So it's just a few days before my math exam and I came across this excercise and I just can't seem to find the right answer. Would really appreciate your help on this one.

It's from a german textbook so please excuse any grammar mistakes:

d 'is celebrating a party. a, b and c are doing a drinking game. They all drink the same from same size glasses:'

I: a and b together drank as much as c.
II: a and c together drank 4 times as much as b
III: a drank n more glasses then b

1) I need to find out the Linear System of Equations.

So I would assume:

I: a + b = c
II: a + c = 4b
III: b + n = a

To write the matrix it in the form of: A*X = B:

=> I: a + b - c = 0 => II: a - 4b + c = 0 => III: -a + b + n = 0

(please correct if I'm wrong)

next: 2) Find the inverse Matrix with the Gauß-Jordan-Algorithm

This is my solution for the inverse Matrix

Is that correct? This is where I had a lot of problems with. Because of that extra variable n in the matrix.

3) From that inverse matrix i have to find universal solution where I just have to put in n to solve.

Don't really know what to do here. If I take my inverse matrix and multiply it by B wich is (0 0 0) all my values for a,b,c would get 0...

So don't really know how to proceed...

4) I just need to try different values for n and show the Equations for different n values.

Would really really appreciate any kind of help. I've been working on this for more than 3 hours and I just can't seem to find the answer.

Thank you in advance !

  • $\begingroup$ say "I came across this excercise." If you say "I came up with this excercise" it means that you wrote the excercise yourself, which isn't the case. :) $\endgroup$ – setholopolus Dec 21 '16 at 0:57
  • $\begingroup$ Edited! Cherrs :) $\endgroup$ – dummker Dec 21 '16 at 0:58
  • $\begingroup$ also, your variables are confusing. You are using 'A' and 'B' as people who are drinking, but you are also using them as a matrix and a vector. You need to have a different name for each thing. $\endgroup$ – setholopolus Dec 21 '16 at 0:59
  • $\begingroup$ This is awkward. Edited as well! Really appreciate your comments! $\endgroup$ – dummker Dec 21 '16 at 1:03

The variables spanning your space are $a,b,c$ , the variable $n$ needs to be treated as a constant $$ \begin{bmatrix} 1 & 1 & -1 \\1 & -4 & 1 \\-1 & 1 & 0 \end{bmatrix} \begin{bmatrix} a \\b \\c \end{bmatrix} = \begin{bmatrix} 0 \\0 \\n \end{bmatrix} $$

There is the matrix you need to take the inverse of - it doesn't depend on $n$

Your general solution will take the form ...

$$ \begin{bmatrix} a \\b \\c \end{bmatrix} = M^{-1} \begin{bmatrix} 0 \\0 \\n \end{bmatrix}$$


$n$ is not a variable to be solved for; it's a condition you are given.

So in Step 1, instead of 'III: $-a + b + n = 0$' you want 'III: $-a + b + 0c = n$'. Then, in step 3, the $B$ in your $A \cdot X = B$ will not be $0$; and so $X = A^{-1} \cdot B$ will give you the answer you want.


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