



 
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,  


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