# Thread: simultaneous equations and inequalities

1. ## simultaneous equations and inequalities

ä-Hi folks,

we are asked to show that the only values of t for which the 3 equations

x + (t - 4)y +2z = 0

2x - 6y + (t + 3)z = 0

(t - 5)x + 12y -8z = 0

have a solution in which x,y and z are not all zero are t = 1 and t = 4.

Putting the equations into a 3 x 3 matrix A, we know that if $det A \ne 0$ then we have a unique solution.

If det A = 0 then we do not. In this case A is singular and we either have no solutions (inconsistent equations) or an infinite number of solutions. i.e. a common line or 3 coincident planes.

So, I evaluate $det A \ne 0$ to get an equation in t that I can solve. This should enable me to find the values of t.

My confusion is to do with the inequality. I am not dealing with det A = 0.

We are told that (t - 1) and (t - 4) are solutions and therefore factors of the equation and when I do the calculations I end up with:

$(t - 1) (t - 1) (t - 4) \ne 0$

I would be fine with this if det A = 0, for then I would get t = 1 and t = 4 as required. But I cannot interpret what the factors not equal to zero mean.

Thanks

2. ## Re: simultaneous equations and inequalities

Suppose your system of equations had constants on the right hand side.

Then you form the matrix equation and would like to determine if a unique solution exists.

You have 3 equations and 3 unknowns so you are guaranteed that a solution exists but it might not be unique.

So you take the determinant of your matrix and check if it is non-zero. A non-zero determinant means you will have a unique solution.

But a zero determinant means you won't. Which means the system matrix does not have full rank, i.e. it has a null space of rank at least 1.

A null space of at least rank 1 means there exist non-zero vectors for which when multiplied by the system matrix results in the zero vector.

This is the situation we are looking for in the given problem.

So in short your understanding of the result of the determinant is backwards. A zero determinant is what will guarantee that there is a non-zero vector satisfying the given equation.

It's clear from your factoring of the determinant that 1 and 4 are the two values of $t$ that result in a zero determinant.

3. ## Re: simultaneous equations and inequalities

Originally Posted by s_ingram
ä-Hi folks,

we are asked to show that the only values of t for which the 3 equations

x + (t - 4)y +2z = 0

2x - 6y + (t + 3)z = 0

(t - 5)x + 12y -8z = 0

have a solution in which x,y and z are not all zero are t = 1 and t = 4.

Putting the equations into a 3 x 3 matrix A, we know that if $det A \ne 0$ then we have a unique solution.

If det A = 0 then we do not. In this case A is singular and we either have no solutions (inconsistent equations) or an infinite number of solutions. i.e. a common line or 3 coincident planes.

So, I evaluate $det A \ne 0$ to get an equation in t that I can solve. This should enable me to find the values of t.

My confusion is to do with the inequality. I am not dealing with det A = 0.

We are told that (t - 1) and (t - 4) are solutions and therefore factors of the equation and when I do the calculations I end up with:

$(t - 1) (t - 1) (t - 4) \ne 0$

I would be fine with this if det A = 0, for then I would get t = 1 and t = 4 as required. But I cannot interpret what the factors not equal to zero mean.

Thanks
The determinant $\left|\begin{array}{ccc}a & b & c \\ d & e & f \\ g & h & i\end{array}\right|$ NOT being 0 means that the system of equations
ax+ by+ cz= p
dx+ ey+ fz= q
gx+ hy+ iz= r

has a unique solution. If that determinant is 0 then either is no solution or an infinite number of solutions. Here, since p = q= r= 0, x= y= z= 0 is an obvious solution. The determinant being no-zero means that is the only solution.

4. ## Re: simultaneous equations and inequalities

Sorry! If det A is not equal to zero, then I assumed our unique solution was something other than x = y = z = 0. Is this correct?

If det A = 0, then there are no solutions OR an infinite number. Since x = y = z is a solution, det A = 0 must mean we have an infinite number of solutions. This makes sense, since when t = 1, we have 3 coincident planes and when t = 4, the planes intersect along a line.

I think the mistake I made was in looking for a unique solution, when I should have been looking for a non unique solution. Is that right? Hope this makes sense!

5. ## Re: simultaneous equations and inequalities

I had never heard of rank and null spaces. I looked them up and thanks to you, I have a better insight into how to understand this stuff. Many thanks Romsek.

6. ## Re: simultaneous equations and inequalities

Originally Posted by s_ingram
Sorry! If det A is not equal to zero, then I assumed our unique solution was something other than x = y = z = 0. Is this correct?
No! "unique" means "only one". x= y= z= 0 obviously does satisfy the system of equations. Since it is unique it is the only solution.

If det A = 0, then there are no solutions OR an infinite number. Since x = y = z is a solution, det A = 0 must mean we have an infinite number of solutions. This makes sense, since when t = 1, we have 3 coincident planes and when t = 4, the planes intersect along a line.

I think the mistake I made was in looking for a unique solution, when I should have been looking for a non unique solution. Is that right? Hope this makes sense!
Yes. Again, the set of equations
ax+ by+ cz= 0
dx+ ey+ fz= 0
gx+ hy+ iz= 0
always has the solution x= y= z= 0. If the determinant is non-zero then that is the only solution. If the determinant is 0 then, essentially, one of the equations is a combination of the other two and we really only have two (or one) independent equations- we can only solve for two of the unknowns in terms of the remaining one so that taking different values for that one gives different solutions.

7. ## Re: simultaneous equations and inequalities

Thanks for your patience. All is clear.