Equations - Number of solutions

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Equations - Number of solutions

by MubbashirAbbas » Thu May 24, 2012 4:56 am
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

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by ronnie1985 » Thu May 24, 2012 11:02 am
The system has trivial solution meaning x=y=z=0

As the ratios are different
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by MubbashirAbbas » Thu May 24, 2012 11:36 am
Ronnie,
How did you come up with the answer. Did you solve the equations or did you apply some shortcut method to arrive at the conclusion. Whatever it is, Please explain the steps..

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by kul512 » Thu May 24, 2012 1:59 pm
I think it will be infinetly many solutions because the third equation can be derived from the other two equations,effectively leaving only two eqautions to solve three unknowns.
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by pemdas » Tue Feb 26, 2013 11:06 am
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
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by Bodhi_Vriksha » Tue Feb 26, 2013 9:07 pm
MubbashirAbbas wrote:

Please explain how this can be solved
Equation b can be obtained using equation a and c. Hence, infinite number of solutions.