Determine the value of the parameters for which the syster 3sx₁ + 5x2 = 3 12x1 + 5sx2 = 2 has a unique solution, and describe the solution. The system has a unique solution when s # 2 and s‡ -2, 3s-2 3(s²-4) D-A = 2(s-6) 5(s²-4) and the unique solution is The system has a unique solution when s -2 and s # 2, s+1 s(s+2) and the unique solution is and the unique solution is x1 [x2] The system has a unique solution when s # -2 and s = 0, s+1 s(s+2) and the unique solution is = x1 []=[ [x2] x1 [x2] 1 2(s+2) The system has a unique solution when s -2 and s = 0, 38-2 3(s²-4) = 2(s+2) 2(s-6) 5(s²-4)

Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
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Solve this linear algebra multile choice question, please show your work.

Determine the value of the parameters for which the syster
3sx15x2 = 3
12x1 + 5sx2 = 2
has a unique solution, and describe the solution.
The system has a unique solution when s # 2 and s‡ -2,
3s-2
3(s²-4)
2(s-6)
5(s²-4)
-4-2
X1
=
x2
and the unique solution is
The system has a unique solution when s -2 and s # 2,
s+1
A-2
x1
s(s+2)
=
and the unique solution is
1
[x2]
2(s+2)
The system has a unique solution when s -2 and s = 0,
s+1
s(s+2)
and the unique solution is
x1
=
Q]-[
[x2]
and the unique solution is
The system has a unique solution when s -2 and s = 0,
3s-2
3(s²-4)
x1
[x2]
2(s+2)
=
2(s-6)
5(s²-4)
Transcribed Image Text:Determine the value of the parameters for which the syster 3sx15x2 = 3 12x1 + 5sx2 = 2 has a unique solution, and describe the solution. The system has a unique solution when s # 2 and s‡ -2, 3s-2 3(s²-4) 2(s-6) 5(s²-4) -4-2 X1 = x2 and the unique solution is The system has a unique solution when s -2 and s # 2, s+1 A-2 x1 s(s+2) = and the unique solution is 1 [x2] 2(s+2) The system has a unique solution when s -2 and s = 0, s+1 s(s+2) and the unique solution is x1 = Q]-[ [x2] and the unique solution is The system has a unique solution when s -2 and s = 0, 3s-2 3(s²-4) x1 [x2] 2(s+2) = 2(s-6) 5(s²-4)
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