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Solving Simultaneous Equations Using Row Reduction and a Matrix Approach Example Step 1: Identify the system of equations x+y+z=6 x–y+z=2 3x + y – z = 2 Step 2: Write in matrix notation x y z b (Create one column for each variable and one column 1 1 1 6 for the right hand sides—referred to as b.) Don’t worry 1 -1 1 2 about where the box around the 1 comes from at this 3 1 -1 2 point—that will come in Steps 3 and 4. Step 3: Select a column to work on (The object is to get Column 1 (Note: I could choose column 1, 2, or 3) a unit vector in each column, and to have a one in each row. Unit vectors will have one one in the column and everything else in that column will be zero.) Step 4: Select a row to make into the pivot row. Row 1 (Note: Although I could select any row, Rows 1 (With steps 3 and 4 you will have identified a specific or 2 look easiest since the entry in the selected column is entry—call this a.) already a 1.) Step 5: To get a one in that position, multiply every a=1 so 1/1 = 1 and the row remains unchanged. number in this row by 1/a 1 1 1 6 (Call the resulting row the pivot row.) Step 6: You want to make the other entries in that Working on the second row: column zero. To do this, work one row at a time. Select Multiply the pivot row by –1 gives: a row. Look at the current entry. Multiply each number -1 -1 -1 -6 in the pivot row by –1 times this entry and add the result Adding this to the current row 2 of: to the corresponding entry in the target row. Repeat this 1 -1 1 2 process until you have finished with the column. Results in the following new row 2: 0 -2 0 -4 Working on the third row: Multiply the pivot row by –3 and add: -3(1)+3 -3(1)+1 -3(1)+(-1) -3(6)+2 0 -2 -4 -16 At this point we have completed one pivot and we have: 1 1 1 6 0 -2 0 -4 0 -2 -4 -16 Step 7: If you have additional columns to transform into Working on the second column and selecting the second unit vectors, return to Step 3. row, make this entry a 1 by multiplying the row by –1/2. 0 1 0 2 Call this the pivot row. Now to make the second entry in the first row a 0, multiply the pivot row by –1 and add: -1(0)+1 -1(1)+1 -1(0)+1 -1(2)+6 1 0 1 4 To transform the third row, multiply the pivot row by 2 and add: 2(0)+0 2(1)+(-2) 2(0)+(-4) 2(2)+(-16) 0 0 -4 -12 At this point we have completed the second pivot and we have: 1 0 1 4 0 1 0 2 0 0 -4 -12 For the next pivot, we must get a one in the third row and third column. We do this by multiplying by –1/4. 0 0 1 3 Call this the pivot row. The second row already has a zero in the third column, so the only other work is to get a 0 in the first row third column. Multiply the pivot row by –1 and add: -1(0)+1 -1(0)+0 -1(1)+1 -1(3)+4 1 0 0 1 After three pivots we have: x y z 1 0 0 1 0 1 0 2 0 0 1 3 Step 8: Read off the answer x y z 1 0 0 1 0 1 0 2 0 0 1 3 Reading off the solution: x=1 y=2 z=3

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differential equation, quadratic equation, linear equation, equal sign, sides of an equation, partial differential equations, Solving Equations, linear equations, literal equation, partial differential equation

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posted: | 3/5/2011 |

language: | English |

pages: | 2 |

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