It is given that "b" has a linear dependency on "a" and "a" has a linear dependency on "c". The equation 3a+2b+4c=5 can be reduced into a linear equation.
Answers
Answer:
Given a, b, c are linearly independent
∴pa+qb+rc=0⇒p=0,q=0,r=0...(1)
Now consider x(a−2b+c)+y(2a−b+c)+z(3a+b+2c)=0 or
(x+2y+3z)a+(−2x−y+z)b+(x+y+2z)c=0
Hence by (1) we have
x+2y+3z=0,−2x−y+z=0,x+y+2z=0 Above is a set of homegeneous equations.
Δ=
∣
∣
∣
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1
−2
1
2
−1
1
3
1
2
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=−3+10−3=4
=0Since Δ
=0, the system of equations has only a trivial solution i.e., x=0,y=0,z=0 Hence linearly independent
Step-by-step explanation:
it's a example
Step-by-step explanation:
Linear dependency signifies that "a","b" and "c" are linearly related, which means that if one of them can be found, the other two can be found.
Hence, the given equation can be reduced to a linear equation if the exact relation between "a","b"and "c" is given.
For a better understanding, let us express the dependencies as equations.
Since b is linearly dependent on a, we can write b = ka + p, where k and p are some constants. Note that in this equation, b and a are the only variables, whereas k and p are known.
Similarly, a can be written as a gc + u, where g and u are constants (known values).
Substituting these values of a and b in the given equation, we get, 3a + 2b + 4c = 5
3(gc + u) + 2(ka + p) + 4c = 5
3gc + 3u + 2p + 2k(gc + u) + 4c = 5
(3g + 2kg + 4)c + (3u + 2p + 2ku - 5) = 0.
Note that in the final equation, the only c is unknown, the others are all constants and hence is a linear equation in one variable.