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# lecture_3 - MA 35100 LECTURE NOTES FRIDAY JANUARY 15...

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Unformatted text preview: MA 35100 LECTURE NOTES: FRIDAY, JANUARY 15 Elimination and Augmented Matrices In the previous lecture, we saw that a system of linear equations a 11 x + a 12 y + a 13 z = a 14 a 21 x + a 22 y + a 23 z = a 24 a 31 x + a 32 y + a 33 z = a 34 may be represented by the augmented matrix a 11 a 12 a 13 a 14 a 21 a 22 a 23 a 24 a 31 a 32 a 33 a 34 . To solve the system above, it’s helpful to work with the augmented matrix instead of the equations themselves. Consider the following system, as introduced at the end of the previous lecture: 2 x + 8 y + 4 z = 2 2 x + 5 y + z = 5 4 x + 10 y- z = 1 ⇐⇒ 2 8 4 2 2 5 1 5 4 10- 1 1 We will find a solution to the system. The idea is to perform operations on the rows of the augmented matrix, since it will be quicker than performing operations on the equations in the system. We want to eliminate x from the second and third equations. To this end, we divide the first equation by 2. This is equivalent to dividing the first row of the augmented matrix by 2: x + 4 y + 2 z = 1 2 x + 5 y + z = 5 4 x + 10 y- z = 1 ⇐⇒ 1 4 2 1 2 5 1 5 4 10- 1 1 Now subtract twice the first equation from the second, and then 4 times the first equation from the third: x + 4 y + 2 z = 1- 3 y- 3 z = 3- 6 y- 9 z =- 3 ⇐⇒ 1 4 2 1- 3- 3 3- 6- 9- 3 We want to eliminate y from the first and third equations. To this end, we divide the second equation by- 3: x + 4 y + 2 z = 1 y + z =- 1- 6 y- 9 z =- 3 ⇐⇒ 1 4 2 1 1 1- 1- 6- 9- 3 Now subtract 4 times the second equation from the first, and then add 6 times the second equation...
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lecture_3 - MA 35100 LECTURE NOTES FRIDAY JANUARY 15...

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