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UB-21 3. Solve the following system of equations using the method of Gauss elimination: x₁ + x2 + x3 = 6 x₁ – x₂ + 2x3 = 1 3x₁ + x₂ + 4x3 = 19 Answer: To solve the given system of equations using the method of Gauss elimination, we begin by writing the corresponding augmented matrix:
[ left[egin{array}{cccc} 1 & -1 & 2 & 1 \ 1 & 1 & 1 & 6 \ 3 & 1 & 4 & 19 end{array}
ight] ]
First, we perform row operations to eliminate the first variable from all rows except the first one. We subtract the first row from the second row and subtract three times the first row from the third row:
[ left[egin{array}{cccc} 1 & 1 & 1 & 6 \ 1 & -1 & 2 & 1 \ 3 & 1 & 4 & 19 end}
ight] -left[egin{array}{cccc} 1 & 1 & 1 & 6 \ 1 - 1 & -1 - 1 & 2 - 1 & 1 - 6 \ 3 - 3 & 1 - 3 & 4 - 3 & 19 - 19 end}
ight] ]
This gives:
[ left[egin{array}{cccc} 1 & 1 & 1 & 6 \ 0 & -2 & 1 & -5 \ 0 & -2 & 1 & -5 end}
ight] ]
Next, we eliminate the third variable from the second and third rows. We subtract the second row from the third row:
[ left[egin{array}{cccc} 1 & 1 & 1 & 6 \ 0 & -2 & 1 & -5 \ 0 & -2 & 1 & -5 end}
ight] -left[egin{array}{cccc} 0 & -2 & 1 & -5 \ 0 & -2 - (-2) & 1 - 1 & -5 - (-5) \ 0 - 0 & -2 - (-2) & 1 - 1 & -5 - (-5) end}
ight] ]
This gives:
[ left[egin{array}{cccc} 1 & 1 & 1 & 6 \ 0 & -2 & 1 & -5 \ 0 & 0 & 0 & 0 end}
ight] ]
Now, we solve for x₁ and x₂. The third row indicates that 0 = 0, which is always true, so we can ignore it. We are left with the following system of equations:
x₁ + x₂ + x₃ = 6 -2x₂ + x₃ = -5
From the second equation, we get:
x₃ = 2x₂ - 5
Substituting this into the first equation:
x₁ + x₂ + (2x₂ - 5) = 6 x₁ + 3x₂ - 5 = 6 x₁ = 11 - 3x₂
Therefore, the solution is:
x₁ = 11 - 3x₂ x₂ = x₂ x₃ = 2x₂ - 5
So, the system has infinitely many solutions, parametrized by x₂:
[egin{cases} x₁ = 11 - 3x₂ \ x₂ = x₂ \ x₃ = 2x₂ - 5 end}$]
12月18日2007年
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