System of linear equations

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?

Borek
Mentor
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.

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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
Mentor
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.

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
Homework Helper
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?

Thanks!

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...

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