To answer the mathematical doubts of three users above with regard to the quoted question, I would use Cramer's rule for solving the system of three linear equations with three unknowns. We have to define whether a determinant (det) of the matrix (underneath shown) is equal to 0 or it's not.
|1 2 3|
|4 5 6| <> det=1*(5*9-8*6)-2(4*9-6*7)+3(4*8-5*7)=-3+12-9=0.
|7 8 9|
So we found that our matrix's determinant is equal to 0, therefore we cannot further apply Cramer's rule.
Choice 2 is out. Instead, we must admit that the trivial solution of this system of linear equations is not the only solution.
Choice 3 is out too. It was correctly noted by user
kul512 in the previous post, that an equation a) can be discounted and/or just dropped, because it has been derived from the equations b) and c); 2*b) - c)=a) ==> 2(4x+5y+6z=0)-(7x+8y+9z=0)=x+2y+3z=0. Many different equations can be derived by using these two equations, but they won't be useful (they are meaningless) as derived equations.
We introduce the value t (it can be any number, any value) and solve the system of two equations
b) 4x+ 5y + 6z =0
c) 7x+ 8y + 9z =0
using already the derived equation a) x+2y+3z=0 and multiplying it by 2 and then subtracting its product from the equation b) 4x+5y+6z=0 we will get 4x+5y+6z - 2(x+2y+3z) = 2x+y=0 and y=-2x. By allowing x=t, we have y=-2t and may rewrite the previous system such as
b) 4t+ 5*(-2t) + 6z =0
c) 7t+ 8*(-2t) + 9z =0
we should be able to find z, a) 6z=6t and b) 9z=9t give us z=t.
Finally, by setting x=t, y=-2t and z=t we see that the solution for this question can be x=y=z=0 when t=0
BUT this question has also IDK, zillion, many different solutions
depending on the value of t.
The correct answer:
The system
1) has infinitely many solutions
MubbashirAbbas wrote:Given the system of linear equations
a) x + 2y + 3z =0
b) 4x+ 5y + 6z =0
c) 7x+ 8y + 9z =0
The system has
1) has Infinietly many solutions
2) has Exactly one solution
3) has trivial solution
4) has no solution
Please explain how this can be solved
