How to solve this system of linear equations using Gaussian elimination?

I'm having trouble with this problem from my linear algebra course. The problem is:

A new restaurant owner decides to have 20 tables for her guests, a certain number of tables with space for 4 people, some with space for 6 people and also one or more tables with space for 8 guests. In total, there are 108 seats available, but if only half of the seats at the 4-tables and 6-tables, and only a quarter of the seats at the 8-tables, are occupied, the restaurant will have 46 guests. How many tables with space for 4, 6 and 8 seats does she need to set up in her premises?

I tried to solve it by writing the system of equations as a matrix:

$$\left( \begin{array}{cccc} 1&1&1&20\\ 2&3&2&46\\ 4&6&8&108 \end{array} \right)$$

Then I applied Gaussian elimination by doing the following steps:

• Subtracting 2 times the first row from the second row
• Swapping the first and third rows
• Multiplying the first row by 1/4
• Subtracting the second row from the third row
• Subtracting 2/3 times the second row from the first row
• Subtracting 2 times the third row from the first row
• Adding the first row to the third row

This gave me the following matrix:

$$\left( \begin{array}{cccc} -1&0&0&-\frac{5}{2} \\ 0&1&0&6\\ 0&0&1&\frac{23}{2} \end{array} \right)$$

However, this does not match the answer given in the textbook, which is:

$$\left(\begin{array}{cccc} 1 & 0 & 0 & 10 \\ 0 & 1 & 0 & 6 \\ 0 & 0 & 1 & 4 \end{array}\right)$$

Can someone please explain where I went wrong and how to get the correct answer? I appreciate any help or hints. Thank you!

A step by step approach.

It is very slow to build but very good in that you can go back to a step and fix things.

Notice that the operation appears besides the row so you know which row was affected exactly.

Starting from $$\left( \begin{array}{cccc} 1&1&1&20\\ 2&3&2&46\\ 4&6&8&108 \end{array} \right)$$ after the fourth step in your list we get $$\left( \begin{array}{cccc} 1&3/2&2&27\\ 0&1&0&6\\ 1&0&1&14 \end{array} \right).$$ Now we subtract $$\large\color{red}{3/2}$$ times the second row from the first row $$\left( \begin{array}{cccc} 1&0&2&18\\ 0&1&0&6\\ 1&0&1&14 \end{array} \right).$$ From here, subtracting the third row from the first row and then subtracting the first row from the third row, we obtain $$\left( \begin{array}{cccc} 0&0&1&4\\ 0&1&0&6\\ 1&0&0&10 \end{array} \right).$$ Finally, after swapping the first and the third row, we get $$\left( \begin{array}{cccc} 1&0&0&10\\ 0&1&0&6\\ 0&0&1&4 \end{array} \right).$$

Simply do the following transformations (R means Row)

R2 becomes R2-2R1

R3 becomes R3-4R1

$$\left( \begin{array}{cccc} 1&1&1&20\\ 0&1&0&6\\ 0&2&4&28 \end{array} \right)$$

R3 becomes (R3-2R2)/4

$$\left( \begin{array}{cccc} 1&1&1&20\\ 0&1&0&6\\ 0&0&1&4 \end{array} \right)$$

R1 becomes R1-R2-R3 $$\left( \begin{array}{cccc} 1&0&0&10\\ 0&1&0&6\\ 0&0&1&4 \end{array} \right)$$

I hope that helps. (In your solution, 2/3 and 3/2 are confused.)