Hi Marian,

I am not sure what method you would use to solve a system with 3 equations and 3 unknowns but you can just apply the same technique to 4 equations with 3 unknowns. I a going to use a matrix approach but with instructions on how to apply my method to an equations approach. The system I want to look at is

$$\begin{array}{llll}x_1 + x_2 = 1\\x_1 + x_3 = 1\\2 x_1 + x_2 = 2\\x_2 -2 x_3 = 0\end{array}$$

The augmented matrix for this system is

$$\left( \begin{array}{rrrr}1 & 1 & 0 & 1\\ 1 & 0 & 1 & 1\\2 & 1 & 0 & 2\\0 & 1 & -2 & 0\end{array} \right).$$

I would begin by adding -1 times the first row to the second row and then adding -2 times the first row to the third row. (If you are using an equations approach then add -1 times the first equation to the second equation and then add -2 times the first equation to the third equation.) The resulting matrix is

$$\left( \begin{array}{rrrr}1 & 1 & 0 & 1\\ 0 & -1 & 1 & 0\\0 & -1 & 0 & 0\\0 & 1 & -2 & 0\end{array} \right).$$

Next multiply the third row by -1 and then interchange the second and third rows to get

$$\left( \begin{array}{rrrr}1 & 1 & 0 & 1\\ 0 & 1 & 0 & 0\\0 & -1 & 1 & 0\\0 & 1 & -2 & 0\end{array} \right).$$

At this point add the second row to the third row and add -1 times the second row to the fourth row to get

$$\left( \begin{array}{rrrr}1 & 1 & 0 & 1\\ 0 & 1 & 0 & 0\\0 & 0 & 1 & 0\\0 & 0 & -2 & 0\end{array} \right).$$

Finally add -1 times the second row to the first row and add twice the third row to the fourth row to get

$$\left( \begin{array}{rrrr}1 & 0 & 0 & 1\\ 0 & 1 & 0 & 0\\0 & 0 & 1 & 0\\0 & 0 & 0 & 0\end{array} \right).$$

The four equations are now

$$\begin{array}{llll}x_1 = 1\\x_2 = 0\\ x_3 = 0\\03 = 0\end{array}$$

and the solution is $x_1 = 1, x_2 = x_3 = 0.$

Things don't always work out as nicely as this example, for example change the original equations to

$$\begin{array}{llll}x_1 + x_2 = 1\\x_1 + x_3 = 1\\2 x_1 + x_2 = 2\\x_2 = 1\end{array}$$

If you substitute $x_2 = 1$ into equations 1 and three and solve for $x_1$ you will see that you get two different values for $x_1$, and thus the system has no solution.

I tried the augmented matrix approach with

$$\left( \begin{array}{rrrr}1 & 1 & 0 & 1\\ 1 & 0 & 1 & 1\\2 & 1 & 0 & 2\\0 & 1 & 0 & 1\end{array} \right).$$

and got to

$$\left( \begin{array}{rrrr}1 & 1 & 0 & 1\\ 0 & 1 & 0 & 0\\0 & 0 & 1 & 0\\0 & 0 & 0 & 1\end{array} \right).$$

This clearly has no solution as the final equation is $0 \times x_1 + 0 \times x_2 + 0 \times x_3 = 1.$

I hope this helps,
Penny