Given the system of equations $$2 x_{1} + x_{2} - x_{3} + x_{4} = 1$$ $$3 x_{2} - x_{4} = 1$$ $$x_{3} - x_{4} = 0$$
We give the system of equations to the canonical form $$2 x_{1} + x_{2} - x_{3} + x_{4} = 1$$ $$3 x_{2} - x_{4} = 1$$ $$x_{3} - x_{4} = 0$$ Rewrite the system of linear equations as the matrix form $$\left[\begin{matrix}2 & 1 & -1 & 1 & 1\\0 & 3 & 0 & -1 & 1\\0 & 0 & 1 & -1 & 0\end{matrix}\right]$$ In 1 -th column $$\left[\begin{matrix}2\\0\\0\end{matrix}\right]$$ let’s convert all the elements, except 1 -th element into zero. - To do this, let’s take 1 -th line $$\left[\begin{matrix}2 & 1 & -1 & 1 & 1\end{matrix}\right]$$ , and subtract it from other lines: In 2 -th column $$\left[\begin{matrix}1\\3\\0\end{matrix}\right]$$ let’s convert all the elements, except 2 -th element into zero. - To do this, let’s take 2 -th line $$\left[\begin{matrix}0 & 3 & 0 & -1 & 1\end{matrix}\right]$$ , and subtract it from other lines: From 1 -th line. Let’s subtract it from this line: $$\left[\begin{matrix}2 & 0 & -1 & \frac{4}{3} & \frac{2}{3}\end{matrix}\right] = \left[\begin{matrix}2 & 0 & -1 & \frac{4}{3} & \frac{2}{3}\end{matrix}\right]$$ you get $$\left[\begin{matrix}2 & 0 & -1 & \frac{4}{3} & \frac{2}{3}\\0 & 3 & 0 & -1 & 1\\0 & 0 & 1 & -1 & 0\end{matrix}\right]$$ In 3 -th column $$\left[\begin{matrix}-1\\0\\1\end{matrix}\right]$$ let’s convert all the elements, except 3 -th element into zero. - To do this, let’s take 3 -th line $$\left[\begin{matrix}0 & 0 & 1 & -1 & 0\end{matrix}\right]$$ , and subtract it from other lines: From 1 -th line. Let’s subtract it from this line: $$\left[\begin{matrix}2 & 0 & 0 & \frac{1}{3} & \frac{2}{3}\end{matrix}\right] = \left[\begin{matrix}2 & 0 & 0 & \frac{1}{3} & \frac{2}{3}\end{matrix}\right]$$ you get $$\left[\begin{matrix}2 & 0 & 0 & \frac{1}{3} & \frac{2}{3}\\0 & 3 & 0 & -1 & 1\\0 & 0 & 1 & -1 & 0\end{matrix}\right]$$ In 1 -th column $$\left[\begin{matrix}2\\0\\0\end{matrix}\right]$$ let’s convert all the elements, except 1 -th element into zero. - To do this, let’s take 1 -th line $$\left[\begin{matrix}2 & 0 & 0 & \frac{1}{3} & \frac{2}{3}\end{matrix}\right]$$ , and subtract it from other lines: In 2 -th column $$\left[\begin{matrix}0\\3\\0\end{matrix}\right]$$ let’s convert all the elements, except 2 -th element into zero. - To do this, let’s take 2 -th line $$\left[\begin{matrix}0 & 3 & 0 & -1 & 1\end{matrix}\right]$$ , and subtract it from other lines: In 3 -th column $$\left[\begin{matrix}0\\0\\1\end{matrix}\right]$$ let’s convert all the elements, except 3 -th element into zero. - To do this, let’s take 3 -th line $$\left[\begin{matrix}0 & 0 & 1 & -1 & 0\end{matrix}\right]$$ , and subtract it from other lines:
Everything is almost ready - it remains only to find the unknowns by solving elementary equations: $$2 x_{1} + \frac{x_{4}}{3} - \frac{2}{3} = 0$$ $$3 x_{2} - x_{4} - 1 = 0$$ $$x_{3} - x_{4} = 0$$ We get the answer: $$x_{1} = - \frac{x_{4}}{6} + \frac{1}{3}$$ $$x_{2} = \frac{x_{4}}{3} + \frac{1}{3}$$ $$x_{3} = x_{4}$$ where x4 - the free variables
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