I understand how to test two equations with two variables but what happens when I get up to 4 variables? I have no idea how to test that. For example with the matrix/system of equations
$x_1 +2 {x_2} + x_3 = 1$
$-2 x_1 - 4 {x_2} - x_3 = 0$
$5 {x_1} + 10 {x_2} + 3 {x_3} = 2$
$3 {x_1} + 6 {x_2} + 3 {x_3} = 4$
How do I find out if it is inconsistent?
$\endgroup$3 Answers
$\begingroup$Hint: take $-3$ times the first equation and add it to the fourth equation. The fourth equation will then become $0 = 1$, which is clearly inconsistent.
You can also express the system of equations as an augmented coefficient matrix, add $-3$ times row 1 + row 4 $\implies$ row 4 becomes $\quad0 \;0\;0\;|\;1$
$\endgroup$ 4 $\begingroup$Try to solve the equation; and check whether the system has the solution or not.
$\begin{pmatrix}1 & 2 & 1 \\ -2 & -4 & -1 \\ 5 & 10 & 3 \\ 3 & 6 & 3 \end{pmatrix}\begin{pmatrix}x_1 \\ x_2 \\ x_3 \end{pmatrix} = \begin{pmatrix}1 \\ 0 \\ 2 \\ 4\end{pmatrix} $, and multiply first row by 3 and substract to 4th row then
$\begin{pmatrix}1 & 2 & 1 \\ -2 & -4 & -1 \\ 5 & 10 & 3 \\ 0&0&0 \end{pmatrix}\begin{pmatrix}x_1 \\ x_2 \\ x_3 \end{pmatrix} = \begin{pmatrix}1 \\ 0 \\ 2 \\ 1\end{pmatrix} $ which says that $0 = 1$ in the 4th row, which is inconsistent.
$\endgroup$ $\begingroup$Yet another way:
Add twice the first equation to the second to get $x_3 = 2$. Now substitute $x_3 = 2$ into the other equations to get $5 x_1 + 10 x_2 = 5 (x_1+2 x_2) = -4$ and $3 x_1 + 6 x_2 = 3 (x_1+2 x_2) = -2$, and you can immediately see that there is no solution.
$\endgroup$
