Given that a = i - j + 2k, b = i + 2j + mk and
c = 3i + nj + k, are linearly dependent. Express m in terms of n in simplest fraction form.
Answer: m = (2n-9)/(n+3)
Can someone pls show their working and explanation? Thx a lot
~C~
+1 1 3
-1 2 n
+2 m 1
determinant = 0
2 + 2n -3m -12 -nm +1 = 0
2n-9 -m(3+n) = 0
m = (2n-9)/(3+n)
To determine if the vectors a, b, and c are linearly dependent, we need to find values for m and n that satisfy the condition:
ka + lb + mc = 0
where k, l, and m are scalars, and at least one of them is nonzero.
Let's substitute the given values of a, b, and c into the equation:
k(i - j + 2k) + l(i + 2j + mk) + m(3i + nj + k) = 0
Distribute the coefficients:
ki - kj + 2k^2 + li + 2lj + mki + mkj + 3mi + mnj + mk = 0
Combine the like terms:
(ki + li + mki + 3mi) - (kj - 2lj - mkj - mnj) + (2k^2 + mk) = 0
Factor out the common terms:
(i(k + l + mk + 3m)) - j(k - 2l - mj - mn) + k(2k + m) = 0
Since all the components need to be zero, we have the following system of equations:
k + l + mk + 3m = 0 (Equation 1)
k - 2l - mj - mn = 0 (Equation 2)
2k + m = 0 (Equation 3)
Now let's solve this system of equations to find m in terms of n.
From Equation 3, we get:
k = -m/2
Substitute this value of k into Equation 1:
(-m/2) + l - m^2/2 + 3m = 0
-l - m + 3m - m^2 = 0
-m^2 + 2m - l = 0
From Equation 2, we get:
m = (k - mn - 2l)/j
m = (-m/2 - mn - 2l)/j
mj = -m - 2mnj - 2lj
mj + m + 2mnj + 2lj = 0
m(j + 1 + 2nj + 2l) = 0
Since we want m in terms of n, we assume that j + 1 + 2nj + 2l ≠ 0.
Divide both sides of the equation by (j + 1 + 2nj + 2l):
m = 0/(j + 1 + 2nj + 2l)
m = 0
So, for any value of n, m is 0. However, since we assumed j + 1 + 2nj + 2l ≠ 0, this is not a valid solution.
Therefore, there is no value of m that satisfies the condition for a linear dependency among vectors a, b, and c.
Hence, the given vectors a, b, and c are linearly independent.