Mister Exam
5x+6y-2z=18; 2x+5y-3z=4; 4x-3y+2z=9
The solution
You have entered
[src]
5*x + 6*y - 2*z = 18
$$- 2 z + \left(5 x + 6 y\right) = 18$$
2*x + 5*y - 3*z = 4
$$- 3 z + \left(2 x + 5 y\right) = 4$$
4*x - 3*y + 2*z = 9
$$2 z + \left(4 x - 3 y\right) = 9$$
2*z + 4*x - 3*y = 9
Rapid solution
$$x_{1} = 2$$
=
$$2$$
=
$$y_{1} = 3$$
=
$$3$$
=
$$z_{1} = 5$$
=
$$5$$
=
=
$$2$$
=
2
$$y_{1} = 3$$
=
$$3$$
=
3
$$z_{1} = 5$$
=
$$5$$
=
5
Gaussian elimination
Given the system of equations
$$- 2 z + \left(5 x + 6 y\right) = 18$$
$$- 3 z + \left(2 x + 5 y\right) = 4$$
$$2 z + \left(4 x - 3 y\right) = 9$$
We give the system of equations to the canonical form
$$5 x + 6 y - 2 z = 18$$
$$2 x + 5 y - 3 z = 4$$
$$4 x - 3 y + 2 z = 9$$
Rewrite the system of linear equations as the matrix form
$$\left[\begin{matrix}5 & 6 & -2 & 18\\2 & 5 & -3 & 4\\4 & -3 & 2 & 9\end{matrix}\right]$$
In 1 -th column
$$\left[\begin{matrix}5\\2\\4\end{matrix}\right]$$
let’s convert all the elements, except
1 -th element into zero.
- To do this, let’s take 1 -th line
$$\left[\begin{matrix}5 & 6 & -2 & 18\end{matrix}\right]$$
,
and subtract it from other lines:
From 2 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}2 - \frac{2 \cdot 5}{5} & 5 - \frac{2 \cdot 6}{5} & -3 - - \frac{4}{5} & 4 - \frac{2 \cdot 18}{5}\end{matrix}\right] = \left[\begin{matrix}0 & \frac{13}{5} & - \frac{11}{5} & - \frac{16}{5}\end{matrix}\right]$$
you get
$$\left[\begin{matrix}5 & 6 & -2 & 18\\0 & \frac{13}{5} & - \frac{11}{5} & - \frac{16}{5}\\4 & -3 & 2 & 9\end{matrix}\right]$$
From 3 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}4 - \frac{4 \cdot 5}{5} & - \frac{4 \cdot 6}{5} - 3 & 2 - - \frac{8}{5} & 9 - \frac{4 \cdot 18}{5}\end{matrix}\right] = \left[\begin{matrix}0 & - \frac{39}{5} & \frac{18}{5} & - \frac{27}{5}\end{matrix}\right]$$
you get
$$\left[\begin{matrix}5 & 6 & -2 & 18\\0 & \frac{13}{5} & - \frac{11}{5} & - \frac{16}{5}\\0 & - \frac{39}{5} & \frac{18}{5} & - \frac{27}{5}\end{matrix}\right]$$
In 2 -th column
$$\left[\begin{matrix}6\\\frac{13}{5}\\- \frac{39}{5}\end{matrix}\right]$$
let’s convert all the elements, except
2 -th element into zero.
- To do this, let’s take 2 -th line
$$\left[\begin{matrix}0 & \frac{13}{5} & - \frac{11}{5} & - \frac{16}{5}\end{matrix}\right]$$
,
and subtract it from other lines:
From 1 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}5 - \frac{0 \cdot 30}{13} & 6 - \frac{13 \cdot 30}{5 \cdot 13} & -2 - - \frac{66}{13} & 18 - - \frac{96}{13}\end{matrix}\right] = \left[\begin{matrix}5 & 0 & \frac{40}{13} & \frac{330}{13}\end{matrix}\right]$$
you get
$$\left[\begin{matrix}5 & 0 & \frac{40}{13} & \frac{330}{13}\\0 & \frac{13}{5} & - \frac{11}{5} & - \frac{16}{5}\\0 & - \frac{39}{5} & \frac{18}{5} & - \frac{27}{5}\end{matrix}\right]$$
From 3 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}- \left(-3\right) 0 & - \frac{39}{5} - - \frac{39}{5} & \frac{18}{5} - - \frac{-33}{5} & - \frac{\left(-16\right) \left(-1\right) 3}{5} - \frac{27}{5}\end{matrix}\right] = \left[\begin{matrix}0 & 0 & -3 & -15\end{matrix}\right]$$
you get
$$\left[\begin{matrix}5 & 0 & \frac{40}{13} & \frac{330}{13}\\0 & \frac{13}{5} & - \frac{11}{5} & - \frac{16}{5}\\0 & 0 & -3 & -15\end{matrix}\right]$$
In 3 -th column
$$\left[\begin{matrix}\frac{40}{13}\\- \frac{11}{5}\\-3\end{matrix}\right]$$
let’s convert all the elements, except
3 -th element into zero.
- To do this, let’s take 3 -th line
$$\left[\begin{matrix}0 & 0 & -3 & -15\end{matrix}\right]$$
,
and subtract it from other lines:
From 1 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}5 - \frac{\left(-40\right) 0}{39} & - \frac{\left(-40\right) 0}{39} & \frac{40}{13} - - \frac{-40}{13} & \frac{330}{13} - - \frac{-200}{13}\end{matrix}\right] = \left[\begin{matrix}5 & 0 & 0 & 10\end{matrix}\right]$$
you get
$$\left[\begin{matrix}5 & 0 & 0 & 10\\0 & \frac{13}{5} & - \frac{11}{5} & - \frac{16}{5}\\0 & 0 & -3 & -15\end{matrix}\right]$$
From 2 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}- \frac{0 \cdot 11}{15} & \frac{13}{5} - \frac{0 \cdot 11}{15} & - \frac{11}{5} - - \frac{11}{5} & - \frac{16}{5} - - 11\end{matrix}\right] = \left[\begin{matrix}0 & \frac{13}{5} & 0 & \frac{39}{5}\end{matrix}\right]$$
you get
$$\left[\begin{matrix}5 & 0 & 0 & 10\\0 & \frac{13}{5} & 0 & \frac{39}{5}\\0 & 0 & -3 & -15\end{matrix}\right]$$
It is almost ready, all we have to do is to find variables, solving the elementary equations:
$$5 x_{1} - 10 = 0$$
$$\frac{13 x_{2}}{5} - \frac{39}{5} = 0$$
$$15 - 3 x_{3} = 0$$
We get the answer:
$$x_{1} = 2$$
$$x_{2} = 3$$
$$x_{3} = 5$$
$$- 2 z + \left(5 x + 6 y\right) = 18$$
$$- 3 z + \left(2 x + 5 y\right) = 4$$
$$2 z + \left(4 x - 3 y\right) = 9$$
We give the system of equations to the canonical form
$$5 x + 6 y - 2 z = 18$$
$$2 x + 5 y - 3 z = 4$$
$$4 x - 3 y + 2 z = 9$$
Rewrite the system of linear equations as the matrix form
$$\left[\begin{matrix}5 & 6 & -2 & 18\\2 & 5 & -3 & 4\\4 & -3 & 2 & 9\end{matrix}\right]$$
In 1 -th column
$$\left[\begin{matrix}5\\2\\4\end{matrix}\right]$$
let’s convert all the elements, except
1 -th element into zero.
- To do this, let’s take 1 -th line
$$\left[\begin{matrix}5 & 6 & -2 & 18\end{matrix}\right]$$
,
and subtract it from other lines:
From 2 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}2 - \frac{2 \cdot 5}{5} & 5 - \frac{2 \cdot 6}{5} & -3 - - \frac{4}{5} & 4 - \frac{2 \cdot 18}{5}\end{matrix}\right] = \left[\begin{matrix}0 & \frac{13}{5} & - \frac{11}{5} & - \frac{16}{5}\end{matrix}\right]$$
you get
$$\left[\begin{matrix}5 & 6 & -2 & 18\\0 & \frac{13}{5} & - \frac{11}{5} & - \frac{16}{5}\\4 & -3 & 2 & 9\end{matrix}\right]$$
From 3 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}4 - \frac{4 \cdot 5}{5} & - \frac{4 \cdot 6}{5} - 3 & 2 - - \frac{8}{5} & 9 - \frac{4 \cdot 18}{5}\end{matrix}\right] = \left[\begin{matrix}0 & - \frac{39}{5} & \frac{18}{5} & - \frac{27}{5}\end{matrix}\right]$$
you get
$$\left[\begin{matrix}5 & 6 & -2 & 18\\0 & \frac{13}{5} & - \frac{11}{5} & - \frac{16}{5}\\0 & - \frac{39}{5} & \frac{18}{5} & - \frac{27}{5}\end{matrix}\right]$$
In 2 -th column
$$\left[\begin{matrix}6\\\frac{13}{5}\\- \frac{39}{5}\end{matrix}\right]$$
let’s convert all the elements, except
2 -th element into zero.
- To do this, let’s take 2 -th line
$$\left[\begin{matrix}0 & \frac{13}{5} & - \frac{11}{5} & - \frac{16}{5}\end{matrix}\right]$$
,
and subtract it from other lines:
From 1 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}5 - \frac{0 \cdot 30}{13} & 6 - \frac{13 \cdot 30}{5 \cdot 13} & -2 - - \frac{66}{13} & 18 - - \frac{96}{13}\end{matrix}\right] = \left[\begin{matrix}5 & 0 & \frac{40}{13} & \frac{330}{13}\end{matrix}\right]$$
you get
$$\left[\begin{matrix}5 & 0 & \frac{40}{13} & \frac{330}{13}\\0 & \frac{13}{5} & - \frac{11}{5} & - \frac{16}{5}\\0 & - \frac{39}{5} & \frac{18}{5} & - \frac{27}{5}\end{matrix}\right]$$
From 3 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}- \left(-3\right) 0 & - \frac{39}{5} - - \frac{39}{5} & \frac{18}{5} - - \frac{-33}{5} & - \frac{\left(-16\right) \left(-1\right) 3}{5} - \frac{27}{5}\end{matrix}\right] = \left[\begin{matrix}0 & 0 & -3 & -15\end{matrix}\right]$$
you get
$$\left[\begin{matrix}5 & 0 & \frac{40}{13} & \frac{330}{13}\\0 & \frac{13}{5} & - \frac{11}{5} & - \frac{16}{5}\\0 & 0 & -3 & -15\end{matrix}\right]$$
In 3 -th column
$$\left[\begin{matrix}\frac{40}{13}\\- \frac{11}{5}\\-3\end{matrix}\right]$$
let’s convert all the elements, except
3 -th element into zero.
- To do this, let’s take 3 -th line
$$\left[\begin{matrix}0 & 0 & -3 & -15\end{matrix}\right]$$
,
and subtract it from other lines:
From 1 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}5 - \frac{\left(-40\right) 0}{39} & - \frac{\left(-40\right) 0}{39} & \frac{40}{13} - - \frac{-40}{13} & \frac{330}{13} - - \frac{-200}{13}\end{matrix}\right] = \left[\begin{matrix}5 & 0 & 0 & 10\end{matrix}\right]$$
you get
$$\left[\begin{matrix}5 & 0 & 0 & 10\\0 & \frac{13}{5} & - \frac{11}{5} & - \frac{16}{5}\\0 & 0 & -3 & -15\end{matrix}\right]$$
From 2 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}- \frac{0 \cdot 11}{15} & \frac{13}{5} - \frac{0 \cdot 11}{15} & - \frac{11}{5} - - \frac{11}{5} & - \frac{16}{5} - - 11\end{matrix}\right] = \left[\begin{matrix}0 & \frac{13}{5} & 0 & \frac{39}{5}\end{matrix}\right]$$
you get
$$\left[\begin{matrix}5 & 0 & 0 & 10\\0 & \frac{13}{5} & 0 & \frac{39}{5}\\0 & 0 & -3 & -15\end{matrix}\right]$$
It is almost ready, all we have to do is to find variables, solving the elementary equations:
$$5 x_{1} - 10 = 0$$
$$\frac{13 x_{2}}{5} - \frac{39}{5} = 0$$
$$15 - 3 x_{3} = 0$$
We get the answer:
$$x_{1} = 2$$
$$x_{2} = 3$$
$$x_{3} = 5$$
Cramer's rule
$$- 2 z + \left(5 x + 6 y\right) = 18$$
$$- 3 z + \left(2 x + 5 y\right) = 4$$
$$2 z + \left(4 x - 3 y\right) = 9$$
We give the system of equations to the canonical form
$$5 x + 6 y - 2 z = 18$$
$$2 x + 5 y - 3 z = 4$$
$$4 x - 3 y + 2 z = 9$$
Rewrite the system of linear equations as the matrix form
$$\left[\begin{matrix}5 x_{1} + 6 x_{2} - 2 x_{3}\\2 x_{1} + 5 x_{2} - 3 x_{3}\\4 x_{1} - 3 x_{2} + 2 x_{3}\end{matrix}\right] = \left[\begin{matrix}18\\4\\9\end{matrix}\right]$$
- 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:
$$A = \operatorname{det}{\left(\left[\begin{matrix}5 & 6 & -2\\2 & 5 & -3\\4 & -3 & 2\end{matrix}\right] \right)} = -39$$
, 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 )
$$x_{1} = - \frac{\operatorname{det}{\left(\left[\begin{matrix}18 & 6 & -2\\4 & 5 & -3\\9 & -3 & 2\end{matrix}\right] \right)}}{39} = 2$$
$$x_{2} = - \frac{\operatorname{det}{\left(\left[\begin{matrix}5 & 18 & -2\\2 & 4 & -3\\4 & 9 & 2\end{matrix}\right] \right)}}{39} = 3$$
$$x_{3} = - \frac{\operatorname{det}{\left(\left[\begin{matrix}5 & 6 & 18\\2 & 5 & 4\\4 & -3 & 9\end{matrix}\right] \right)}}{39} = 5$$
$$- 3 z + \left(2 x + 5 y\right) = 4$$
$$2 z + \left(4 x - 3 y\right) = 9$$
We give the system of equations to the canonical form
$$5 x + 6 y - 2 z = 18$$
$$2 x + 5 y - 3 z = 4$$
$$4 x - 3 y + 2 z = 9$$
Rewrite the system of linear equations as the matrix form
$$\left[\begin{matrix}5 x_{1} + 6 x_{2} - 2 x_{3}\\2 x_{1} + 5 x_{2} - 3 x_{3}\\4 x_{1} - 3 x_{2} + 2 x_{3}\end{matrix}\right] = \left[\begin{matrix}18\\4\\9\end{matrix}\right]$$
- 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:
$$A = \operatorname{det}{\left(\left[\begin{matrix}5 & 6 & -2\\2 & 5 & -3\\4 & -3 & 2\end{matrix}\right] \right)} = -39$$
, 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 )
$$x_{1} = - \frac{\operatorname{det}{\left(\left[\begin{matrix}18 & 6 & -2\\4 & 5 & -3\\9 & -3 & 2\end{matrix}\right] \right)}}{39} = 2$$
$$x_{2} = - \frac{\operatorname{det}{\left(\left[\begin{matrix}5 & 18 & -2\\2 & 4 & -3\\4 & 9 & 2\end{matrix}\right] \right)}}{39} = 3$$
$$x_{3} = - \frac{\operatorname{det}{\left(\left[\begin{matrix}5 & 6 & 18\\2 & 5 & 4\\4 & -3 & 9\end{matrix}\right] \right)}}{39} = 5$$