I am new into Linear Algebra and I am trying to solve the following system of equation:
$\begin{cases} \begin{matrix} x+2y-z-t=0 \\ \frac { 1 }{ 4 } x-y-\frac { 1 }{ 2 } z+2t=1 \\ 2x-2y-3z+7t=4 \end{matrix} \end{cases}$
So far, I've represented the system of equations in its respective matrix and obtained the echelon form as follows:
$[A|B]= \left[\begin{array}{rrrr|r} 1 & 2 & -1 &-1& 0 \\ 1/4 & -1 & -1/2 & 2 & 1 \\ 2 & -2 & -3 & 7 & 4 \\ \end{array} \right] \xrightarrow[R_3-2R_1 \rightarrow R_3]{4R_2-R_1 \rightarrow R_2} \left[\begin{array}{rrrr|r} 1 & 2 & -1 &-1& 0 \\ 0 & -6 & -1 & 9 & 4 \\ 0 & -6 & -1 & 9 & 4 \\ \end{array} \right] \xrightarrow{R_3-R_2 \rightarrow R_3} \left[\begin{array}{rrrr|r} 1 & 2 & -1 &-1& 0 \\ 0 & -6 & -1 & 9 & 4 \\ 0 & 0 & 0 & 0 & 0 \\ \end{array} \right]$
But I don't know how to proceed in solving the system of equations. Can someone also explain in detail the steps?
$$[A|B]= \left[\begin{array}{rrrr|r} 1 & 2 & -1 &-1& 0 \\ 1/4 & -1 & -1/2 & 2 & 1 \\ 2 & -2 & -3 & 7 & 4 \\ \end{array} \right] \xrightarrow[R_3-2R_1 \rightarrow R_3]{4R_2-R_1 \rightarrow R_2} \left[\begin{array}{rrrr|r} 1 & 2 & -1 &-1& 0 \\ 0 & -6 & -1 & 9 & 4 \\ 0 & -6 & -1 & 9 & 4 \\ \end{array} \right] \xrightarrow{R_3-R_2 \rightarrow R_3} \left[\begin{array}{rrrr|r} 1 & 2 & -1 &-1& 0 \\ 0 & -6 & -1 & 9 & 4 \\ 0 & 0 & 0 & 0 & 0 \\ \end{array} \right]$$
You get two equations namely $$x+2y-z-t=0$$ and $$-6x-y+9z=4$$ Consider the row reduced matrix, $$\left[\begin{array}{rrrr} 1 & 2 & -1 &-1 \\ 0 & -6 & -1 & 9 \\ 0 & 0 & 0 & 0 \\ \end{array} \right]$$ This is a $3\times 4$ matrix with rank with rank $2$. By rank -nullity theorem, we conclude that nullity of $A$ is $2$. So number of free parameters in the system is two. Let us take $z=k_1$ and $t=k_2$, where $k_1,k_2$ are variables. In other words you are free to chose $z$ and $t$.
Our two equation reduces to $$x+2y=k_1+k_2\tag{1}$$ and $$6x+y=9k_1-4\tag{2}$$ Multiply equation $(2)$ by $2$ and subtract from $(1)$ from $(2)$ you get $$x=\frac{17k_1-k_2-8}{11}$$
Put the vale of $x$ in $(1)$ to get the value of $y$