System of Linear Equations: Infinitely Many Solutions

[Mano Jow]

Homework Statement

Exercise 1: For what value of m will the following system of linear equations have infinitely many solutions?
mx + 3y = 12
2x + (1/2)y = 5

Exercise 2: For what value of k will the following system of linear equations
x + 2y + kz = 1
2x + ky + 8z = 3

have
a) unique solution?
b) no solution?
c) more than one solution?

3. Discuss the following system:
x + z + w = 0
x + ky + k²w = 1
x + (k + 1)z + w = 1
x + z + kw = 2

2. The attempt at a solution
I've put this questions together because they are all related... I hope it's ok to do so.

Honestly I don't know what to do. I know how to do the Gauss' Elimination and Crammer's Rule but I don't see how to apply them. I mean, for the second one, can I just say that if k = 4 the system will have no solution because I'll get 1 = 3? Or do I have to show it by using one of those techniques?

Can you guys point me out the right direction? I mean, help me to understand what to do?

Thanks in advance.

 

[Borek]

Do you know how to use determinants to solve system of equations?

Edit: OK, I didn't know it is called Cramer's rule. You are close.
Edit 2: OK, once I knew, but it was long ago and I forgot. In Polish it is called "wzory Kramera" - Cramer's formulas.

Think about division by zero.

 

[Mano Jow]

Ok, let me see if I'm doing right. For the first one let A be the coeficient matrix:

A = \left(\begin{array}{cc}m&3\\2&\frac{1}{2}\end{array}\right)

so

det(A) = \frac{m}{2} - 6

Ax = \left(\begin{array}{cc}m&12\\2&5\end{array}\right)

and

Ay = \left(\begin{array}{cc}3&12\\\frac{1}{2}&5\end{array}\right)

We have that

x = \frac{Dx}{D} and y = \frac{Dy}{D}

If I need D to be 0, m = 12 right?

But how does the system have infinitely many solutions if Dy is never 0? Or does the solutions being like \frac{a}{0} with a being a real number means this system has infinitely many solutions?


Also, for the second one, how can I calculate D if I have a 2x3 coeficient matrix?


Thanks!

 

[Borek]

You need to check out what det A = 0 means, that a basic property.

2nd - you have a 4x4 matrix as far as I can tell.

 

[Mano Jow]

Well, I know that if det A = 0 means the two slopes are equal, but they can be parallel or identical. I want them to be identical so the system will have infinitely many solutions.

I have the following equations:
y = 4 - \frac{m}{3}x
y = 10 - 4x

For the slopes to be identical,
-\frac{m}{3} = -4
and so m = 12.

But the slopes are parallel, because I have 4 and 10 as the point where the slopes cross the y axis. So how can I make them identical?

If I say that det(Ax) = 0, I have
5m - 24 = 0 => m = 4.5
Which is different from that previous m... what am I doing wrong?

About that 4x4 matrix, are you talking about the 2nd ou 3rd exercise?


Thanks again!

 

[HallsofIvy]

Ok, let me see if I'm doing right. For the first one let A be the coeficient matrix:

A = \left(\begin{array}{cc}m&3\\2&\frac{1}{2}\end{array}\right)

so

det(A) = \frac{m}{2} - 6

Ax = \left(\begin{array}{cc}m&12\\2&5\end{array}\right)

and

Ay = \left(\begin{array}{cc}3&12\\\frac{1}{2}&5\end{array}\right)

We have that

x = \frac{Dx}{D} and y = \frac{Dy}{D}

If I need D to be 0, m = 12 right?

But how does the system have infinitely many solutions if Dy is never 0? Or does the solutions being like \frac{a}{0} with a being a real number means this system has infinitely many solutions?

No it doesn't. \frac{a}{0} with a NOT EQUAL TO 0 does not exist and so there is no solution. \frac{0}{0} means there is an infinite number of solutions.

Look at 0x= b with x and b numbers. 0x= 0 for all x. If b\ne 0 that is never true- there is no solution. If b= 0, 0x= b= 0 is true for all x.

Also, for the second one, how can I calculate D if I have a 2x3 coeficient matrix?

You don't. Use elimination instead.

Thanks!

 

[Mano Jow]

Ok, so for the first one there's no way I can find m so the system will have infinitely many solutions?

Edit: For the second, I did the following:

1. multiplied the first line by -2 and added to the second;
2. multiplied the second line by \frac{1}{k-4}

So I got:
x + 2y + kz = 1
0x + y - 2z = \frac{1}{k-4}

This means that k-4 must not be zero and so if k = 4 then the system is impossible, right? Now I can't see what else I can get from here, I mean, how to get the system to have one single solution and infinitely many solutions...