Mister Exam
7х-5у=31; 4х+11z=-43; 2x+3y+4z=-20
The solution
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7*x - 5*y = 31
$$7 x - 5 y = 31$$
4*x + 11*z = -43
$$4 x + 11 z = -43$$
2*x + 3*y + 4*z = -20
$$4 z + \left(2 x + 3 y\right) = -20$$
4*z + 2*x + 3*y = -20
Rapid solution
$$x_{1} = 3$$
=
$$3$$
=
$$y_{1} = -2$$
=
$$-2$$
=
$$z_{1} = -5$$
=
$$-5$$
=
=
$$3$$
=
3
$$y_{1} = -2$$
=
$$-2$$
=
-2
$$z_{1} = -5$$
=
$$-5$$
=
-5
Gaussian elimination
Given the system of equations
$$7 x - 5 y = 31$$
$$4 x + 11 z = -43$$
$$4 z + \left(2 x + 3 y\right) = -20$$
We give the system of equations to the canonical form
$$7 x - 5 y = 31$$
$$4 x + 11 z = -43$$
$$2 x + 3 y + 4 z = -20$$
Rewrite the system of linear equations as the matrix form
$$\left[\begin{matrix}7 & -5 & 0 & 31\\4 & 0 & 11 & -43\\2 & 3 & 4 & -20\end{matrix}\right]$$
In 1 -th column
$$\left[\begin{matrix}7\\4\\2\end{matrix}\right]$$
let’s convert all the elements, except
1 -th element into zero.
- To do this, let’s take 1 -th line
$$\left[\begin{matrix}7 & -5 & 0 & 31\end{matrix}\right]$$
,
and subtract it from other lines:
From 2 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}4 - \frac{4 \cdot 7}{7} & - \frac{\left(-1\right) 4 \cdot 5}{7} & 11 - \frac{0 \cdot 4}{7} & -43 - \frac{4 \cdot 31}{7}\end{matrix}\right] = \left[\begin{matrix}0 & \frac{20}{7} & 11 & - \frac{425}{7}\end{matrix}\right]$$
you get
$$\left[\begin{matrix}7 & -5 & 0 & 31\\0 & \frac{20}{7} & 11 & - \frac{425}{7}\\2 & 3 & 4 & -20\end{matrix}\right]$$
From 3 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}2 - \frac{2 \cdot 7}{7} & 3 - - \frac{10}{7} & 4 - \frac{0 \cdot 2}{7} & -20 - \frac{2 \cdot 31}{7}\end{matrix}\right] = \left[\begin{matrix}0 & \frac{31}{7} & 4 & - \frac{202}{7}\end{matrix}\right]$$
you get
$$\left[\begin{matrix}7 & -5 & 0 & 31\\0 & \frac{20}{7} & 11 & - \frac{425}{7}\\0 & \frac{31}{7} & 4 & - \frac{202}{7}\end{matrix}\right]$$
In 2 -th column
$$\left[\begin{matrix}-5\\\frac{20}{7}\\\frac{31}{7}\end{matrix}\right]$$
let’s convert all the elements, except
1 -th element into zero.
- To do this, let’s take 1 -th line
$$\left[\begin{matrix}7 & -5 & 0 & 31\end{matrix}\right]$$
,
and subtract it from other lines:
From 2 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}- \frac{\left(-4\right) 7}{7} & \frac{20}{7} - - \frac{-20}{7} & 11 - \frac{\left(-4\right) 0}{7} & - \frac{425}{7} - \frac{\left(-4\right) 31}{7}\end{matrix}\right] = \left[\begin{matrix}4 & 0 & 11 & -43\end{matrix}\right]$$
you get
$$\left[\begin{matrix}7 & -5 & 0 & 31\\4 & 0 & 11 & -43\\0 & \frac{31}{7} & 4 & - \frac{202}{7}\end{matrix}\right]$$
From 3 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}- \frac{\left(-31\right) 7}{35} & \frac{31}{7} - - \frac{-31}{7} & 4 - \frac{\left(-31\right) 0}{35} & - \frac{202}{7} - \frac{\left(-31\right) 31}{35}\end{matrix}\right] = \left[\begin{matrix}\frac{31}{5} & 0 & 4 & - \frac{7}{5}\end{matrix}\right]$$
you get
$$\left[\begin{matrix}7 & -5 & 0 & 31\\4 & 0 & 11 & -43\\\frac{31}{5} & 0 & 4 & - \frac{7}{5}\end{matrix}\right]$$
In 3 -th column
$$\left[\begin{matrix}0\\11\\4\end{matrix}\right]$$
let’s convert all the elements, except
2 -th element into zero.
- To do this, let’s take 2 -th line
$$\left[\begin{matrix}4 & 0 & 11 & -43\end{matrix}\right]$$
,
and subtract it from other lines:
From 3 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}\frac{31}{5} - \frac{4 \cdot 4}{11} & - \frac{0 \cdot 4}{11} & 4 - \frac{4 \cdot 11}{11} & - \frac{7}{5} - - \frac{172}{11}\end{matrix}\right] = \left[\begin{matrix}\frac{261}{55} & 0 & 0 & \frac{783}{55}\end{matrix}\right]$$
you get
$$\left[\begin{matrix}7 & -5 & 0 & 31\\4 & 0 & 11 & -43\\\frac{261}{55} & 0 & 0 & \frac{783}{55}\end{matrix}\right]$$
In 1 -th column
$$\left[\begin{matrix}7\\4\\\frac{261}{55}\end{matrix}\right]$$
let’s convert all the elements, except
3 -th element into zero.
- To do this, let’s take 3 -th line
$$\left[\begin{matrix}\frac{261}{55} & 0 & 0 & \frac{783}{55}\end{matrix}\right]$$
,
and subtract it from other lines:
From 1 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}7 - \frac{261 \cdot 385}{55 \cdot 261} & -5 - \frac{0 \cdot 385}{261} & - \frac{0 \cdot 385}{261} & 31 - \frac{385 \cdot 783}{55 \cdot 261}\end{matrix}\right] = \left[\begin{matrix}0 & -5 & 0 & 10\end{matrix}\right]$$
you get
$$\left[\begin{matrix}0 & -5 & 0 & 10\\4 & 0 & 11 & -43\\\frac{261}{55} & 0 & 0 & \frac{783}{55}\end{matrix}\right]$$
From 2 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}4 - \frac{220 \cdot 261}{55 \cdot 261} & - \frac{0 \cdot 220}{261} & 11 - \frac{0 \cdot 220}{261} & -43 - \frac{220 \cdot 783}{55 \cdot 261}\end{matrix}\right] = \left[\begin{matrix}0 & 0 & 11 & -55\end{matrix}\right]$$
you get
$$\left[\begin{matrix}0 & -5 & 0 & 10\\0 & 0 & 11 & -55\\\frac{261}{55} & 0 & 0 & \frac{783}{55}\end{matrix}\right]$$
It is almost ready, all we have to do is to find variables, solving the elementary equations:
$$- 5 x_{2} - 10 = 0$$
$$11 x_{3} + 55 = 0$$
$$\frac{261 x_{1}}{55} - \frac{783}{55} = 0$$
We get the answer:
$$x_{2} = -2$$
$$x_{3} = -5$$
$$x_{1} = 3$$
$$7 x - 5 y = 31$$
$$4 x + 11 z = -43$$
$$4 z + \left(2 x + 3 y\right) = -20$$
We give the system of equations to the canonical form
$$7 x - 5 y = 31$$
$$4 x + 11 z = -43$$
$$2 x + 3 y + 4 z = -20$$
Rewrite the system of linear equations as the matrix form
$$\left[\begin{matrix}7 & -5 & 0 & 31\\4 & 0 & 11 & -43\\2 & 3 & 4 & -20\end{matrix}\right]$$
In 1 -th column
$$\left[\begin{matrix}7\\4\\2\end{matrix}\right]$$
let’s convert all the elements, except
1 -th element into zero.
- To do this, let’s take 1 -th line
$$\left[\begin{matrix}7 & -5 & 0 & 31\end{matrix}\right]$$
,
and subtract it from other lines:
From 2 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}4 - \frac{4 \cdot 7}{7} & - \frac{\left(-1\right) 4 \cdot 5}{7} & 11 - \frac{0 \cdot 4}{7} & -43 - \frac{4 \cdot 31}{7}\end{matrix}\right] = \left[\begin{matrix}0 & \frac{20}{7} & 11 & - \frac{425}{7}\end{matrix}\right]$$
you get
$$\left[\begin{matrix}7 & -5 & 0 & 31\\0 & \frac{20}{7} & 11 & - \frac{425}{7}\\2 & 3 & 4 & -20\end{matrix}\right]$$
From 3 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}2 - \frac{2 \cdot 7}{7} & 3 - - \frac{10}{7} & 4 - \frac{0 \cdot 2}{7} & -20 - \frac{2 \cdot 31}{7}\end{matrix}\right] = \left[\begin{matrix}0 & \frac{31}{7} & 4 & - \frac{202}{7}\end{matrix}\right]$$
you get
$$\left[\begin{matrix}7 & -5 & 0 & 31\\0 & \frac{20}{7} & 11 & - \frac{425}{7}\\0 & \frac{31}{7} & 4 & - \frac{202}{7}\end{matrix}\right]$$
In 2 -th column
$$\left[\begin{matrix}-5\\\frac{20}{7}\\\frac{31}{7}\end{matrix}\right]$$
let’s convert all the elements, except
1 -th element into zero.
- To do this, let’s take 1 -th line
$$\left[\begin{matrix}7 & -5 & 0 & 31\end{matrix}\right]$$
,
and subtract it from other lines:
From 2 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}- \frac{\left(-4\right) 7}{7} & \frac{20}{7} - - \frac{-20}{7} & 11 - \frac{\left(-4\right) 0}{7} & - \frac{425}{7} - \frac{\left(-4\right) 31}{7}\end{matrix}\right] = \left[\begin{matrix}4 & 0 & 11 & -43\end{matrix}\right]$$
you get
$$\left[\begin{matrix}7 & -5 & 0 & 31\\4 & 0 & 11 & -43\\0 & \frac{31}{7} & 4 & - \frac{202}{7}\end{matrix}\right]$$
From 3 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}- \frac{\left(-31\right) 7}{35} & \frac{31}{7} - - \frac{-31}{7} & 4 - \frac{\left(-31\right) 0}{35} & - \frac{202}{7} - \frac{\left(-31\right) 31}{35}\end{matrix}\right] = \left[\begin{matrix}\frac{31}{5} & 0 & 4 & - \frac{7}{5}\end{matrix}\right]$$
you get
$$\left[\begin{matrix}7 & -5 & 0 & 31\\4 & 0 & 11 & -43\\\frac{31}{5} & 0 & 4 & - \frac{7}{5}\end{matrix}\right]$$
In 3 -th column
$$\left[\begin{matrix}0\\11\\4\end{matrix}\right]$$
let’s convert all the elements, except
2 -th element into zero.
- To do this, let’s take 2 -th line
$$\left[\begin{matrix}4 & 0 & 11 & -43\end{matrix}\right]$$
,
and subtract it from other lines:
From 3 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}\frac{31}{5} - \frac{4 \cdot 4}{11} & - \frac{0 \cdot 4}{11} & 4 - \frac{4 \cdot 11}{11} & - \frac{7}{5} - - \frac{172}{11}\end{matrix}\right] = \left[\begin{matrix}\frac{261}{55} & 0 & 0 & \frac{783}{55}\end{matrix}\right]$$
you get
$$\left[\begin{matrix}7 & -5 & 0 & 31\\4 & 0 & 11 & -43\\\frac{261}{55} & 0 & 0 & \frac{783}{55}\end{matrix}\right]$$
In 1 -th column
$$\left[\begin{matrix}7\\4\\\frac{261}{55}\end{matrix}\right]$$
let’s convert all the elements, except
3 -th element into zero.
- To do this, let’s take 3 -th line
$$\left[\begin{matrix}\frac{261}{55} & 0 & 0 & \frac{783}{55}\end{matrix}\right]$$
,
and subtract it from other lines:
From 1 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}7 - \frac{261 \cdot 385}{55 \cdot 261} & -5 - \frac{0 \cdot 385}{261} & - \frac{0 \cdot 385}{261} & 31 - \frac{385 \cdot 783}{55 \cdot 261}\end{matrix}\right] = \left[\begin{matrix}0 & -5 & 0 & 10\end{matrix}\right]$$
you get
$$\left[\begin{matrix}0 & -5 & 0 & 10\\4 & 0 & 11 & -43\\\frac{261}{55} & 0 & 0 & \frac{783}{55}\end{matrix}\right]$$
From 2 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}4 - \frac{220 \cdot 261}{55 \cdot 261} & - \frac{0 \cdot 220}{261} & 11 - \frac{0 \cdot 220}{261} & -43 - \frac{220 \cdot 783}{55 \cdot 261}\end{matrix}\right] = \left[\begin{matrix}0 & 0 & 11 & -55\end{matrix}\right]$$
you get
$$\left[\begin{matrix}0 & -5 & 0 & 10\\0 & 0 & 11 & -55\\\frac{261}{55} & 0 & 0 & \frac{783}{55}\end{matrix}\right]$$
It is almost ready, all we have to do is to find variables, solving the elementary equations:
$$- 5 x_{2} - 10 = 0$$
$$11 x_{3} + 55 = 0$$
$$\frac{261 x_{1}}{55} - \frac{783}{55} = 0$$
We get the answer:
$$x_{2} = -2$$
$$x_{3} = -5$$
$$x_{1} = 3$$
Cramer's rule
$$7 x - 5 y = 31$$
$$4 x + 11 z = -43$$
$$4 z + \left(2 x + 3 y\right) = -20$$
We give the system of equations to the canonical form
$$7 x - 5 y = 31$$
$$4 x + 11 z = -43$$
$$2 x + 3 y + 4 z = -20$$
Rewrite the system of linear equations as the matrix form
$$\left[\begin{matrix}7 x_{1} - 5 x_{2} + 0 x_{3}\\4 x_{1} + 0 x_{2} + 11 x_{3}\\2 x_{1} + 3 x_{2} + 4 x_{3}\end{matrix}\right] = \left[\begin{matrix}31\\-43\\-20\end{matrix}\right]$$
- 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:
$$A = \operatorname{det}{\left(\left[\begin{matrix}7 & -5 & 0\\4 & 0 & 11\\2 & 3 & 4\end{matrix}\right] \right)} = -261$$
, 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 )
$$x_{1} = - \frac{\operatorname{det}{\left(\left[\begin{matrix}31 & -5 & 0\\-43 & 0 & 11\\-20 & 3 & 4\end{matrix}\right] \right)}}{261} = 3$$
$$x_{2} = - \frac{\operatorname{det}{\left(\left[\begin{matrix}7 & 31 & 0\\4 & -43 & 11\\2 & -20 & 4\end{matrix}\right] \right)}}{261} = -2$$
$$x_{3} = - \frac{\operatorname{det}{\left(\left[\begin{matrix}7 & -5 & 31\\4 & 0 & -43\\2 & 3 & -20\end{matrix}\right] \right)}}{261} = -5$$
$$4 x + 11 z = -43$$
$$4 z + \left(2 x + 3 y\right) = -20$$
We give the system of equations to the canonical form
$$7 x - 5 y = 31$$
$$4 x + 11 z = -43$$
$$2 x + 3 y + 4 z = -20$$
Rewrite the system of linear equations as the matrix form
$$\left[\begin{matrix}7 x_{1} - 5 x_{2} + 0 x_{3}\\4 x_{1} + 0 x_{2} + 11 x_{3}\\2 x_{1} + 3 x_{2} + 4 x_{3}\end{matrix}\right] = \left[\begin{matrix}31\\-43\\-20\end{matrix}\right]$$
- 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:
$$A = \operatorname{det}{\left(\left[\begin{matrix}7 & -5 & 0\\4 & 0 & 11\\2 & 3 & 4\end{matrix}\right] \right)} = -261$$
, 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 )
$$x_{1} = - \frac{\operatorname{det}{\left(\left[\begin{matrix}31 & -5 & 0\\-43 & 0 & 11\\-20 & 3 & 4\end{matrix}\right] \right)}}{261} = 3$$
$$x_{2} = - \frac{\operatorname{det}{\left(\left[\begin{matrix}7 & 31 & 0\\4 & -43 & 11\\2 & -20 & 4\end{matrix}\right] \right)}}{261} = -2$$
$$x_{3} = - \frac{\operatorname{det}{\left(\left[\begin{matrix}7 & -5 & 31\\4 & 0 & -43\\2 & 3 & -20\end{matrix}\right] \right)}}{261} = -5$$