Linear Algebra Solutions 43

Linear Algebra Solutions 43 - Hence the coefficient...

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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 32 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. 46 ...
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