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Proof. By definition The first and the second determinants are zero because they have identical rows. Thus, Because y , y , y are solutions of the homogeneous portion of the equation, we have
1 2 3 1 2 3 1 2 3 Therefore
1 2 3 1 2 3 1 1 Multiplying Row1 by 1 2 2 and Row2 by 2 3 3 and add them to Row3, we have
1 1 1 1 Thus,
1 Q33 (7.5 Points): Use Substitution Method and Operator Method to solve the following
system of equations (3.75 Points for each method):
2
3
2 Solution:
3 3 (1) Substitution Method. With
2
The characteristic equation is
3
We have 2 , we have
2 3
4 0. 1
2 2
1, 4 . 4 3 2 4 3
2 (2) Operator Method. We have 1
2
0 1
. We use
2
1
3
2
0 2
3
4
0. The characteristic equation is
3
4 0. 1,
4. We have
. 2 2, and we have and Therefore, we have the same solutions as found earlier 3 2 4...
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 Spring '08
 Staff
 Derivative, Elementary algebra, Characteristic polynomial, Homogeneity

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