Mister Exam
2x-3y+z=2; 2x+y-4z=9; 6x-5y+2x=17
The solution
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2*x - 3*y + z = 2
$$z + \left(2 x - 3 y\right) = 2$$
2*x + y - 4*z = 9
$$- 4 z + \left(2 x + y\right) = 9$$
6*x - 5*y + 2*x = 17
$$2 x + \left(6 x - 5 y\right) = 17$$
2*x + 6*x - 5*y = 17
Rapid solution
$$x_{1} = \frac{51}{19}$$
=
$$\frac{51}{19}$$
=
$$y_{1} = \frac{17}{19}$$
=
$$\frac{17}{19}$$
=
$$z_{1} = - \frac{13}{19}$$
=
$$- \frac{13}{19}$$
=
=
$$\frac{51}{19}$$
=
2.68421052631579
$$y_{1} = \frac{17}{19}$$
=
$$\frac{17}{19}$$
=
0.894736842105263
$$z_{1} = - \frac{13}{19}$$
=
$$- \frac{13}{19}$$
=
-0.684210526315789
Cramer's rule
$$z + \left(2 x - 3 y\right) = 2$$
$$- 4 z + \left(2 x + y\right) = 9$$
$$2 x + \left(6 x - 5 y\right) = 17$$
We give the system of equations to the canonical form
$$2 x - 3 y + z = 2$$
$$2 x + y - 4 z = 9$$
$$8 x - 5 y = 17$$
Rewrite the system of linear equations as the matrix form
$$\left[\begin{matrix}2 x_{1} - 3 x_{2} + x_{3}\\2 x_{1} + x_{2} - 4 x_{3}\\8 x_{1} - 5 x_{2} + 0 x_{3}\end{matrix}\right] = \left[\begin{matrix}2\\9\\17\end{matrix}\right]$$
- 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:
$$A = \operatorname{det}{\left(\left[\begin{matrix}2 & -3 & 1\\2 & 1 & -4\\8 & -5 & 0\end{matrix}\right] \right)} = 38$$
, 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 )
$$x_{1} = \frac{\operatorname{det}{\left(\left[\begin{matrix}2 & -3 & 1\\9 & 1 & -4\\17 & -5 & 0\end{matrix}\right] \right)}}{38} = \frac{51}{19}$$
$$x_{2} = \frac{\operatorname{det}{\left(\left[\begin{matrix}2 & 2 & 1\\2 & 9 & -4\\8 & 17 & 0\end{matrix}\right] \right)}}{38} = \frac{17}{19}$$
$$x_{3} = \frac{\operatorname{det}{\left(\left[\begin{matrix}2 & -3 & 2\\2 & 1 & 9\\8 & -5 & 17\end{matrix}\right] \right)}}{38} = - \frac{13}{19}$$
$$- 4 z + \left(2 x + y\right) = 9$$
$$2 x + \left(6 x - 5 y\right) = 17$$
We give the system of equations to the canonical form
$$2 x - 3 y + z = 2$$
$$2 x + y - 4 z = 9$$
$$8 x - 5 y = 17$$
Rewrite the system of linear equations as the matrix form
$$\left[\begin{matrix}2 x_{1} - 3 x_{2} + x_{3}\\2 x_{1} + x_{2} - 4 x_{3}\\8 x_{1} - 5 x_{2} + 0 x_{3}\end{matrix}\right] = \left[\begin{matrix}2\\9\\17\end{matrix}\right]$$
- 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:
$$A = \operatorname{det}{\left(\left[\begin{matrix}2 & -3 & 1\\2 & 1 & -4\\8 & -5 & 0\end{matrix}\right] \right)} = 38$$
, 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 )
$$x_{1} = \frac{\operatorname{det}{\left(\left[\begin{matrix}2 & -3 & 1\\9 & 1 & -4\\17 & -5 & 0\end{matrix}\right] \right)}}{38} = \frac{51}{19}$$
$$x_{2} = \frac{\operatorname{det}{\left(\left[\begin{matrix}2 & 2 & 1\\2 & 9 & -4\\8 & 17 & 0\end{matrix}\right] \right)}}{38} = \frac{17}{19}$$
$$x_{3} = \frac{\operatorname{det}{\left(\left[\begin{matrix}2 & -3 & 2\\2 & 1 & 9\\8 & -5 & 17\end{matrix}\right] \right)}}{38} = - \frac{13}{19}$$
Gaussian elimination
Given the system of equations
$$z + \left(2 x - 3 y\right) = 2$$
$$- 4 z + \left(2 x + y\right) = 9$$
$$2 x + \left(6 x - 5 y\right) = 17$$
We give the system of equations to the canonical form
$$2 x - 3 y + z = 2$$
$$2 x + y - 4 z = 9$$
$$8 x - 5 y = 17$$
Rewrite the system of linear equations as the matrix form
$$\left[\begin{matrix}2 & -3 & 1 & 2\\2 & 1 & -4 & 9\\8 & -5 & 0 & 17\end{matrix}\right]$$
In 1 -th column
$$\left[\begin{matrix}2\\2\\8\end{matrix}\right]$$
let’s convert all the elements, except
3 -th element into zero.
- To do this, let’s take 3 -th line
$$\left[\begin{matrix}8 & -5 & 0 & 17\end{matrix}\right]$$
,
and subtract it from other lines:
From 1 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}2 - \frac{8}{4} & -3 - - \frac{5}{4} & 1 - \frac{0}{4} & 2 - \frac{17}{4}\end{matrix}\right] = \left[\begin{matrix}0 & - \frac{7}{4} & 1 & - \frac{9}{4}\end{matrix}\right]$$
you get
$$\left[\begin{matrix}0 & - \frac{7}{4} & 1 & - \frac{9}{4}\\2 & 1 & -4 & 9\\8 & -5 & 0 & 17\end{matrix}\right]$$
From 2 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}2 - \frac{8}{4} & 1 - - \frac{5}{4} & -4 - \frac{0}{4} & 9 - \frac{17}{4}\end{matrix}\right] = \left[\begin{matrix}0 & \frac{9}{4} & -4 & \frac{19}{4}\end{matrix}\right]$$
you get
$$\left[\begin{matrix}0 & - \frac{7}{4} & 1 & - \frac{9}{4}\\0 & \frac{9}{4} & -4 & \frac{19}{4}\\8 & -5 & 0 & 17\end{matrix}\right]$$
In 2 -th column
$$\left[\begin{matrix}- \frac{7}{4}\\\frac{9}{4}\\-5\end{matrix}\right]$$
let’s convert all the elements, except
1 -th element into zero.
- To do this, let’s take 1 -th line
$$\left[\begin{matrix}0 & - \frac{7}{4} & 1 & - \frac{9}{4}\end{matrix}\right]$$
,
and subtract it from other lines:
From 2 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}- \frac{\left(-9\right) 0}{7} & \frac{9}{4} - - \frac{-9}{4} & -4 - - \frac{9}{7} & \frac{19}{4} - - \frac{-81}{28}\end{matrix}\right] = \left[\begin{matrix}0 & 0 & - \frac{19}{7} & \frac{13}{7}\end{matrix}\right]$$
you get
$$\left[\begin{matrix}0 & - \frac{7}{4} & 1 & - \frac{9}{4}\\0 & 0 & - \frac{19}{7} & \frac{13}{7}\\8 & -5 & 0 & 17\end{matrix}\right]$$
From 3 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}8 - \frac{0 \cdot 20}{7} & -5 - - 5 & \frac{\left(-1\right) 20}{7} & 17 - - \frac{45}{7}\end{matrix}\right] = \left[\begin{matrix}8 & 0 & - \frac{20}{7} & \frac{164}{7}\end{matrix}\right]$$
you get
$$\left[\begin{matrix}0 & - \frac{7}{4} & 1 & - \frac{9}{4}\\0 & 0 & - \frac{19}{7} & \frac{13}{7}\\8 & 0 & - \frac{20}{7} & \frac{164}{7}\end{matrix}\right]$$
In 3 -th column
$$\left[\begin{matrix}1\\- \frac{19}{7}\\- \frac{20}{7}\end{matrix}\right]$$
let’s convert all the elements, except
2 -th element into zero.
- To do this, let’s take 2 -th line
$$\left[\begin{matrix}0 & 0 & - \frac{19}{7} & \frac{13}{7}\end{matrix}\right]$$
,
and subtract it from other lines:
From 1 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}- \frac{\left(-7\right) 0}{19} & - \frac{7}{4} - \frac{\left(-7\right) 0}{19} & 1 - - -1 & - \frac{9}{4} - \frac{\left(-7\right) 13}{7 \cdot 19}\end{matrix}\right] = \left[\begin{matrix}0 & - \frac{7}{4} & 0 & - \frac{119}{76}\end{matrix}\right]$$
you get
$$\left[\begin{matrix}0 & - \frac{7}{4} & 0 & - \frac{119}{76}\\0 & 0 & - \frac{19}{7} & \frac{13}{7}\\8 & 0 & - \frac{20}{7} & \frac{164}{7}\end{matrix}\right]$$
From 3 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}8 - \frac{0 \cdot 20}{19} & - \frac{0 \cdot 20}{19} & - \frac{20}{7} - - \frac{20}{7} & \frac{164}{7} - \frac{13 \cdot 20}{7 \cdot 19}\end{matrix}\right] = \left[\begin{matrix}8 & 0 & 0 & \frac{408}{19}\end{matrix}\right]$$
you get
$$\left[\begin{matrix}0 & - \frac{7}{4} & 0 & - \frac{119}{76}\\0 & 0 & - \frac{19}{7} & \frac{13}{7}\\8 & 0 & 0 & \frac{408}{19}\end{matrix}\right]$$
It is almost ready, all we have to do is to find variables, solving the elementary equations:
$$\frac{119}{76} - \frac{7 x_{2}}{4} = 0$$
$$- \frac{19 x_{3}}{7} - \frac{13}{7} = 0$$
$$8 x_{1} - \frac{408}{19} = 0$$
We get the answer:
$$x_{2} = \frac{17}{19}$$
$$x_{3} = - \frac{13}{19}$$
$$x_{1} = \frac{51}{19}$$
$$z + \left(2 x - 3 y\right) = 2$$
$$- 4 z + \left(2 x + y\right) = 9$$
$$2 x + \left(6 x - 5 y\right) = 17$$
We give the system of equations to the canonical form
$$2 x - 3 y + z = 2$$
$$2 x + y - 4 z = 9$$
$$8 x - 5 y = 17$$
Rewrite the system of linear equations as the matrix form
$$\left[\begin{matrix}2 & -3 & 1 & 2\\2 & 1 & -4 & 9\\8 & -5 & 0 & 17\end{matrix}\right]$$
In 1 -th column
$$\left[\begin{matrix}2\\2\\8\end{matrix}\right]$$
let’s convert all the elements, except
3 -th element into zero.
- To do this, let’s take 3 -th line
$$\left[\begin{matrix}8 & -5 & 0 & 17\end{matrix}\right]$$
,
and subtract it from other lines:
From 1 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}2 - \frac{8}{4} & -3 - - \frac{5}{4} & 1 - \frac{0}{4} & 2 - \frac{17}{4}\end{matrix}\right] = \left[\begin{matrix}0 & - \frac{7}{4} & 1 & - \frac{9}{4}\end{matrix}\right]$$
you get
$$\left[\begin{matrix}0 & - \frac{7}{4} & 1 & - \frac{9}{4}\\2 & 1 & -4 & 9\\8 & -5 & 0 & 17\end{matrix}\right]$$
From 2 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}2 - \frac{8}{4} & 1 - - \frac{5}{4} & -4 - \frac{0}{4} & 9 - \frac{17}{4}\end{matrix}\right] = \left[\begin{matrix}0 & \frac{9}{4} & -4 & \frac{19}{4}\end{matrix}\right]$$
you get
$$\left[\begin{matrix}0 & - \frac{7}{4} & 1 & - \frac{9}{4}\\0 & \frac{9}{4} & -4 & \frac{19}{4}\\8 & -5 & 0 & 17\end{matrix}\right]$$
In 2 -th column
$$\left[\begin{matrix}- \frac{7}{4}\\\frac{9}{4}\\-5\end{matrix}\right]$$
let’s convert all the elements, except
1 -th element into zero.
- To do this, let’s take 1 -th line
$$\left[\begin{matrix}0 & - \frac{7}{4} & 1 & - \frac{9}{4}\end{matrix}\right]$$
,
and subtract it from other lines:
From 2 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}- \frac{\left(-9\right) 0}{7} & \frac{9}{4} - - \frac{-9}{4} & -4 - - \frac{9}{7} & \frac{19}{4} - - \frac{-81}{28}\end{matrix}\right] = \left[\begin{matrix}0 & 0 & - \frac{19}{7} & \frac{13}{7}\end{matrix}\right]$$
you get
$$\left[\begin{matrix}0 & - \frac{7}{4} & 1 & - \frac{9}{4}\\0 & 0 & - \frac{19}{7} & \frac{13}{7}\\8 & -5 & 0 & 17\end{matrix}\right]$$
From 3 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}8 - \frac{0 \cdot 20}{7} & -5 - - 5 & \frac{\left(-1\right) 20}{7} & 17 - - \frac{45}{7}\end{matrix}\right] = \left[\begin{matrix}8 & 0 & - \frac{20}{7} & \frac{164}{7}\end{matrix}\right]$$
you get
$$\left[\begin{matrix}0 & - \frac{7}{4} & 1 & - \frac{9}{4}\\0 & 0 & - \frac{19}{7} & \frac{13}{7}\\8 & 0 & - \frac{20}{7} & \frac{164}{7}\end{matrix}\right]$$
In 3 -th column
$$\left[\begin{matrix}1\\- \frac{19}{7}\\- \frac{20}{7}\end{matrix}\right]$$
let’s convert all the elements, except
2 -th element into zero.
- To do this, let’s take 2 -th line
$$\left[\begin{matrix}0 & 0 & - \frac{19}{7} & \frac{13}{7}\end{matrix}\right]$$
,
and subtract it from other lines:
From 1 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}- \frac{\left(-7\right) 0}{19} & - \frac{7}{4} - \frac{\left(-7\right) 0}{19} & 1 - - -1 & - \frac{9}{4} - \frac{\left(-7\right) 13}{7 \cdot 19}\end{matrix}\right] = \left[\begin{matrix}0 & - \frac{7}{4} & 0 & - \frac{119}{76}\end{matrix}\right]$$
you get
$$\left[\begin{matrix}0 & - \frac{7}{4} & 0 & - \frac{119}{76}\\0 & 0 & - \frac{19}{7} & \frac{13}{7}\\8 & 0 & - \frac{20}{7} & \frac{164}{7}\end{matrix}\right]$$
From 3 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}8 - \frac{0 \cdot 20}{19} & - \frac{0 \cdot 20}{19} & - \frac{20}{7} - - \frac{20}{7} & \frac{164}{7} - \frac{13 \cdot 20}{7 \cdot 19}\end{matrix}\right] = \left[\begin{matrix}8 & 0 & 0 & \frac{408}{19}\end{matrix}\right]$$
you get
$$\left[\begin{matrix}0 & - \frac{7}{4} & 0 & - \frac{119}{76}\\0 & 0 & - \frac{19}{7} & \frac{13}{7}\\8 & 0 & 0 & \frac{408}{19}\end{matrix}\right]$$
It is almost ready, all we have to do is to find variables, solving the elementary equations:
$$\frac{119}{76} - \frac{7 x_{2}}{4} = 0$$
$$- \frac{19 x_{3}}{7} - \frac{13}{7} = 0$$
$$8 x_{1} - \frac{408}{19} = 0$$
We get the answer:
$$x_{2} = \frac{17}{19}$$
$$x_{3} = - \frac{13}{19}$$
$$x_{1} = \frac{51}{19}$$
Numerical answer
[src]
x1 = 2.684210526315789 y1 = 0.8947368421052632 z1 = -0.6842105263157895
x1 = 2.684210526315789 y1 = 0.8947368421052632 z1 = -0.6842105263157895