By algebraic handling we find that the unique solution of the system of linear equations is: (x, y, z) = (- 2, 4, 3). (Correct choice: A)
What is the nature of a system of linear equations?
If the system of equations has no solutions, then the determinant of the dependent coefficients of the system of linear equations must be zero. Let see:
[tex]D = \left|\begin{array}{ccc}1&1&1\\4&-1&-8\\-1&-1&7\end{array}\right|[/tex]
This determinant can be determined by Sarrus' rule:
D = (1) · (- 1) · (7) + 4 · (- 1) · 1 + (- 1) · 1 · (- 8) - (- 1) · (- 1) · 1 + 4 · 1 · 7 - 1 · (- 1) · (- 8)
D = - 7 - 4 + 8 - 1 + 28 - 8
D = 16
The system of linear equations have at least one solution. This system has only one solution any of the three equations is not a function of the other two. By algebraic handling we find that the unique solution of the system is: (x, y, z) = (- 2, 4, 3).
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