniqueness of the Reduced Echelon Form Using different sequences of row operations, any nonzero matrix can be transformed into more than one matrix in echelon form. However, the reduced echelon form that we obtain from a matrix can be proven to be unique. Pivot Positions apple apple * * * * * * * 0 0 apple * * * * * apple * * * * apple * Pivots apple are used in subsequent row operations to create above and below them in the process of transforming the matrix to the reduced echelon form. General Steps for Solving Systems by Row Reduction Step 1. Begin with the leftmost non-zero column. The required position is at the top. Step 2. Select a non-zero entry in this column as a. If necessary, interchange rows to move this entry into the position at the top of the column. Step 3. Use row operations to create in all column positions below the.

5 General Steps for Solving Systems by Row Reduction (continued) Step 4. Cover the s row and return to Step 1 for the sub-matrix that remains. Repeat as needed. Step 5. Beginning with the rightmost and working upward and to the left, use row operations to create in all column positions above each. Use scaling operations to make each equal to 1. What If the Reduced Echelon Form Looks Like: s x 1-5x 3 = 1 x 2 + x 3 = 4 0x 3 = Variables x 1 and x 2 corresponding to the columns are called basic variables; the rest (in this case x 3 ) are called free variables. We are free to choose any value of x Then, We Do Not Have a Unique Solution: x 1 = 1 + 5x 3 x 2 = 4 - x 3 x 3 = free We say that this system is a parametric description of a solution set in which the variable x 3 acts as a free parameter. In this and similar cases, there exists an infinity of solutions. What If the Reduced Echelon Form Looks Like: x 1-5x 3 = 1 x 2 + x 3 = 4 0x 3 = 2? This system has a built-in contradiction. There is no value ofx 3 that can satisfy the third equation. The original system has no solution, i.e., is inconsistent. 173