Mister Exam
3х-2у+z=10; x+5y-2z=-13; 2x-2y-z=3
The solution
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3*x - 2*y + z = 10
$$z + \left(3 x - 2 y\right) = 10$$
x + 5*y - 2*z = -13
$$- 2 z + \left(x + 5 y\right) = -13$$
2*x - 2*y - z = 3
$$- z + \left(2 x - 2 y\right) = 3$$
-z + 2*x - 2*y = 3
Rapid solution
$$x_{1} = \frac{41}{33}$$
=
$$\frac{41}{33}$$
=
$$y_{1} = - \frac{56}{33}$$
=
$$- \frac{56}{33}$$
=
$$z_{1} = \frac{95}{33}$$
=
$$\frac{95}{33}$$
=
=
$$\frac{41}{33}$$
=
1.24242424242424
$$y_{1} = - \frac{56}{33}$$
=
$$- \frac{56}{33}$$
=
-1.6969696969697
$$z_{1} = \frac{95}{33}$$
=
$$\frac{95}{33}$$
=
2.87878787878788
Gaussian elimination
Given the system of equations
$$z + \left(3 x - 2 y\right) = 10$$
$$- 2 z + \left(x + 5 y\right) = -13$$
$$- z + \left(2 x - 2 y\right) = 3$$
We give the system of equations to the canonical form
$$3 x - 2 y + z = 10$$
$$x + 5 y - 2 z = -13$$
$$2 x - 2 y - z = 3$$
Rewrite the system of linear equations as the matrix form
$$\left[\begin{matrix}3 & -2 & 1 & 10\\1 & 5 & -2 & -13\\2 & -2 & -1 & 3\end{matrix}\right]$$
In 1 -th column
$$\left[\begin{matrix}3\\1\\2\end{matrix}\right]$$
let’s convert all the elements, except
1 -th element into zero.
- To do this, let’s take 1 -th line
$$\left[\begin{matrix}3 & -2 & 1 & 10\end{matrix}\right]$$
,
and subtract it from other lines:
From 2 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}1 - \frac{3}{3} & 5 - - \frac{2}{3} & -2 + \frac{-1}{3} & -13 - \frac{10}{3}\end{matrix}\right] = \left[\begin{matrix}0 & \frac{17}{3} & - \frac{7}{3} & - \frac{49}{3}\end{matrix}\right]$$
you get
$$\left[\begin{matrix}3 & -2 & 1 & 10\\0 & \frac{17}{3} & - \frac{7}{3} & - \frac{49}{3}\\2 & -2 & -1 & 3\end{matrix}\right]$$
From 3 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}2 - \frac{2 \cdot 3}{3} & -2 - - \frac{4}{3} & -1 + \frac{\left(-1\right) 2}{3} & 3 - \frac{2 \cdot 10}{3}\end{matrix}\right] = \left[\begin{matrix}0 & - \frac{2}{3} & - \frac{5}{3} & - \frac{11}{3}\end{matrix}\right]$$
you get
$$\left[\begin{matrix}3 & -2 & 1 & 10\\0 & \frac{17}{3} & - \frac{7}{3} & - \frac{49}{3}\\0 & - \frac{2}{3} & - \frac{5}{3} & - \frac{11}{3}\end{matrix}\right]$$
In 2 -th column
$$\left[\begin{matrix}-2\\\frac{17}{3}\\- \frac{2}{3}\end{matrix}\right]$$
let’s convert all the elements, except
2 -th element into zero.
- To do this, let’s take 2 -th line
$$\left[\begin{matrix}0 & \frac{17}{3} & - \frac{7}{3} & - \frac{49}{3}\end{matrix}\right]$$
,
and subtract it from other lines:
From 1 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}3 - \frac{\left(-6\right) 0}{17} & -2 - \frac{\left(-6\right) 17}{3 \cdot 17} & 1 - - \frac{-14}{17} & 10 - - \frac{-98}{17}\end{matrix}\right] = \left[\begin{matrix}3 & 0 & \frac{3}{17} & \frac{72}{17}\end{matrix}\right]$$
you get
$$\left[\begin{matrix}3 & 0 & \frac{3}{17} & \frac{72}{17}\\0 & \frac{17}{3} & - \frac{7}{3} & - \frac{49}{3}\\0 & - \frac{2}{3} & - \frac{5}{3} & - \frac{11}{3}\end{matrix}\right]$$
From 3 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}- \frac{\left(-2\right) 0}{17} & - \frac{2}{3} - \frac{\left(-2\right) 17}{3 \cdot 17} & - \frac{5}{3} - - \frac{-14}{51} & - \frac{11}{3} - - \frac{-98}{51}\end{matrix}\right] = \left[\begin{matrix}0 & 0 & - \frac{33}{17} & - \frac{95}{17}\end{matrix}\right]$$
you get
$$\left[\begin{matrix}3 & 0 & \frac{3}{17} & \frac{72}{17}\\0 & \frac{17}{3} & - \frac{7}{3} & - \frac{49}{3}\\0 & 0 & - \frac{33}{17} & - \frac{95}{17}\end{matrix}\right]$$
In 3 -th column
$$\left[\begin{matrix}\frac{3}{17}\\- \frac{7}{3}\\- \frac{33}{17}\end{matrix}\right]$$
let’s convert all the elements, except
3 -th element into zero.
- To do this, let’s take 3 -th line
$$\left[\begin{matrix}0 & 0 & - \frac{33}{17} & - \frac{95}{17}\end{matrix}\right]$$
,
and subtract it from other lines:
From 1 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}3 - \frac{\left(-1\right) 0}{11} & - \frac{\left(-1\right) 0}{11} & \frac{3}{17} - - \frac{-3}{17} & \frac{72}{17} - - \frac{-95}{187}\end{matrix}\right] = \left[\begin{matrix}3 & 0 & 0 & \frac{41}{11}\end{matrix}\right]$$
you get
$$\left[\begin{matrix}3 & 0 & 0 & \frac{41}{11}\\0 & \frac{17}{3} & - \frac{7}{3} & - \frac{49}{3}\\0 & 0 & - \frac{33}{17} & - \frac{95}{17}\end{matrix}\right]$$
From 2 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}- \frac{0 \cdot 119}{99} & \frac{17}{3} - \frac{0 \cdot 119}{99} & - \frac{7}{3} - - \frac{7}{3} & - \frac{49}{3} - - \frac{665}{99}\end{matrix}\right] = \left[\begin{matrix}0 & \frac{17}{3} & 0 & - \frac{952}{99}\end{matrix}\right]$$
you get
$$\left[\begin{matrix}3 & 0 & 0 & \frac{41}{11}\\0 & \frac{17}{3} & 0 & - \frac{952}{99}\\0 & 0 & - \frac{33}{17} & - \frac{95}{17}\end{matrix}\right]$$
It is almost ready, all we have to do is to find variables, solving the elementary equations:
$$3 x_{1} - \frac{41}{11} = 0$$
$$\frac{17 x_{2}}{3} + \frac{952}{99} = 0$$
$$\frac{95}{17} - \frac{33 x_{3}}{17} = 0$$
We get the answer:
$$x_{1} = \frac{41}{33}$$
$$x_{2} = - \frac{56}{33}$$
$$x_{3} = \frac{95}{33}$$
$$z + \left(3 x - 2 y\right) = 10$$
$$- 2 z + \left(x + 5 y\right) = -13$$
$$- z + \left(2 x - 2 y\right) = 3$$
We give the system of equations to the canonical form
$$3 x - 2 y + z = 10$$
$$x + 5 y - 2 z = -13$$
$$2 x - 2 y - z = 3$$
Rewrite the system of linear equations as the matrix form
$$\left[\begin{matrix}3 & -2 & 1 & 10\\1 & 5 & -2 & -13\\2 & -2 & -1 & 3\end{matrix}\right]$$
In 1 -th column
$$\left[\begin{matrix}3\\1\\2\end{matrix}\right]$$
let’s convert all the elements, except
1 -th element into zero.
- To do this, let’s take 1 -th line
$$\left[\begin{matrix}3 & -2 & 1 & 10\end{matrix}\right]$$
,
and subtract it from other lines:
From 2 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}1 - \frac{3}{3} & 5 - - \frac{2}{3} & -2 + \frac{-1}{3} & -13 - \frac{10}{3}\end{matrix}\right] = \left[\begin{matrix}0 & \frac{17}{3} & - \frac{7}{3} & - \frac{49}{3}\end{matrix}\right]$$
you get
$$\left[\begin{matrix}3 & -2 & 1 & 10\\0 & \frac{17}{3} & - \frac{7}{3} & - \frac{49}{3}\\2 & -2 & -1 & 3\end{matrix}\right]$$
From 3 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}2 - \frac{2 \cdot 3}{3} & -2 - - \frac{4}{3} & -1 + \frac{\left(-1\right) 2}{3} & 3 - \frac{2 \cdot 10}{3}\end{matrix}\right] = \left[\begin{matrix}0 & - \frac{2}{3} & - \frac{5}{3} & - \frac{11}{3}\end{matrix}\right]$$
you get
$$\left[\begin{matrix}3 & -2 & 1 & 10\\0 & \frac{17}{3} & - \frac{7}{3} & - \frac{49}{3}\\0 & - \frac{2}{3} & - \frac{5}{3} & - \frac{11}{3}\end{matrix}\right]$$
In 2 -th column
$$\left[\begin{matrix}-2\\\frac{17}{3}\\- \frac{2}{3}\end{matrix}\right]$$
let’s convert all the elements, except
2 -th element into zero.
- To do this, let’s take 2 -th line
$$\left[\begin{matrix}0 & \frac{17}{3} & - \frac{7}{3} & - \frac{49}{3}\end{matrix}\right]$$
,
and subtract it from other lines:
From 1 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}3 - \frac{\left(-6\right) 0}{17} & -2 - \frac{\left(-6\right) 17}{3 \cdot 17} & 1 - - \frac{-14}{17} & 10 - - \frac{-98}{17}\end{matrix}\right] = \left[\begin{matrix}3 & 0 & \frac{3}{17} & \frac{72}{17}\end{matrix}\right]$$
you get
$$\left[\begin{matrix}3 & 0 & \frac{3}{17} & \frac{72}{17}\\0 & \frac{17}{3} & - \frac{7}{3} & - \frac{49}{3}\\0 & - \frac{2}{3} & - \frac{5}{3} & - \frac{11}{3}\end{matrix}\right]$$
From 3 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}- \frac{\left(-2\right) 0}{17} & - \frac{2}{3} - \frac{\left(-2\right) 17}{3 \cdot 17} & - \frac{5}{3} - - \frac{-14}{51} & - \frac{11}{3} - - \frac{-98}{51}\end{matrix}\right] = \left[\begin{matrix}0 & 0 & - \frac{33}{17} & - \frac{95}{17}\end{matrix}\right]$$
you get
$$\left[\begin{matrix}3 & 0 & \frac{3}{17} & \frac{72}{17}\\0 & \frac{17}{3} & - \frac{7}{3} & - \frac{49}{3}\\0 & 0 & - \frac{33}{17} & - \frac{95}{17}\end{matrix}\right]$$
In 3 -th column
$$\left[\begin{matrix}\frac{3}{17}\\- \frac{7}{3}\\- \frac{33}{17}\end{matrix}\right]$$
let’s convert all the elements, except
3 -th element into zero.
- To do this, let’s take 3 -th line
$$\left[\begin{matrix}0 & 0 & - \frac{33}{17} & - \frac{95}{17}\end{matrix}\right]$$
,
and subtract it from other lines:
From 1 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}3 - \frac{\left(-1\right) 0}{11} & - \frac{\left(-1\right) 0}{11} & \frac{3}{17} - - \frac{-3}{17} & \frac{72}{17} - - \frac{-95}{187}\end{matrix}\right] = \left[\begin{matrix}3 & 0 & 0 & \frac{41}{11}\end{matrix}\right]$$
you get
$$\left[\begin{matrix}3 & 0 & 0 & \frac{41}{11}\\0 & \frac{17}{3} & - \frac{7}{3} & - \frac{49}{3}\\0 & 0 & - \frac{33}{17} & - \frac{95}{17}\end{matrix}\right]$$
From 2 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}- \frac{0 \cdot 119}{99} & \frac{17}{3} - \frac{0 \cdot 119}{99} & - \frac{7}{3} - - \frac{7}{3} & - \frac{49}{3} - - \frac{665}{99}\end{matrix}\right] = \left[\begin{matrix}0 & \frac{17}{3} & 0 & - \frac{952}{99}\end{matrix}\right]$$
you get
$$\left[\begin{matrix}3 & 0 & 0 & \frac{41}{11}\\0 & \frac{17}{3} & 0 & - \frac{952}{99}\\0 & 0 & - \frac{33}{17} & - \frac{95}{17}\end{matrix}\right]$$
It is almost ready, all we have to do is to find variables, solving the elementary equations:
$$3 x_{1} - \frac{41}{11} = 0$$
$$\frac{17 x_{2}}{3} + \frac{952}{99} = 0$$
$$\frac{95}{17} - \frac{33 x_{3}}{17} = 0$$
We get the answer:
$$x_{1} = \frac{41}{33}$$
$$x_{2} = - \frac{56}{33}$$
$$x_{3} = \frac{95}{33}$$
Cramer's rule
$$z + \left(3 x - 2 y\right) = 10$$
$$- 2 z + \left(x + 5 y\right) = -13$$
$$- z + \left(2 x - 2 y\right) = 3$$
We give the system of equations to the canonical form
$$3 x - 2 y + z = 10$$
$$x + 5 y - 2 z = -13$$
$$2 x - 2 y - z = 3$$
Rewrite the system of linear equations as the matrix form
$$\left[\begin{matrix}3 x_{1} - 2 x_{2} + x_{3}\\x_{1} + 5 x_{2} - 2 x_{3}\\2 x_{1} - 2 x_{2} - x_{3}\end{matrix}\right] = \left[\begin{matrix}10\\-13\\3\end{matrix}\right]$$
- 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:
$$A = \operatorname{det}{\left(\left[\begin{matrix}3 & -2 & 1\\1 & 5 & -2\\2 & -2 & -1\end{matrix}\right] \right)} = -33$$
, 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 )
$$x_{1} = - \frac{\operatorname{det}{\left(\left[\begin{matrix}10 & -2 & 1\\-13 & 5 & -2\\3 & -2 & -1\end{matrix}\right] \right)}}{33} = \frac{41}{33}$$
$$x_{2} = - \frac{\operatorname{det}{\left(\left[\begin{matrix}3 & 10 & 1\\1 & -13 & -2\\2 & 3 & -1\end{matrix}\right] \right)}}{33} = - \frac{56}{33}$$
$$x_{3} = - \frac{\operatorname{det}{\left(\left[\begin{matrix}3 & -2 & 10\\1 & 5 & -13\\2 & -2 & 3\end{matrix}\right] \right)}}{33} = \frac{95}{33}$$
$$- 2 z + \left(x + 5 y\right) = -13$$
$$- z + \left(2 x - 2 y\right) = 3$$
We give the system of equations to the canonical form
$$3 x - 2 y + z = 10$$
$$x + 5 y - 2 z = -13$$
$$2 x - 2 y - z = 3$$
Rewrite the system of linear equations as the matrix form
$$\left[\begin{matrix}3 x_{1} - 2 x_{2} + x_{3}\\x_{1} + 5 x_{2} - 2 x_{3}\\2 x_{1} - 2 x_{2} - x_{3}\end{matrix}\right] = \left[\begin{matrix}10\\-13\\3\end{matrix}\right]$$
- 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:
$$A = \operatorname{det}{\left(\left[\begin{matrix}3 & -2 & 1\\1 & 5 & -2\\2 & -2 & -1\end{matrix}\right] \right)} = -33$$
, 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 )
$$x_{1} = - \frac{\operatorname{det}{\left(\left[\begin{matrix}10 & -2 & 1\\-13 & 5 & -2\\3 & -2 & -1\end{matrix}\right] \right)}}{33} = \frac{41}{33}$$
$$x_{2} = - \frac{\operatorname{det}{\left(\left[\begin{matrix}3 & 10 & 1\\1 & -13 & -2\\2 & 3 & -1\end{matrix}\right] \right)}}{33} = - \frac{56}{33}$$
$$x_{3} = - \frac{\operatorname{det}{\left(\left[\begin{matrix}3 & -2 & 10\\1 & 5 & -13\\2 & -2 & 3\end{matrix}\right] \right)}}{33} = \frac{95}{33}$$
Numerical answer
[src]
x1 = 1.242424242424242 y1 = -1.696969696969697 z1 = 2.878787878787879
x1 = 1.242424242424242 y1 = -1.696969696969697 z1 = 2.878787878787879