For no solutions, put a = b = c = -1, for exactly one solution, put a = 1 and b = c = 0 and for infinite solutions, put a = b = c = 5 in the system of linear equations.
We have the system of equations:
x + y + z = 2 …(1)
a x + b y + c z = 10 …(2)
x - 2 y + z = 4 …(3)
a. The system has no solution.
Put a = b = c = -1 in (2)
We get:
- x - y - z = 10
Now add (1) and (2)
0 = 12
which is not possible.
So, no solution.
b. The system has exactly one solution.
a = 1 and b = c = 0
We have the equations as:
x + y + z = 2 …(1)
x = 10 …(2)
x - 2 y + z = 4 …(3)
c. The system has infinitely many solutions.
a = b = c = 5
We will have the equations:
x + y + z = 2 …(1)
5 x + 5 y + 5 z = 10 …(2)
x - 2 y + z = 4 …(3)
Therefore, we get that, for no solutions, put a = b = c = -1, for exactly one solution, put a = 1 and b = c = 0 and for infinite solutions, put a = b = c = 5 in the system of linear equations.
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