The correct option is B Is linearly independent
To examine the dependence or independence of vectors first form the equation x¯a+y¯b+z¯c=¯0, Then find the scalars x, y and z. If there exists a non-zero solution for x, y and z, the system will be linearly dependent. If the only solution comes out to be x = y = z = 0, the system will be linearly independent. Thus, x¯a+y¯b+z¯c=¯0,
⇒x(2^i−3^j+^k)+y(^i−^j+3^k)+z(^i+^j−^k)=0
Equating the coefficients of ^i,^jand^k
We get,
2x + y + z = 0 - Equating coefficients of^i (1)
-3x - y + z = 0 - Equating coefficients of ^j (2)
x + 3y - z = 0 - Equating coefficients of ^k (3)
Adding (1) and (3) we get
3x + 4y = 0 ….(4)
Adding (2) and (3) we get
-2x + 2y = 0 ….(5)
or -4x + 4y = 0 ….(6)
Equation (4) – equation (6) gives 7x = 0
⇒ x = 0
Putting x = 0 in (4) we get
y = 0
Put x = y = 0 in (3)
We get
z = 0.
So x = y = z = 0 is the only solution. So the system of the given vectors is linearly independent.