We'll use the matrix to solve the system. We'll form the
matrix of the system:
1
1
A =
2
-1
We'll calculate the determinant of the
system:
detA = -1 - 2 =
-3
Since det A is not cancelling, the system is
determinated and it will have only one solution.
x = det
X/detA
5
1
det X =
1
-1
detX = -5 - 1 = -6
x = det
X/detA
x =
-6/-3
x =
2
We'll calculate
y:
1
5
det Y =
2
1
det Y = 1 - 10
det Y =
-9
y = detY/detA
y =
-9/-3
y =
3
The solution of the system
is: (2 , 3).
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