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Unformatted text preview: Hence the coeﬃcient determinant is non–zero and by Cramer’s rule, there
is a unique solution for a, b, c.
8. Let ∆ = det A = ∆= 1 1 −1
23
k . Then
1k
3 C3 → C 3 + C 1
C2 → C 2 − C 1 1
0
0
2
1
k+2
1 k−1
4 = 1
k+2
k−1
4 = 4 − (k − 1)(k + 2) = −(k 2 − k − 6) = −(k + 3)(k − 2).
Hence det A = 0 if and only if k = −3 or k = 2.
Consequently if k = −3 and k = 2, then det A = 0 and the given system
x+y−z 2x + 3y + kz = 1 = 3 x + ky + 3z = 2
has a unique solution. We consider the cases k = −3 and k = 2 separately.
k = −3 : 1
1 −1 1
1
1 −1 1
R2 → R2 − 2R1 3 −3 3 0
1 −1 1 AM = 2
R3 → R3 − R1
1 −3
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0 −4
41
R3 → R3 + 4R2 1 1 −1 1 0 1 −1 1 ,
00
05 from which we read oﬀ inconsistency.
k=2: 1 1 −1 1
1 1 −1 1
R → R2 − 2R1 2 3 2
01
4 1
AM = 2 3
R3 → R3 − R1
12
32
01
41
R3 → R3 − R2 1 0 −5 0
0 1
4 1 .
00
00 We read oﬀ the complete solution x = 5z, y = 1 − 4z , where z is arbitrary.
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 Fall '10
 Dreibelbis

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