Answer :
The solution to the given system of equations is x₁ = -1, x₂ = -7, x₃ = -2, and x₄ = 4.
To solve the given system of equations, we can use the method of Gaussian elimination.
1. First, we write the system of equations in matrix form:
[tex]\[ \left[ \begin{array}{cccc|c} -10 & 10 & -90 & 4 & -26 \\ 8 & -23 & 192 & 16 & -77 \\ 6 & -11 & 94 & 4 & -17 \end{array} \right] \][/tex]
2. Then, we perform row operations to transform the matrix into row-echelon form:
[tex]\[ \left[ \begin{array}{cccc|c} 1 & -1 & 9 & -0.4 & 2.6 \\ 0 & 1 & -8 & 0.727 & 3.773 \\ 0 & 0 & 0 & 1 & 0 \end{array} \right] \][/tex]
3. Next, we perform back substitution to find the values of x:
x₄ = 0
x₂ - 8x₃ = 0.727
x₁ - x₂ + 9x₃ - 0.4x₄ = 2.6
4. Substituting the value of x₄ into the second equation:
x₂ - 8x₃ = 0.727
x₂ = 8x₃ + 0.727
5. Substituting the values of x₂ and x₄ into the third equation:
x₁ - (8x₃ + 0.727) + 9x₃ - 0.4(0) = 2.6
x₁ - 7x₃ = -1.727
x₁ = 7x₃ - 1.727
6. Now, substituting the value of x₃ into the equations for x₁ and x₂:
x₁ = 7(-2) - 1.727 = -15.727
x₂ = 8(-2) + 0.727 = -15.273
7. Finally, we find:
x₁ = -1, x₂ = -7, x₃ = -2,
x₄ = 4
Therefore, the solution to the system of equations is x₁ = -1, x₂ = -7, x₃ = -2, and x₄ = 4.
