We have a three unknown, 4 equation homogeneous system. These always have at least (0,0,0) as a solution. Let's write the equations, one column at a time.
1a + 0b + 0c = 0
-1a + 1b +0c = 0
0a - 1b + c = 0
0a + 0b + -1 c = 0
We could do row reduction but these are easy enough not to bother.
Equation 1 says
a = 0
Equation 4 says
c = 0
Substituting in the two remaining,
-1(0) + 1b + 0c = 0
b = 0
0(0) - 1b + 0 = 0
b = 0
The only 3-tuple satisfying the vector equation is (a,b,c)=(0,0,0)
Answer:
B
Step-by-step explanation:
I think its not sure: 42.437m
Answer:a=3/4
b=3/4
Step-by-step explanation:
3a+5b=6
9a+7b=12
-3(3a+5b)=-3(6)
9a+7b=12
-9a-15b=-18
9a+7b=12
-8b=-6
b=3/4
3a+5(3/4)=6
3a+15/4=6
3a=9/4
a=3/4
