Mister Exam
Other calculators
- Inequality Systems step by step
- Complex numbers step by step
- Differential equations Step by Step
- The system of differential equations in steps
-
How to use it?
- X+y-z=1; 8x+3y-6z=-1; 4x-y+3z=2
- 5х-7у=9; 6х+5у=-16
- x2+7xy=-6; 9y2-xy=10
-
х^2-xy=-8; y^2-xy=24
-
Identical expressions
- X+y-z= one ; 8x+3y-6z=- one ; 4x-y+3z= two
- X plus y minus z equally 1; 8x plus 3y minus 6z equally minus 1; 4x minus y plus 3z equally 2
- X plus y minus z equally one ; 8x plus 3y minus 6z equally minus one ; 4x minus y plus 3z equally two
-
Similar expressions
- X+y+z=1; 8x+3y-6z=-1; 4x-y+3z=2
- X+y-z=1; 8x+3y-6z=-1; 4x-y-3z=2
- X+y-z=1; 8x+3y+6z=-1; 4x-y+3z=2
- X+y-z=1; 8x+3y-6z=-1; 4x+y+3z=2
- X+y-z=1; 8x-3y-6z=-1; 4x-y+3z=2
- X-y-z=1; 8x+3y-6z=-1; 4x-y+3z=2
- X+y-z=1; 8x+3y-6z=+1; 4x-y+3z=2
X+y-z=1; 8x+3y-6z=-1; 4x-y+3z=2
The solution
You have entered
[src]
x + y - z = 1
$$- z + \left(x + y\right) = 1$$
8*x + 3*y - 6*z = -1
$$- 6 z + \left(8 x + 3 y\right) = -1$$
4*x - y + 3*z = 2
$$3 z + \left(4 x - y\right) = 2$$
3*z + 4*x - y = 2
Rapid solution
$$x_{1} = \frac{1}{25}$$
=
$$\frac{1}{25}$$
=
$$y_{1} = \frac{59}{25}$$
=
$$\frac{59}{25}$$
=
$$z_{1} = \frac{7}{5}$$
=
$$\frac{7}{5}$$
=
=
$$\frac{1}{25}$$
=
0.0400000000000000
$$y_{1} = \frac{59}{25}$$
=
$$\frac{59}{25}$$
=
2.36
$$z_{1} = \frac{7}{5}$$
=
$$\frac{7}{5}$$
=
1.4
Gaussian elimination
Given the system of equations
$$- z + \left(x + y\right) = 1$$
$$- 6 z + \left(8 x + 3 y\right) = -1$$
$$3 z + \left(4 x - y\right) = 2$$
We give the system of equations to the canonical form
$$x + y - z = 1$$
$$8 x + 3 y - 6 z = -1$$
$$4 x - y + 3 z = 2$$
Rewrite the system of linear equations as the matrix form
$$\left[\begin{matrix}1 & 1 & -1 & 1\\8 & 3 & -6 & -1\\4 & -1 & 3 & 2\end{matrix}\right]$$
In 1 -th column
$$\left[\begin{matrix}1\\8\\4\end{matrix}\right]$$
let’s convert all the elements, except
1 -th element into zero.
- To do this, let’s take 1 -th line
$$\left[\begin{matrix}1 & 1 & -1 & 1\end{matrix}\right]$$
,
and subtract it from other lines:
From 2 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}\left(-1\right) 8 + 8 & \left(-1\right) 8 + 3 & -6 - - 8 & \left(-1\right) 8 - 1\end{matrix}\right] = \left[\begin{matrix}0 & -5 & 2 & -9\end{matrix}\right]$$
you get
$$\left[\begin{matrix}1 & 1 & -1 & 1\\0 & -5 & 2 & -9\\4 & -1 & 3 & 2\end{matrix}\right]$$
From 3 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}\left(-1\right) 4 + 4 & \left(-1\right) 4 - 1 & 3 - - 4 & \left(-1\right) 4 + 2\end{matrix}\right] = \left[\begin{matrix}0 & -5 & 7 & -2\end{matrix}\right]$$
you get
$$\left[\begin{matrix}1 & 1 & -1 & 1\\0 & -5 & 2 & -9\\0 & -5 & 7 & -2\end{matrix}\right]$$
In 2 -th column
$$\left[\begin{matrix}1\\-5\\-5\end{matrix}\right]$$
let’s convert all the elements, except
2 -th element into zero.
- To do this, let’s take 2 -th line
$$\left[\begin{matrix}0 & -5 & 2 & -9\end{matrix}\right]$$
,
and subtract it from other lines:
From 1 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}1 - \frac{\left(-1\right) 0}{5} & 1 - - -1 & -1 - \frac{\left(-1\right) 2}{5} & 1 - - \frac{-9}{5}\end{matrix}\right] = \left[\begin{matrix}1 & 0 & - \frac{3}{5} & - \frac{4}{5}\end{matrix}\right]$$
you get
$$\left[\begin{matrix}1 & 0 & - \frac{3}{5} & - \frac{4}{5}\\0 & -5 & 2 & -9\\0 & -5 & 7 & -2\end{matrix}\right]$$
From 3 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}\left(-1\right) 0 & -5 - -5 & \left(-1\right) 2 + 7 & -2 - -9\end{matrix}\right] = \left[\begin{matrix}0 & 0 & 5 & 7\end{matrix}\right]$$
you get
$$\left[\begin{matrix}1 & 0 & - \frac{3}{5} & - \frac{4}{5}\\0 & -5 & 2 & -9\\0 & 0 & 5 & 7\end{matrix}\right]$$
In 3 -th column
$$\left[\begin{matrix}- \frac{3}{5}\\2\\5\end{matrix}\right]$$
let’s convert all the elements, except
3 -th element into zero.
- To do this, let’s take 3 -th line
$$\left[\begin{matrix}0 & 0 & 5 & 7\end{matrix}\right]$$
,
and subtract it from other lines:
From 1 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}1 - \frac{\left(-3\right) 0}{25} & - \frac{\left(-3\right) 0}{25} & - \frac{3}{5} - \frac{\left(-3\right) 5}{25} & - \frac{4}{5} - \frac{\left(-3\right) 7}{25}\end{matrix}\right] = \left[\begin{matrix}1 & 0 & 0 & \frac{1}{25}\end{matrix}\right]$$
you get
$$\left[\begin{matrix}1 & 0 & 0 & \frac{1}{25}\\0 & -5 & 2 & -9\\0 & 0 & 5 & 7\end{matrix}\right]$$
From 2 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}- \frac{0 \cdot 2}{5} & -5 - \frac{0 \cdot 2}{5} & 2 - \frac{2 \cdot 5}{5} & -9 - \frac{2 \cdot 7}{5}\end{matrix}\right] = \left[\begin{matrix}0 & -5 & 0 & - \frac{59}{5}\end{matrix}\right]$$
you get
$$\left[\begin{matrix}1 & 0 & 0 & \frac{1}{25}\\0 & -5 & 0 & - \frac{59}{5}\\0 & 0 & 5 & 7\end{matrix}\right]$$
It is almost ready, all we have to do is to find variables, solving the elementary equations:
$$x_{1} - \frac{1}{25} = 0$$
$$\frac{59}{5} - 5 x_{2} = 0$$
$$5 x_{3} - 7 = 0$$
We get the answer:
$$x_{1} = \frac{1}{25}$$
$$x_{2} = \frac{59}{25}$$
$$x_{3} = \frac{7}{5}$$
$$- z + \left(x + y\right) = 1$$
$$- 6 z + \left(8 x + 3 y\right) = -1$$
$$3 z + \left(4 x - y\right) = 2$$
We give the system of equations to the canonical form
$$x + y - z = 1$$
$$8 x + 3 y - 6 z = -1$$
$$4 x - y + 3 z = 2$$
Rewrite the system of linear equations as the matrix form
$$\left[\begin{matrix}1 & 1 & -1 & 1\\8 & 3 & -6 & -1\\4 & -1 & 3 & 2\end{matrix}\right]$$
In 1 -th column
$$\left[\begin{matrix}1\\8\\4\end{matrix}\right]$$
let’s convert all the elements, except
1 -th element into zero.
- To do this, let’s take 1 -th line
$$\left[\begin{matrix}1 & 1 & -1 & 1\end{matrix}\right]$$
,
and subtract it from other lines:
From 2 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}\left(-1\right) 8 + 8 & \left(-1\right) 8 + 3 & -6 - - 8 & \left(-1\right) 8 - 1\end{matrix}\right] = \left[\begin{matrix}0 & -5 & 2 & -9\end{matrix}\right]$$
you get
$$\left[\begin{matrix}1 & 1 & -1 & 1\\0 & -5 & 2 & -9\\4 & -1 & 3 & 2\end{matrix}\right]$$
From 3 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}\left(-1\right) 4 + 4 & \left(-1\right) 4 - 1 & 3 - - 4 & \left(-1\right) 4 + 2\end{matrix}\right] = \left[\begin{matrix}0 & -5 & 7 & -2\end{matrix}\right]$$
you get
$$\left[\begin{matrix}1 & 1 & -1 & 1\\0 & -5 & 2 & -9\\0 & -5 & 7 & -2\end{matrix}\right]$$
In 2 -th column
$$\left[\begin{matrix}1\\-5\\-5\end{matrix}\right]$$
let’s convert all the elements, except
2 -th element into zero.
- To do this, let’s take 2 -th line
$$\left[\begin{matrix}0 & -5 & 2 & -9\end{matrix}\right]$$
,
and subtract it from other lines:
From 1 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}1 - \frac{\left(-1\right) 0}{5} & 1 - - -1 & -1 - \frac{\left(-1\right) 2}{5} & 1 - - \frac{-9}{5}\end{matrix}\right] = \left[\begin{matrix}1 & 0 & - \frac{3}{5} & - \frac{4}{5}\end{matrix}\right]$$
you get
$$\left[\begin{matrix}1 & 0 & - \frac{3}{5} & - \frac{4}{5}\\0 & -5 & 2 & -9\\0 & -5 & 7 & -2\end{matrix}\right]$$
From 3 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}\left(-1\right) 0 & -5 - -5 & \left(-1\right) 2 + 7 & -2 - -9\end{matrix}\right] = \left[\begin{matrix}0 & 0 & 5 & 7\end{matrix}\right]$$
you get
$$\left[\begin{matrix}1 & 0 & - \frac{3}{5} & - \frac{4}{5}\\0 & -5 & 2 & -9\\0 & 0 & 5 & 7\end{matrix}\right]$$
In 3 -th column
$$\left[\begin{matrix}- \frac{3}{5}\\2\\5\end{matrix}\right]$$
let’s convert all the elements, except
3 -th element into zero.
- To do this, let’s take 3 -th line
$$\left[\begin{matrix}0 & 0 & 5 & 7\end{matrix}\right]$$
,
and subtract it from other lines:
From 1 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}1 - \frac{\left(-3\right) 0}{25} & - \frac{\left(-3\right) 0}{25} & - \frac{3}{5} - \frac{\left(-3\right) 5}{25} & - \frac{4}{5} - \frac{\left(-3\right) 7}{25}\end{matrix}\right] = \left[\begin{matrix}1 & 0 & 0 & \frac{1}{25}\end{matrix}\right]$$
you get
$$\left[\begin{matrix}1 & 0 & 0 & \frac{1}{25}\\0 & -5 & 2 & -9\\0 & 0 & 5 & 7\end{matrix}\right]$$
From 2 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}- \frac{0 \cdot 2}{5} & -5 - \frac{0 \cdot 2}{5} & 2 - \frac{2 \cdot 5}{5} & -9 - \frac{2 \cdot 7}{5}\end{matrix}\right] = \left[\begin{matrix}0 & -5 & 0 & - \frac{59}{5}\end{matrix}\right]$$
you get
$$\left[\begin{matrix}1 & 0 & 0 & \frac{1}{25}\\0 & -5 & 0 & - \frac{59}{5}\\0 & 0 & 5 & 7\end{matrix}\right]$$
It is almost ready, all we have to do is to find variables, solving the elementary equations:
$$x_{1} - \frac{1}{25} = 0$$
$$\frac{59}{5} - 5 x_{2} = 0$$
$$5 x_{3} - 7 = 0$$
We get the answer:
$$x_{1} = \frac{1}{25}$$
$$x_{2} = \frac{59}{25}$$
$$x_{3} = \frac{7}{5}$$
Cramer's rule
$$- z + \left(x + y\right) = 1$$
$$- 6 z + \left(8 x + 3 y\right) = -1$$
$$3 z + \left(4 x - y\right) = 2$$
We give the system of equations to the canonical form
$$x + y - z = 1$$
$$8 x + 3 y - 6 z = -1$$
$$4 x - y + 3 z = 2$$
Rewrite the system of linear equations as the matrix form
$$\left[\begin{matrix}x_{1} + x_{2} - x_{3}\\8 x_{1} + 3 x_{2} - 6 x_{3}\\4 x_{1} - x_{2} + 3 x_{3}\end{matrix}\right] = \left[\begin{matrix}1\\-1\\2\end{matrix}\right]$$
- 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:
$$A = \operatorname{det}{\left(\left[\begin{matrix}1 & 1 & -1\\8 & 3 & -6\\4 & -1 & 3\end{matrix}\right] \right)} = -25$$
, 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 )
$$x_{1} = - \frac{\operatorname{det}{\left(\left[\begin{matrix}1 & 1 & -1\\-1 & 3 & -6\\2 & -1 & 3\end{matrix}\right] \right)}}{25} = \frac{1}{25}$$
$$x_{2} = - \frac{\operatorname{det}{\left(\left[\begin{matrix}1 & 1 & -1\\8 & -1 & -6\\4 & 2 & 3\end{matrix}\right] \right)}}{25} = \frac{59}{25}$$
$$x_{3} = - \frac{\operatorname{det}{\left(\left[\begin{matrix}1 & 1 & 1\\8 & 3 & -1\\4 & -1 & 2\end{matrix}\right] \right)}}{25} = \frac{7}{5}$$
$$- 6 z + \left(8 x + 3 y\right) = -1$$
$$3 z + \left(4 x - y\right) = 2$$
We give the system of equations to the canonical form
$$x + y - z = 1$$
$$8 x + 3 y - 6 z = -1$$
$$4 x - y + 3 z = 2$$
Rewrite the system of linear equations as the matrix form
$$\left[\begin{matrix}x_{1} + x_{2} - x_{3}\\8 x_{1} + 3 x_{2} - 6 x_{3}\\4 x_{1} - x_{2} + 3 x_{3}\end{matrix}\right] = \left[\begin{matrix}1\\-1\\2\end{matrix}\right]$$
- 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:
$$A = \operatorname{det}{\left(\left[\begin{matrix}1 & 1 & -1\\8 & 3 & -6\\4 & -1 & 3\end{matrix}\right] \right)} = -25$$
, 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 )
$$x_{1} = - \frac{\operatorname{det}{\left(\left[\begin{matrix}1 & 1 & -1\\-1 & 3 & -6\\2 & -1 & 3\end{matrix}\right] \right)}}{25} = \frac{1}{25}$$
$$x_{2} = - \frac{\operatorname{det}{\left(\left[\begin{matrix}1 & 1 & -1\\8 & -1 & -6\\4 & 2 & 3\end{matrix}\right] \right)}}{25} = \frac{59}{25}$$
$$x_{3} = - \frac{\operatorname{det}{\left(\left[\begin{matrix}1 & 1 & 1\\8 & 3 & -1\\4 & -1 & 2\end{matrix}\right] \right)}}{25} = \frac{7}{5}$$