Mister Exam

х1+х2+х3+х4+х5=15; 2х1-х2+х3-х4+х5

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The solution

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x1 + x2 + x3 + x4 + x5 = 15
$$x_{5} + \left(x_{4} + \left(x_{3} + \left(x_{1} + x_{2}\right)\right)\right) = 15$$
2*x1 - x2 + x3 - x4 + x5 = 0
$$x_{5} + \left(- x_{4} + \left(x_{3} + \left(2 x_{1} - x_{2}\right)\right)\right) = 0$$
x5 - x4 + x3 + 2*x1 - x2 = 0
Rapid solution
$$x_{21} = - \frac{x_{3}}{3} - x_{4} - \frac{x_{5}}{3} + 10$$
=
$$- \frac{x_{3}}{3} - x_{4} - \frac{x_{5}}{3} + 10$$
=
10 - x4 - 0.333333333333333*x3 - 0.333333333333333*x5

$$x_{11} = - \frac{2 x_{3}}{3} - \frac{2 x_{5}}{3} + 5$$
=
$$- \frac{2 x_{3}}{3} - \frac{2 x_{5}}{3} + 5$$
=
5 - 0.666666666666667*x3 - 0.666666666666667*x5
Gaussian elimination
Given the system of equations
$$x_{5} + \left(x_{4} + \left(x_{3} + \left(x_{1} + x_{2}\right)\right)\right) = 15$$
$$x_{5} + \left(- x_{4} + \left(x_{3} + \left(2 x_{1} - x_{2}\right)\right)\right) = 0$$

We give the system of equations to the canonical form
$$x_{1} + x_{2} + x_{3} + x_{4} + x_{5} = 15$$
$$2 x_{1} - x_{2} + x_{3} - x_{4} + x_{5} = 0$$
Rewrite the system of linear equations as the matrix form
$$\left[\begin{matrix}1 & 1 & 1 & 1 & 1 & 15\\2 & -1 & 1 & -1 & 1 & 0\end{matrix}\right]$$
In 1 -th column
$$\left[\begin{matrix}1\\2\end{matrix}\right]$$
let’s convert all the elements, except
1 -th element into zero.
- To do this, let’s take 1 -th line
$$\left[\begin{matrix}1 & 1 & 1 & 1 & 1 & 15\end{matrix}\right]$$
,
and subtract it from other lines:
From 2 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}\left(-1\right) 2 + 2 & \left(-1\right) 2 - 1 & \left(-1\right) 2 + 1 & \left(-1\right) 2 - 1 & \left(-1\right) 2 + 1 & - 2 \cdot 15\end{matrix}\right] = \left[\begin{matrix}0 & -3 & -1 & -3 & -1 & -30\end{matrix}\right]$$
you get
$$\left[\begin{matrix}1 & 1 & 1 & 1 & 1 & 15\\0 & -3 & -1 & -3 & -1 & -30\end{matrix}\right]$$
In 2 -th column
$$\left[\begin{matrix}1\\-3\end{matrix}\right]$$
let’s convert all the elements, except
2 -th element into zero.
- To do this, let’s take 2 -th line
$$\left[\begin{matrix}0 & -3 & -1 & -3 & -1 & -30\end{matrix}\right]$$
,
and subtract it from other lines:
From 1 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}1 - \frac{\left(-1\right) 0}{3} & 1 - - -1 & 1 - - \frac{-1}{3} & 1 - - -1 & 1 - - \frac{-1}{3} & 15 - - -10\end{matrix}\right] = \left[\begin{matrix}1 & 0 & \frac{2}{3} & 0 & \frac{2}{3} & 5\end{matrix}\right]$$
you get
$$\left[\begin{matrix}1 & 0 & \frac{2}{3} & 0 & \frac{2}{3} & 5\\0 & -3 & -1 & -3 & -1 & -30\end{matrix}\right]$$
In 1 -th column
$$\left[\begin{matrix}1\\0\end{matrix}\right]$$
let’s convert all the elements, except
1 -th element into zero.
- To do this, let’s take 1 -th line
$$\left[\begin{matrix}1 & 0 & \frac{2}{3} & 0 & \frac{2}{3} & 5\end{matrix}\right]$$
,
and subtract it from other lines:
In 2 -th column
$$\left[\begin{matrix}0\\-3\end{matrix}\right]$$
let’s convert all the elements, except
2 -th element into zero.
- To do this, let’s take 2 -th line
$$\left[\begin{matrix}0 & -3 & -1 & -3 & -1 & -30\end{matrix}\right]$$
,
and subtract it from other lines:

It is almost ready, all we have to do is to find variables, solving the elementary equations:
$$x_{1} + \frac{2 x_{3}}{3} + \frac{2 x_{5}}{3} - 5 = 0$$
$$- 3 x_{2} - x_{3} - 3 x_{4} - x_{5} + 30 = 0$$
We get the answer:
$$x_{1} = - \frac{2 x_{3}}{3} - \frac{2 x_{5}}{3} + 5$$
$$x_{2} = - \frac{x_{3}}{3} - x_{4} - \frac{x_{5}}{3} + 10$$
where x3, x4, x5 - the free variables