Mister Exam
x=y+z; y*20+x*20=141,42+j*141,42; x*20+j*20*z=141,42+j*141,42
The solution
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x = y + z
$$x = y + z$$
7071 I*7071
y*20 + x*20 = ---- + ------
50 50
$$20 x + 20 y = \frac{7071}{50} + \frac{7071 i}{50}$$
7071 I*7071
x*20 + I*20*z = ---- + ------
50 50
$$20 x + z 20 i = \frac{7071}{50} + \frac{7071 i}{50}$$
20*x + z*(20*i) = 7071/50 + 7071*i/50
Rapid solution
$$x_{1} = \frac{7071}{1250} + \frac{7071 i}{2500}$$
=
$$\frac{7071}{1250} + \frac{7071 i}{2500}$$
=
$$y_{1} = \frac{7071}{5000} + \frac{21213 i}{5000}$$
=
$$\frac{7071}{5000} + \frac{21213 i}{5000}$$
=
$$z_{1} = \frac{21213}{5000} - \frac{7071 i}{5000}$$
=
$$\frac{21213}{5000} - \frac{7071 i}{5000}$$
=
=
$$\frac{7071}{1250} + \frac{7071 i}{2500}$$
=
5.6568 + 2.8284*i
$$y_{1} = \frac{7071}{5000} + \frac{21213 i}{5000}$$
=
$$\frac{7071}{5000} + \frac{21213 i}{5000}$$
=
1.4142 + 4.2426*i
$$z_{1} = \frac{21213}{5000} - \frac{7071 i}{5000}$$
=
$$\frac{21213}{5000} - \frac{7071 i}{5000}$$
=
4.2426 - 1.4142*i
Cramer's rule
$$x = y + z$$
$$20 x + 20 y = \frac{7071}{50} + \frac{7071 i}{50}$$
$$20 x + z 20 i = \frac{7071}{50} + \frac{7071 i}{50}$$
We give the system of equations to the canonical form
$$x - y - z = 0$$
$$20 x + 20 y - \frac{7071}{50} - \frac{7071 i}{50} = 0$$
$$20 x + 20 i z - \frac{7071}{50} - \frac{7071 i}{50} = 0$$
Rewrite the system of linear equations as the matrix form
$$\left[\begin{matrix}x_{1} - x_{2} - x_{3}\\20 x_{1} + 20 x_{2} + 0 x_{3}\\20 x_{1} + 0 x_{2} + 20 i x_{3}\end{matrix}\right] = \left[\begin{matrix}0\\\frac{7071}{50} + \frac{7071 i}{50}\\\frac{7071}{50} + \frac{7071 i}{50}\end{matrix}\right]$$
- this is the system of equations that has the form
A*x = B
Let´s find a solution of this matrix equations using Cramer´s rule:
Since the determinant of the matrix:
$$A = \operatorname{det}{\left(\left[\begin{matrix}1 & -1 & -1\\20 & 20 & 0\\20 & 0 & 20 i\end{matrix}\right] \right)} = 400 + 800 i$$
, then
The root xi is obtained by dividing the determinant of the matrix Ai. by the determinant of the matrix A.
( Ai we get it by replacement in the matrix A i-th column with column B )
$$x_{1} = \frac{\left(400 - 800 i\right) \operatorname{det}{\left(\left[\begin{matrix}0 & -1 & -1\\\frac{7071}{50} + \frac{7071 i}{50} & 20 & 0\\\frac{7071}{50} + \frac{7071 i}{50} & 0 & 20 i\end{matrix}\right] \right)}}{800000} = \frac{7071}{2000} + \frac{\left(10 - 20 i\right) \left(\frac{7071}{100} + \frac{7071 i}{100}\right)}{1000} + \frac{7071 i}{2000}$$
=
$$\frac{7071}{1250} + \frac{7071 i}{2500}$$
$$x_{2} = \frac{\left(400 - 800 i\right) \operatorname{det}{\left(\left[\begin{matrix}1 & 0 & -1\\20 & \frac{7071}{50} + \frac{7071 i}{50} & 0\\20 & \frac{7071}{50} + \frac{7071 i}{50} & 20 i\end{matrix}\right] \right)}}{800000} = \frac{7071}{2000} - \frac{\left(10 - 20 i\right) \left(\frac{7071}{100} + \frac{7071 i}{100}\right)}{1000} + \frac{7071 i}{2000}$$
=
$$\frac{7071}{5000} + \frac{21213 i}{5000}$$
$$x_{3} = \frac{\left(400 - 800 i\right) \operatorname{det}{\left(\left[\begin{matrix}1 & -1 & 0\\20 & 20 & \frac{7071}{50} + \frac{7071 i}{50}\\20 & 0 & \frac{7071}{50} + \frac{7071 i}{50}\end{matrix}\right] \right)}}{800000} = \frac{\left(10 - 20 i\right) \left(\frac{7071}{100} + \frac{7071 i}{100}\right)}{500}$$
=
$$\frac{21213}{5000} - \frac{7071 i}{5000}$$
$$20 x + 20 y = \frac{7071}{50} + \frac{7071 i}{50}$$
$$20 x + z 20 i = \frac{7071}{50} + \frac{7071 i}{50}$$
We give the system of equations to the canonical form
$$x - y - z = 0$$
$$20 x + 20 y - \frac{7071}{50} - \frac{7071 i}{50} = 0$$
$$20 x + 20 i z - \frac{7071}{50} - \frac{7071 i}{50} = 0$$
Rewrite the system of linear equations as the matrix form
$$\left[\begin{matrix}x_{1} - x_{2} - x_{3}\\20 x_{1} + 20 x_{2} + 0 x_{3}\\20 x_{1} + 0 x_{2} + 20 i x_{3}\end{matrix}\right] = \left[\begin{matrix}0\\\frac{7071}{50} + \frac{7071 i}{50}\\\frac{7071}{50} + \frac{7071 i}{50}\end{matrix}\right]$$
- this is the system of equations that has the form
A*x = B
Let´s find a solution of this matrix equations using Cramer´s rule:
Since the determinant of the matrix:
$$A = \operatorname{det}{\left(\left[\begin{matrix}1 & -1 & -1\\20 & 20 & 0\\20 & 0 & 20 i\end{matrix}\right] \right)} = 400 + 800 i$$
, then
The root xi is obtained by dividing the determinant of the matrix Ai. by the determinant of the matrix A.
( Ai we get it by replacement in the matrix A i-th column with column B )
$$x_{1} = \frac{\left(400 - 800 i\right) \operatorname{det}{\left(\left[\begin{matrix}0 & -1 & -1\\\frac{7071}{50} + \frac{7071 i}{50} & 20 & 0\\\frac{7071}{50} + \frac{7071 i}{50} & 0 & 20 i\end{matrix}\right] \right)}}{800000} = \frac{7071}{2000} + \frac{\left(10 - 20 i\right) \left(\frac{7071}{100} + \frac{7071 i}{100}\right)}{1000} + \frac{7071 i}{2000}$$
=
$$\frac{7071}{1250} + \frac{7071 i}{2500}$$
$$x_{2} = \frac{\left(400 - 800 i\right) \operatorname{det}{\left(\left[\begin{matrix}1 & 0 & -1\\20 & \frac{7071}{50} + \frac{7071 i}{50} & 0\\20 & \frac{7071}{50} + \frac{7071 i}{50} & 20 i\end{matrix}\right] \right)}}{800000} = \frac{7071}{2000} - \frac{\left(10 - 20 i\right) \left(\frac{7071}{100} + \frac{7071 i}{100}\right)}{1000} + \frac{7071 i}{2000}$$
=
$$\frac{7071}{5000} + \frac{21213 i}{5000}$$
$$x_{3} = \frac{\left(400 - 800 i\right) \operatorname{det}{\left(\left[\begin{matrix}1 & -1 & 0\\20 & 20 & \frac{7071}{50} + \frac{7071 i}{50}\\20 & 0 & \frac{7071}{50} + \frac{7071 i}{50}\end{matrix}\right] \right)}}{800000} = \frac{\left(10 - 20 i\right) \left(\frac{7071}{100} + \frac{7071 i}{100}\right)}{500}$$
=
$$\frac{21213}{5000} - \frac{7071 i}{5000}$$
Gaussian elimination
Given the system of equations
$$x = y + z$$
$$20 x + 20 y = \frac{7071}{50} + \frac{7071 i}{50}$$
$$20 x + z 20 i = \frac{7071}{50} + \frac{7071 i}{50}$$
We give the system of equations to the canonical form
$$x - y - z = 0$$
$$20 x + 20 y - \frac{7071}{50} - \frac{7071 i}{50} = 0$$
$$20 x + 20 i z - \frac{7071}{50} - \frac{7071 i}{50} = 0$$
Rewrite the system of linear equations as the matrix form
$$\left[\begin{matrix}1 & -1 & -1 & 0\\20 & 20 & 0 & \frac{7071}{50} + \frac{7071 i}{50}\\20 & 0 & 20 i & \frac{7071}{50} + \frac{7071 i}{50}\end{matrix}\right]$$
In 1 -th column
$$\left[\begin{matrix}1\\20\\20\end{matrix}\right]$$
let’s convert all the elements, except
2 -th element into zero.
- To do this, let’s take 2 -th line
$$\left[\begin{matrix}20 & 20 & 0 & \frac{7071}{50} + \frac{7071 i}{50}\end{matrix}\right]$$
,
and subtract it from other lines:
From 1 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}1 - \frac{20}{20} & -1 - \frac{20}{20} & -1 - \frac{0}{20} & - \frac{\frac{7071}{50} + \frac{7071 i}{50}}{20}\end{matrix}\right] = \left[\begin{matrix}0 & -2 & -1 & - \frac{7071}{1000} - \frac{7071 i}{1000}\end{matrix}\right]$$
you get
$$\left[\begin{matrix}0 & -2 & -1 & - \frac{7071}{1000} - \frac{7071 i}{1000}\\20 & 20 & 0 & \frac{7071}{50} + \frac{7071 i}{50}\\20 & 0 & 20 i & \frac{7071}{50} + \frac{7071 i}{50}\end{matrix}\right]$$
From 3 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}\left(-1\right) 20 + 20 & \left(-1\right) 20 & \left(-1\right) 0 + 20 i & - (\frac{7071}{50} + \frac{7071 i}{50}) + \left(\frac{7071}{50} + \frac{7071 i}{50}\right)\end{matrix}\right] = \left[\begin{matrix}0 & -20 & 20 i & 0\end{matrix}\right]$$
you get
$$\left[\begin{matrix}0 & -2 & -1 & - \frac{7071}{1000} - \frac{7071 i}{1000}\\20 & 20 & 0 & \frac{7071}{50} + \frac{7071 i}{50}\\0 & -20 & 20 i & 0\end{matrix}\right]$$
In 2 -th column
$$\left[\begin{matrix}-2\\20\\-20\end{matrix}\right]$$
let’s convert all the elements, except
1 -th element into zero.
- To do this, let’s take 1 -th line
$$\left[\begin{matrix}0 & -2 & -1 & - \frac{7071}{1000} - \frac{7071 i}{1000}\end{matrix}\right]$$
,
and subtract it from other lines:
From 2 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}20 - \left(-10\right) 0 & 20 - - -20 & - \left(-10\right) \left(-1\right) & - \left(-10\right) \left(- \frac{7071}{1000} - \frac{7071 i}{1000}\right) + \left(\frac{7071}{50} + \frac{7071 i}{50}\right)\end{matrix}\right] = \left[\begin{matrix}20 & 0 & -10 & \frac{7071}{100} + \frac{7071 i}{100}\end{matrix}\right]$$
you get
$$\left[\begin{matrix}0 & -2 & -1 & - \frac{7071}{1000} - \frac{7071 i}{1000}\\20 & 0 & -10 & \frac{7071}{100} + \frac{7071 i}{100}\\0 & -20 & 20 i & 0\end{matrix}\right]$$
From 3 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}- 0 \cdot 10 & -20 - - 20 & - \left(-1\right) 10 + 20 i & - 10 \left(- \frac{7071}{1000} - \frac{7071 i}{1000}\right)\end{matrix}\right] = \left[\begin{matrix}0 & 0 & 10 + 20 i & \frac{7071}{100} + \frac{7071 i}{100}\end{matrix}\right]$$
you get
$$\left[\begin{matrix}0 & -2 & -1 & - \frac{7071}{1000} - \frac{7071 i}{1000}\\20 & 0 & -10 & \frac{7071}{100} + \frac{7071 i}{100}\\0 & 0 & 10 + 20 i & \frac{7071}{100} + \frac{7071 i}{100}\end{matrix}\right]$$
In 3 -th column
$$\left[\begin{matrix}-1\\-10\\10 + 20 i\end{matrix}\right]$$
let’s convert all the elements, except
3 -th element into zero.
- To do this, let’s take 3 -th line
$$\left[\begin{matrix}0 & 0 & 10 + 20 i & \frac{7071}{100} + \frac{7071 i}{100}\end{matrix}\right]$$
,
and subtract it from other lines:
From 1 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}- 0 \left(- \frac{10 - 20 i}{500}\right) & -2 - 0 \left(- \frac{10 - 20 i}{500}\right) & -1 - - \frac{10 - 20 i}{500} \left(10 + 20 i\right) & \left(- \frac{7071}{1000} - \frac{7071 i}{1000}\right) - - \frac{10 - 20 i}{500} \left(\frac{7071}{100} + \frac{7071 i}{100}\right)\end{matrix}\right] = \left[\begin{matrix}0 & -2 & -1 + \frac{\left(10 - 20 i\right) \left(10 + 20 i\right)}{500} & - \frac{7071}{1000} - \frac{7071 i}{1000} + \frac{\left(10 - 20 i\right) \left(\frac{7071}{100} + \frac{7071 i}{100}\right)}{500}\end{matrix}\right]$$
you get
$$\left[\begin{matrix}0 & -2 & -1 + \frac{\left(10 - 20 i\right) \left(10 + 20 i\right)}{500} & - \frac{7071}{1000} - \frac{7071 i}{1000} + \frac{\left(10 - 20 i\right) \left(\frac{7071}{100} + \frac{7071 i}{100}\right)}{500}\\20 & 0 & -10 & \frac{7071}{100} + \frac{7071 i}{100}\\0 & 0 & 10 + 20 i & \frac{7071}{100} + \frac{7071 i}{100}\end{matrix}\right]$$
From 2 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}20 - 0 \left(- \frac{10 - 20 i}{50}\right) & - 0 \left(- \frac{10 - 20 i}{50}\right) & -10 - - \frac{10 - 20 i}{50} \left(10 + 20 i\right) & - - \frac{10 - 20 i}{50} \left(\frac{7071}{100} + \frac{7071 i}{100}\right) + \left(\frac{7071}{100} + \frac{7071 i}{100}\right)\end{matrix}\right] = \left[\begin{matrix}20 & 0 & -10 + \frac{\left(10 - 20 i\right) \left(10 + 20 i\right)}{50} & \frac{7071}{100} + \frac{\left(10 - 20 i\right) \left(\frac{7071}{100} + \frac{7071 i}{100}\right)}{50} + \frac{7071 i}{100}\end{matrix}\right]$$
you get
$$\left[\begin{matrix}0 & -2 & -1 + \frac{\left(10 - 20 i\right) \left(10 + 20 i\right)}{500} & - \frac{7071}{1000} - \frac{7071 i}{1000} + \frac{\left(10 - 20 i\right) \left(\frac{7071}{100} + \frac{7071 i}{100}\right)}{500}\\20 & 0 & -10 + \frac{\left(10 - 20 i\right) \left(10 + 20 i\right)}{50} & \frac{7071}{100} + \frac{\left(10 - 20 i\right) \left(\frac{7071}{100} + \frac{7071 i}{100}\right)}{50} + \frac{7071 i}{100}\\0 & 0 & 10 + 20 i & \frac{7071}{100} + \frac{7071 i}{100}\end{matrix}\right]$$
In 1 -th column
$$\left[\begin{matrix}0\\20\\0\end{matrix}\right]$$
let’s convert all the elements, except
2 -th element into zero.
- To do this, let’s take 2 -th line
$$\left[\begin{matrix}20 & 0 & -10 + \frac{\left(10 - 20 i\right) \left(10 + 20 i\right)}{50} & \frac{7071}{100} + \frac{\left(10 - 20 i\right) \left(\frac{7071}{100} + \frac{7071 i}{100}\right)}{50} + \frac{7071 i}{100}\end{matrix}\right]$$
,
and subtract it from other lines:
In 2 -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
$$\left[\begin{matrix}0 & -2 & -1 + \frac{\left(10 - 20 i\right) \left(10 + 20 i\right)}{500} & - \frac{7071}{1000} - \frac{7071 i}{1000} + \frac{\left(10 - 20 i\right) \left(\frac{7071}{100} + \frac{7071 i}{100}\right)}{500}\end{matrix}\right]$$
,
and subtract it from other lines:
In 3 -th column
$$\left[\begin{matrix}-1 + \frac{\left(10 - 20 i\right) \left(10 + 20 i\right)}{500}\\-10 + \frac{\left(10 - 20 i\right) \left(10 + 20 i\right)}{50}\\10 + 20 i\end{matrix}\right]$$
let’s convert all the elements, except
3 -th element into zero.
- To do this, let’s take 3 -th line
$$\left[\begin{matrix}0 & 0 & 10 + 20 i & \frac{7071}{100} + \frac{7071 i}{100}\end{matrix}\right]$$
,
and subtract it from other lines:
From 1 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}- 0 \frac{\left(-1 + \frac{\left(10 - 20 i\right) \left(10 + 20 i\right)}{500}\right) \left(10 - 20 i\right)}{500} & -2 - 0 \frac{\left(-1 + \frac{\left(10 - 20 i\right) \left(10 + 20 i\right)}{500}\right) \left(10 - 20 i\right)}{500} & - \frac{\left(-1 + \frac{\left(10 - 20 i\right) \left(10 + 20 i\right)}{500}\right) \left(10 - 20 i\right)}{500} \left(10 + 20 i\right) + \left(-1 + \frac{\left(10 - 20 i\right) \left(10 + 20 i\right)}{500}\right) & \left(- \frac{7071}{1000} - \frac{7071 i}{1000} + \frac{\left(10 - 20 i\right) \left(\frac{7071}{100} + \frac{7071 i}{100}\right)}{500}\right) - \frac{\left(-1 + \frac{\left(10 - 20 i\right) \left(10 + 20 i\right)}{500}\right) \left(10 - 20 i\right)}{500} \left(\frac{7071}{100} + \frac{7071 i}{100}\right)\end{matrix}\right] = \left[\begin{matrix}0 & -2 & -1 - \frac{\left(-1 + \frac{\left(10 - 20 i\right) \left(10 + 20 i\right)}{500}\right) \left(10 - 20 i\right) \left(10 + 20 i\right)}{500} + \frac{\left(10 - 20 i\right) \left(10 + 20 i\right)}{500} & - \frac{7071}{1000} - \frac{7071 i}{1000} + \frac{\left(10 - 20 i\right) \left(\frac{7071}{100} + \frac{7071 i}{100}\right)}{500} - \frac{\left(-1 + \frac{\left(10 - 20 i\right) \left(10 + 20 i\right)}{500}\right) \left(10 - 20 i\right) \left(\frac{7071}{100} + \frac{7071 i}{100}\right)}{500}\end{matrix}\right]$$
you get
$$\left[\begin{matrix}0 & -2 & -1 - \frac{\left(-1 + \frac{\left(10 - 20 i\right) \left(10 + 20 i\right)}{500}\right) \left(10 - 20 i\right) \left(10 + 20 i\right)}{500} + \frac{\left(10 - 20 i\right) \left(10 + 20 i\right)}{500} & - \frac{7071}{1000} - \frac{7071 i}{1000} + \frac{\left(10 - 20 i\right) \left(\frac{7071}{100} + \frac{7071 i}{100}\right)}{500} - \frac{\left(-1 + \frac{\left(10 - 20 i\right) \left(10 + 20 i\right)}{500}\right) \left(10 - 20 i\right) \left(\frac{7071}{100} + \frac{7071 i}{100}\right)}{500}\\20 & 0 & -10 + \frac{\left(10 - 20 i\right) \left(10 + 20 i\right)}{50} & \frac{7071}{100} + \frac{\left(10 - 20 i\right) \left(\frac{7071}{100} + \frac{7071 i}{100}\right)}{50} + \frac{7071 i}{100}\\0 & 0 & 10 + 20 i & \frac{7071}{100} + \frac{7071 i}{100}\end{matrix}\right]$$
From 2 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}20 - 0 \frac{\left(-10 + \frac{\left(10 - 20 i\right) \left(10 + 20 i\right)}{50}\right) \left(10 - 20 i\right)}{500} & - 0 \frac{\left(-10 + \frac{\left(10 - 20 i\right) \left(10 + 20 i\right)}{50}\right) \left(10 - 20 i\right)}{500} & - \frac{\left(-10 + \frac{\left(10 - 20 i\right) \left(10 + 20 i\right)}{50}\right) \left(10 - 20 i\right)}{500} \left(10 + 20 i\right) + \left(-10 + \frac{\left(10 - 20 i\right) \left(10 + 20 i\right)}{50}\right) & - \frac{\left(-10 + \frac{\left(10 - 20 i\right) \left(10 + 20 i\right)}{50}\right) \left(10 - 20 i\right)}{500} \left(\frac{7071}{100} + \frac{7071 i}{100}\right) + \left(\frac{7071}{100} + \frac{\left(10 - 20 i\right) \left(\frac{7071}{100} + \frac{7071 i}{100}\right)}{50} + \frac{7071 i}{100}\right)\end{matrix}\right] = \left[\begin{matrix}20 & 0 & -10 - \frac{\left(-10 + \frac{\left(10 - 20 i\right) \left(10 + 20 i\right)}{50}\right) \left(10 - 20 i\right) \left(10 + 20 i\right)}{500} + \frac{\left(10 - 20 i\right) \left(10 + 20 i\right)}{50} & \frac{7071}{100} + \frac{\left(10 - 20 i\right) \left(\frac{7071}{100} + \frac{7071 i}{100}\right)}{50} - \frac{\left(-10 + \frac{\left(10 - 20 i\right) \left(10 + 20 i\right)}{50}\right) \left(10 - 20 i\right) \left(\frac{7071}{100} + \frac{7071 i}{100}\right)}{500} + \frac{7071 i}{100}\end{matrix}\right]$$
you get
$$\left[\begin{matrix}0 & -2 & -1 - \frac{\left(-1 + \frac{\left(10 - 20 i\right) \left(10 + 20 i\right)}{500}\right) \left(10 - 20 i\right) \left(10 + 20 i\right)}{500} + \frac{\left(10 - 20 i\right) \left(10 + 20 i\right)}{500} & - \frac{7071}{1000} - \frac{7071 i}{1000} + \frac{\left(10 - 20 i\right) \left(\frac{7071}{100} + \frac{7071 i}{100}\right)}{500} - \frac{\left(-1 + \frac{\left(10 - 20 i\right) \left(10 + 20 i\right)}{500}\right) \left(10 - 20 i\right) \left(\frac{7071}{100} + \frac{7071 i}{100}\right)}{500}\\20 & 0 & -10 - \frac{\left(-10 + \frac{\left(10 - 20 i\right) \left(10 + 20 i\right)}{50}\right) \left(10 - 20 i\right) \left(10 + 20 i\right)}{500} + \frac{\left(10 - 20 i\right) \left(10 + 20 i\right)}{50} & \frac{7071}{100} + \frac{\left(10 - 20 i\right) \left(\frac{7071}{100} + \frac{7071 i}{100}\right)}{50} - \frac{\left(-10 + \frac{\left(10 - 20 i\right) \left(10 + 20 i\right)}{50}\right) \left(10 - 20 i\right) \left(\frac{7071}{100} + \frac{7071 i}{100}\right)}{500} + \frac{7071 i}{100}\\0 & 0 & 10 + 20 i & \frac{7071}{100} + \frac{7071 i}{100}\end{matrix}\right]$$
It is almost ready, all we have to do is to find variables, solving the elementary equations:
$$- 2 x_{2} + x_{3} \left(-1 - \frac{\left(-1 + \frac{\left(10 - 20 i\right) \left(10 + 20 i\right)}{500}\right) \left(10 - 20 i\right) \left(10 + 20 i\right)}{500} + \frac{\left(10 - 20 i\right) \left(10 + 20 i\right)}{500}\right) + \frac{7071}{1000} + \frac{\left(-1 + \frac{\left(10 - 20 i\right) \left(10 + 20 i\right)}{500}\right) \left(10 - 20 i\right) \left(\frac{7071}{100} + \frac{7071 i}{100}\right)}{500} - \frac{\left(10 - 20 i\right) \left(\frac{7071}{100} + \frac{7071 i}{100}\right)}{500} + \frac{7071 i}{1000} = 0$$
$$20 x_{1} + x_{3} \left(-10 - \frac{\left(-10 + \frac{\left(10 - 20 i\right) \left(10 + 20 i\right)}{50}\right) \left(10 - 20 i\right) \left(10 + 20 i\right)}{500} + \frac{\left(10 - 20 i\right) \left(10 + 20 i\right)}{50}\right) - \frac{7071}{100} - \frac{7071 i}{100} + \frac{\left(-10 + \frac{\left(10 - 20 i\right) \left(10 + 20 i\right)}{50}\right) \left(10 - 20 i\right) \left(\frac{7071}{100} + \frac{7071 i}{100}\right)}{500} - \frac{\left(10 - 20 i\right) \left(\frac{7071}{100} + \frac{7071 i}{100}\right)}{50} = 0$$
$$x_{3} \left(10 + 20 i\right) - \frac{7071}{100} - \frac{7071 i}{100} = 0$$
We get the answer:
$$x_{2} = \frac{7071}{5000} + \frac{21213 i}{5000}$$
$$x_{1} = \frac{7071}{1250} + \frac{7071 i}{2500}$$
$$x_{3} = \frac{21213}{5000} - \frac{7071 i}{5000}$$
where x3 - the free variables
$$x = y + z$$
$$20 x + 20 y = \frac{7071}{50} + \frac{7071 i}{50}$$
$$20 x + z 20 i = \frac{7071}{50} + \frac{7071 i}{50}$$
We give the system of equations to the canonical form
$$x - y - z = 0$$
$$20 x + 20 y - \frac{7071}{50} - \frac{7071 i}{50} = 0$$
$$20 x + 20 i z - \frac{7071}{50} - \frac{7071 i}{50} = 0$$
Rewrite the system of linear equations as the matrix form
$$\left[\begin{matrix}1 & -1 & -1 & 0\\20 & 20 & 0 & \frac{7071}{50} + \frac{7071 i}{50}\\20 & 0 & 20 i & \frac{7071}{50} + \frac{7071 i}{50}\end{matrix}\right]$$
In 1 -th column
$$\left[\begin{matrix}1\\20\\20\end{matrix}\right]$$
let’s convert all the elements, except
2 -th element into zero.
- To do this, let’s take 2 -th line
$$\left[\begin{matrix}20 & 20 & 0 & \frac{7071}{50} + \frac{7071 i}{50}\end{matrix}\right]$$
,
and subtract it from other lines:
From 1 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}1 - \frac{20}{20} & -1 - \frac{20}{20} & -1 - \frac{0}{20} & - \frac{\frac{7071}{50} + \frac{7071 i}{50}}{20}\end{matrix}\right] = \left[\begin{matrix}0 & -2 & -1 & - \frac{7071}{1000} - \frac{7071 i}{1000}\end{matrix}\right]$$
you get
$$\left[\begin{matrix}0 & -2 & -1 & - \frac{7071}{1000} - \frac{7071 i}{1000}\\20 & 20 & 0 & \frac{7071}{50} + \frac{7071 i}{50}\\20 & 0 & 20 i & \frac{7071}{50} + \frac{7071 i}{50}\end{matrix}\right]$$
From 3 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}\left(-1\right) 20 + 20 & \left(-1\right) 20 & \left(-1\right) 0 + 20 i & - (\frac{7071}{50} + \frac{7071 i}{50}) + \left(\frac{7071}{50} + \frac{7071 i}{50}\right)\end{matrix}\right] = \left[\begin{matrix}0 & -20 & 20 i & 0\end{matrix}\right]$$
you get
$$\left[\begin{matrix}0 & -2 & -1 & - \frac{7071}{1000} - \frac{7071 i}{1000}\\20 & 20 & 0 & \frac{7071}{50} + \frac{7071 i}{50}\\0 & -20 & 20 i & 0\end{matrix}\right]$$
In 2 -th column
$$\left[\begin{matrix}-2\\20\\-20\end{matrix}\right]$$
let’s convert all the elements, except
1 -th element into zero.
- To do this, let’s take 1 -th line
$$\left[\begin{matrix}0 & -2 & -1 & - \frac{7071}{1000} - \frac{7071 i}{1000}\end{matrix}\right]$$
,
and subtract it from other lines:
From 2 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}20 - \left(-10\right) 0 & 20 - - -20 & - \left(-10\right) \left(-1\right) & - \left(-10\right) \left(- \frac{7071}{1000} - \frac{7071 i}{1000}\right) + \left(\frac{7071}{50} + \frac{7071 i}{50}\right)\end{matrix}\right] = \left[\begin{matrix}20 & 0 & -10 & \frac{7071}{100} + \frac{7071 i}{100}\end{matrix}\right]$$
you get
$$\left[\begin{matrix}0 & -2 & -1 & - \frac{7071}{1000} - \frac{7071 i}{1000}\\20 & 0 & -10 & \frac{7071}{100} + \frac{7071 i}{100}\\0 & -20 & 20 i & 0\end{matrix}\right]$$
From 3 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}- 0 \cdot 10 & -20 - - 20 & - \left(-1\right) 10 + 20 i & - 10 \left(- \frac{7071}{1000} - \frac{7071 i}{1000}\right)\end{matrix}\right] = \left[\begin{matrix}0 & 0 & 10 + 20 i & \frac{7071}{100} + \frac{7071 i}{100}\end{matrix}\right]$$
you get
$$\left[\begin{matrix}0 & -2 & -1 & - \frac{7071}{1000} - \frac{7071 i}{1000}\\20 & 0 & -10 & \frac{7071}{100} + \frac{7071 i}{100}\\0 & 0 & 10 + 20 i & \frac{7071}{100} + \frac{7071 i}{100}\end{matrix}\right]$$
In 3 -th column
$$\left[\begin{matrix}-1\\-10\\10 + 20 i\end{matrix}\right]$$
let’s convert all the elements, except
3 -th element into zero.
- To do this, let’s take 3 -th line
$$\left[\begin{matrix}0 & 0 & 10 + 20 i & \frac{7071}{100} + \frac{7071 i}{100}\end{matrix}\right]$$
,
and subtract it from other lines:
From 1 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}- 0 \left(- \frac{10 - 20 i}{500}\right) & -2 - 0 \left(- \frac{10 - 20 i}{500}\right) & -1 - - \frac{10 - 20 i}{500} \left(10 + 20 i\right) & \left(- \frac{7071}{1000} - \frac{7071 i}{1000}\right) - - \frac{10 - 20 i}{500} \left(\frac{7071}{100} + \frac{7071 i}{100}\right)\end{matrix}\right] = \left[\begin{matrix}0 & -2 & -1 + \frac{\left(10 - 20 i\right) \left(10 + 20 i\right)}{500} & - \frac{7071}{1000} - \frac{7071 i}{1000} + \frac{\left(10 - 20 i\right) \left(\frac{7071}{100} + \frac{7071 i}{100}\right)}{500}\end{matrix}\right]$$
you get
$$\left[\begin{matrix}0 & -2 & -1 + \frac{\left(10 - 20 i\right) \left(10 + 20 i\right)}{500} & - \frac{7071}{1000} - \frac{7071 i}{1000} + \frac{\left(10 - 20 i\right) \left(\frac{7071}{100} + \frac{7071 i}{100}\right)}{500}\\20 & 0 & -10 & \frac{7071}{100} + \frac{7071 i}{100}\\0 & 0 & 10 + 20 i & \frac{7071}{100} + \frac{7071 i}{100}\end{matrix}\right]$$
From 2 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}20 - 0 \left(- \frac{10 - 20 i}{50}\right) & - 0 \left(- \frac{10 - 20 i}{50}\right) & -10 - - \frac{10 - 20 i}{50} \left(10 + 20 i\right) & - - \frac{10 - 20 i}{50} \left(\frac{7071}{100} + \frac{7071 i}{100}\right) + \left(\frac{7071}{100} + \frac{7071 i}{100}\right)\end{matrix}\right] = \left[\begin{matrix}20 & 0 & -10 + \frac{\left(10 - 20 i\right) \left(10 + 20 i\right)}{50} & \frac{7071}{100} + \frac{\left(10 - 20 i\right) \left(\frac{7071}{100} + \frac{7071 i}{100}\right)}{50} + \frac{7071 i}{100}\end{matrix}\right]$$
you get
$$\left[\begin{matrix}0 & -2 & -1 + \frac{\left(10 - 20 i\right) \left(10 + 20 i\right)}{500} & - \frac{7071}{1000} - \frac{7071 i}{1000} + \frac{\left(10 - 20 i\right) \left(\frac{7071}{100} + \frac{7071 i}{100}\right)}{500}\\20 & 0 & -10 + \frac{\left(10 - 20 i\right) \left(10 + 20 i\right)}{50} & \frac{7071}{100} + \frac{\left(10 - 20 i\right) \left(\frac{7071}{100} + \frac{7071 i}{100}\right)}{50} + \frac{7071 i}{100}\\0 & 0 & 10 + 20 i & \frac{7071}{100} + \frac{7071 i}{100}\end{matrix}\right]$$
In 1 -th column
$$\left[\begin{matrix}0\\20\\0\end{matrix}\right]$$
let’s convert all the elements, except
2 -th element into zero.
- To do this, let’s take 2 -th line
$$\left[\begin{matrix}20 & 0 & -10 + \frac{\left(10 - 20 i\right) \left(10 + 20 i\right)}{50} & \frac{7071}{100} + \frac{\left(10 - 20 i\right) \left(\frac{7071}{100} + \frac{7071 i}{100}\right)}{50} + \frac{7071 i}{100}\end{matrix}\right]$$
,
and subtract it from other lines:
In 2 -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
$$\left[\begin{matrix}0 & -2 & -1 + \frac{\left(10 - 20 i\right) \left(10 + 20 i\right)}{500} & - \frac{7071}{1000} - \frac{7071 i}{1000} + \frac{\left(10 - 20 i\right) \left(\frac{7071}{100} + \frac{7071 i}{100}\right)}{500}\end{matrix}\right]$$
,
and subtract it from other lines:
In 3 -th column
$$\left[\begin{matrix}-1 + \frac{\left(10 - 20 i\right) \left(10 + 20 i\right)}{500}\\-10 + \frac{\left(10 - 20 i\right) \left(10 + 20 i\right)}{50}\\10 + 20 i\end{matrix}\right]$$
let’s convert all the elements, except
3 -th element into zero.
- To do this, let’s take 3 -th line
$$\left[\begin{matrix}0 & 0 & 10 + 20 i & \frac{7071}{100} + \frac{7071 i}{100}\end{matrix}\right]$$
,
and subtract it from other lines:
From 1 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}- 0 \frac{\left(-1 + \frac{\left(10 - 20 i\right) \left(10 + 20 i\right)}{500}\right) \left(10 - 20 i\right)}{500} & -2 - 0 \frac{\left(-1 + \frac{\left(10 - 20 i\right) \left(10 + 20 i\right)}{500}\right) \left(10 - 20 i\right)}{500} & - \frac{\left(-1 + \frac{\left(10 - 20 i\right) \left(10 + 20 i\right)}{500}\right) \left(10 - 20 i\right)}{500} \left(10 + 20 i\right) + \left(-1 + \frac{\left(10 - 20 i\right) \left(10 + 20 i\right)}{500}\right) & \left(- \frac{7071}{1000} - \frac{7071 i}{1000} + \frac{\left(10 - 20 i\right) \left(\frac{7071}{100} + \frac{7071 i}{100}\right)}{500}\right) - \frac{\left(-1 + \frac{\left(10 - 20 i\right) \left(10 + 20 i\right)}{500}\right) \left(10 - 20 i\right)}{500} \left(\frac{7071}{100} + \frac{7071 i}{100}\right)\end{matrix}\right] = \left[\begin{matrix}0 & -2 & -1 - \frac{\left(-1 + \frac{\left(10 - 20 i\right) \left(10 + 20 i\right)}{500}\right) \left(10 - 20 i\right) \left(10 + 20 i\right)}{500} + \frac{\left(10 - 20 i\right) \left(10 + 20 i\right)}{500} & - \frac{7071}{1000} - \frac{7071 i}{1000} + \frac{\left(10 - 20 i\right) \left(\frac{7071}{100} + \frac{7071 i}{100}\right)}{500} - \frac{\left(-1 + \frac{\left(10 - 20 i\right) \left(10 + 20 i\right)}{500}\right) \left(10 - 20 i\right) \left(\frac{7071}{100} + \frac{7071 i}{100}\right)}{500}\end{matrix}\right]$$
you get
$$\left[\begin{matrix}0 & -2 & -1 - \frac{\left(-1 + \frac{\left(10 - 20 i\right) \left(10 + 20 i\right)}{500}\right) \left(10 - 20 i\right) \left(10 + 20 i\right)}{500} + \frac{\left(10 - 20 i\right) \left(10 + 20 i\right)}{500} & - \frac{7071}{1000} - \frac{7071 i}{1000} + \frac{\left(10 - 20 i\right) \left(\frac{7071}{100} + \frac{7071 i}{100}\right)}{500} - \frac{\left(-1 + \frac{\left(10 - 20 i\right) \left(10 + 20 i\right)}{500}\right) \left(10 - 20 i\right) \left(\frac{7071}{100} + \frac{7071 i}{100}\right)}{500}\\20 & 0 & -10 + \frac{\left(10 - 20 i\right) \left(10 + 20 i\right)}{50} & \frac{7071}{100} + \frac{\left(10 - 20 i\right) \left(\frac{7071}{100} + \frac{7071 i}{100}\right)}{50} + \frac{7071 i}{100}\\0 & 0 & 10 + 20 i & \frac{7071}{100} + \frac{7071 i}{100}\end{matrix}\right]$$
From 2 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}20 - 0 \frac{\left(-10 + \frac{\left(10 - 20 i\right) \left(10 + 20 i\right)}{50}\right) \left(10 - 20 i\right)}{500} & - 0 \frac{\left(-10 + \frac{\left(10 - 20 i\right) \left(10 + 20 i\right)}{50}\right) \left(10 - 20 i\right)}{500} & - \frac{\left(-10 + \frac{\left(10 - 20 i\right) \left(10 + 20 i\right)}{50}\right) \left(10 - 20 i\right)}{500} \left(10 + 20 i\right) + \left(-10 + \frac{\left(10 - 20 i\right) \left(10 + 20 i\right)}{50}\right) & - \frac{\left(-10 + \frac{\left(10 - 20 i\right) \left(10 + 20 i\right)}{50}\right) \left(10 - 20 i\right)}{500} \left(\frac{7071}{100} + \frac{7071 i}{100}\right) + \left(\frac{7071}{100} + \frac{\left(10 - 20 i\right) \left(\frac{7071}{100} + \frac{7071 i}{100}\right)}{50} + \frac{7071 i}{100}\right)\end{matrix}\right] = \left[\begin{matrix}20 & 0 & -10 - \frac{\left(-10 + \frac{\left(10 - 20 i\right) \left(10 + 20 i\right)}{50}\right) \left(10 - 20 i\right) \left(10 + 20 i\right)}{500} + \frac{\left(10 - 20 i\right) \left(10 + 20 i\right)}{50} & \frac{7071}{100} + \frac{\left(10 - 20 i\right) \left(\frac{7071}{100} + \frac{7071 i}{100}\right)}{50} - \frac{\left(-10 + \frac{\left(10 - 20 i\right) \left(10 + 20 i\right)}{50}\right) \left(10 - 20 i\right) \left(\frac{7071}{100} + \frac{7071 i}{100}\right)}{500} + \frac{7071 i}{100}\end{matrix}\right]$$
you get
$$\left[\begin{matrix}0 & -2 & -1 - \frac{\left(-1 + \frac{\left(10 - 20 i\right) \left(10 + 20 i\right)}{500}\right) \left(10 - 20 i\right) \left(10 + 20 i\right)}{500} + \frac{\left(10 - 20 i\right) \left(10 + 20 i\right)}{500} & - \frac{7071}{1000} - \frac{7071 i}{1000} + \frac{\left(10 - 20 i\right) \left(\frac{7071}{100} + \frac{7071 i}{100}\right)}{500} - \frac{\left(-1 + \frac{\left(10 - 20 i\right) \left(10 + 20 i\right)}{500}\right) \left(10 - 20 i\right) \left(\frac{7071}{100} + \frac{7071 i}{100}\right)}{500}\\20 & 0 & -10 - \frac{\left(-10 + \frac{\left(10 - 20 i\right) \left(10 + 20 i\right)}{50}\right) \left(10 - 20 i\right) \left(10 + 20 i\right)}{500} + \frac{\left(10 - 20 i\right) \left(10 + 20 i\right)}{50} & \frac{7071}{100} + \frac{\left(10 - 20 i\right) \left(\frac{7071}{100} + \frac{7071 i}{100}\right)}{50} - \frac{\left(-10 + \frac{\left(10 - 20 i\right) \left(10 + 20 i\right)}{50}\right) \left(10 - 20 i\right) \left(\frac{7071}{100} + \frac{7071 i}{100}\right)}{500} + \frac{7071 i}{100}\\0 & 0 & 10 + 20 i & \frac{7071}{100} + \frac{7071 i}{100}\end{matrix}\right]$$
It is almost ready, all we have to do is to find variables, solving the elementary equations:
$$- 2 x_{2} + x_{3} \left(-1 - \frac{\left(-1 + \frac{\left(10 - 20 i\right) \left(10 + 20 i\right)}{500}\right) \left(10 - 20 i\right) \left(10 + 20 i\right)}{500} + \frac{\left(10 - 20 i\right) \left(10 + 20 i\right)}{500}\right) + \frac{7071}{1000} + \frac{\left(-1 + \frac{\left(10 - 20 i\right) \left(10 + 20 i\right)}{500}\right) \left(10 - 20 i\right) \left(\frac{7071}{100} + \frac{7071 i}{100}\right)}{500} - \frac{\left(10 - 20 i\right) \left(\frac{7071}{100} + \frac{7071 i}{100}\right)}{500} + \frac{7071 i}{1000} = 0$$
$$20 x_{1} + x_{3} \left(-10 - \frac{\left(-10 + \frac{\left(10 - 20 i\right) \left(10 + 20 i\right)}{50}\right) \left(10 - 20 i\right) \left(10 + 20 i\right)}{500} + \frac{\left(10 - 20 i\right) \left(10 + 20 i\right)}{50}\right) - \frac{7071}{100} - \frac{7071 i}{100} + \frac{\left(-10 + \frac{\left(10 - 20 i\right) \left(10 + 20 i\right)}{50}\right) \left(10 - 20 i\right) \left(\frac{7071}{100} + \frac{7071 i}{100}\right)}{500} - \frac{\left(10 - 20 i\right) \left(\frac{7071}{100} + \frac{7071 i}{100}\right)}{50} = 0$$
$$x_{3} \left(10 + 20 i\right) - \frac{7071}{100} - \frac{7071 i}{100} = 0$$
We get the answer:
$$x_{2} = \frac{7071}{5000} + \frac{21213 i}{5000}$$
$$x_{1} = \frac{7071}{1250} + \frac{7071 i}{2500}$$
$$x_{3} = \frac{21213}{5000} - \frac{7071 i}{5000}$$
where x3 - the free variables
Numerical answer
[src]
x1 = 5.6568 + 2.8284*i y1 = 1.4142 + 4.2426*i z1 = 4.2426 - 1.4142*i
x1 = 5.6568 + 2.8284*i y1 = 1.4142 + 4.2426*i z1 = 4.2426 - 1.4142*i