# Solve linear system of equations ?

Sep 30, 2016

${x}_{1} = 0 , {x}_{2} = - 1 , {x}_{3} = - 1 , {x}_{4} = 2$

#### Explanation:

Supposing that my understanding of the problem is the right one,

Calling

${M}_{1} = \left(\begin{matrix}1 & 1 & 1 & 1 \\ 2 & - 1 & - 1 & - 1\end{matrix}\right)$ and ${b}_{1} = \left(\begin{matrix}0 \\ 0\end{matrix}\right)$

${M}_{2} = \left(\begin{matrix}1 & 1 & 2 & 2 \\ 1 & 2 & 2 & 1\end{matrix}\right)$ and ${b}_{2} = \left(\begin{matrix}7 \\ 4\end{matrix}\right)$

and

$w = {\left(1 , 1 , 1 , 1\right)}^{T}$
$V = {\left({x}_{1} , {x}_{2} , {x}_{3} , {x}_{4}\right)}^{T}$ we have

${S}_{1} \to {M}_{1} \cdot X = {b}_{1}$ and
${S}_{2} \to {M}_{2} \cdot \left(X + w\right) = {b}_{2}$

Solving the system

$\left(\begin{matrix}{M}_{1} \\ {M}_{2}\end{matrix}\right) \cdot X = \left(\begin{matrix}{b}_{1} \\ {b}_{2} - {M}_{2} \cdot w\end{matrix}\right)$ we obtain

${x}_{1} = 0 , {x}_{2} = - 1 , {x}_{3} = - 1 , {x}_{4} = 2$