Mister Exam
5X+y-3z=-2; 4x+3y+2z=16; 2X-3y+z=17
The solution
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5*x + y - 3*z = -2
$$- 3 z + \left(5 x + y\right) = -2$$
4*x + 3*y + 2*z = 16
$$2 z + \left(4 x + 3 y\right) = 16$$
2*x - 3*y + z = 17
$$z + \left(2 x - 3 y\right) = 17$$
z + 2*x - 3*y = 17
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
$$- 3 z + \left(5 x + y\right) = -2$$
$$2 z + \left(4 x + 3 y\right) = 16$$
$$z + \left(2 x - 3 y\right) = 17$$
We give the system of equations to the canonical form
$$5 x + y - 3 z = -2$$
$$4 x + 3 y + 2 z = 16$$
$$2 x - 3 y + z = 17$$
Rewrite the system of linear equations as the matrix form
$$\left[\begin{matrix}5 & 1 & -3 & -2\\4 & 3 & 2 & 16\\2 & -3 & 1 & 17\end{matrix}\right]$$
In 1 -th column
$$\left[\begin{matrix}5\\4\\2\end{matrix}\right]$$
let’s convert all the elements, except
1 -th element into zero.
- To do this, let’s take 1 -th line
$$\left[\begin{matrix}5 & 1 & -3 & -2\end{matrix}\right]$$
,
and subtract it from other lines:
From 2 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}4 - \frac{4 \cdot 5}{5} & \frac{\left(-1\right) 4}{5} + 3 & 2 - - \frac{12}{5} & 16 - - \frac{8}{5}\end{matrix}\right] = \left[\begin{matrix}0 & \frac{11}{5} & \frac{22}{5} & \frac{88}{5}\end{matrix}\right]$$
you get
$$\left[\begin{matrix}5 & 1 & -3 & -2\\0 & \frac{11}{5} & \frac{22}{5} & \frac{88}{5}\\2 & -3 & 1 & 17\end{matrix}\right]$$
From 3 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}2 - \frac{2 \cdot 5}{5} & -3 + \frac{\left(-1\right) 2}{5} & 1 - - \frac{6}{5} & 17 - - \frac{4}{5}\end{matrix}\right] = \left[\begin{matrix}0 & - \frac{17}{5} & \frac{11}{5} & \frac{89}{5}\end{matrix}\right]$$
you get
$$\left[\begin{matrix}5 & 1 & -3 & -2\\0 & \frac{11}{5} & \frac{22}{5} & \frac{88}{5}\\0 & - \frac{17}{5} & \frac{11}{5} & \frac{89}{5}\end{matrix}\right]$$
In 2 -th column
$$\left[\begin{matrix}1\\\frac{11}{5}\\- \frac{17}{5}\end{matrix}\right]$$
let’s convert all the elements, except
2 -th element into zero.
- To do this, let’s take 2 -th line
$$\left[\begin{matrix}0 & \frac{11}{5} & \frac{22}{5} & \frac{88}{5}\end{matrix}\right]$$
,
and subtract it from other lines:
From 1 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}5 - \frac{0 \cdot 5}{11} & 1 - \frac{5 \cdot 11}{5 \cdot 11} & -3 - \frac{5 \cdot 22}{5 \cdot 11} & - \frac{5 \cdot 88}{5 \cdot 11} - 2\end{matrix}\right] = \left[\begin{matrix}5 & 0 & -5 & -10\end{matrix}\right]$$
you get
$$\left[\begin{matrix}5 & 0 & -5 & -10\\0 & \frac{11}{5} & \frac{22}{5} & \frac{88}{5}\\0 & - \frac{17}{5} & \frac{11}{5} & \frac{89}{5}\end{matrix}\right]$$
From 3 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}- \frac{\left(-17\right) 0}{11} & - \frac{17}{5} - \frac{\left(-17\right) 11}{5 \cdot 11} & \frac{11}{5} - \frac{\left(-17\right) 22}{5 \cdot 11} & \frac{89}{5} - \frac{\left(-17\right) 88}{5 \cdot 11}\end{matrix}\right] = \left[\begin{matrix}0 & 0 & 9 & 45\end{matrix}\right]$$
you get
$$\left[\begin{matrix}5 & 0 & -5 & -10\\0 & \frac{11}{5} & \frac{22}{5} & \frac{88}{5}\\0 & 0 & 9 & 45\end{matrix}\right]$$
In 3 -th column
$$\left[\begin{matrix}-5\\\frac{22}{5}\\9\end{matrix}\right]$$
let’s convert all the elements, except
3 -th element into zero.
- To do this, let’s take 3 -th line
$$\left[\begin{matrix}0 & 0 & 9 & 45\end{matrix}\right]$$
,
and subtract it from other lines:
From 1 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}5 - \frac{\left(-5\right) 0}{9} & - \frac{\left(-5\right) 0}{9} & -5 - \frac{\left(-5\right) 9}{9} & -10 - \frac{\left(-5\right) 45}{9}\end{matrix}\right] = \left[\begin{matrix}5 & 0 & 0 & 15\end{matrix}\right]$$
you get
$$\left[\begin{matrix}5 & 0 & 0 & 15\\0 & \frac{11}{5} & \frac{22}{5} & \frac{88}{5}\\0 & 0 & 9 & 45\end{matrix}\right]$$
From 2 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}- \frac{0 \cdot 22}{45} & \frac{11}{5} - \frac{0 \cdot 22}{45} & \frac{22}{5} - \frac{9 \cdot 22}{45} & \frac{88}{5} - \frac{22 \cdot 45}{45}\end{matrix}\right] = \left[\begin{matrix}0 & \frac{11}{5} & 0 & - \frac{22}{5}\end{matrix}\right]$$
you get
$$\left[\begin{matrix}5 & 0 & 0 & 15\\0 & \frac{11}{5} & 0 & - \frac{22}{5}\\0 & 0 & 9 & 45\end{matrix}\right]$$
It is almost ready, all we have to do is to find variables, solving the elementary equations:
$$5 x_{1} - 15 = 0$$
$$\frac{11 x_{2}}{5} + \frac{22}{5} = 0$$
$$9 x_{3} - 45 = 0$$
We get the answer:
$$x_{1} = 3$$
$$x_{2} = -2$$
$$x_{3} = 5$$
$$- 3 z + \left(5 x + y\right) = -2$$
$$2 z + \left(4 x + 3 y\right) = 16$$
$$z + \left(2 x - 3 y\right) = 17$$
We give the system of equations to the canonical form
$$5 x + y - 3 z = -2$$
$$4 x + 3 y + 2 z = 16$$
$$2 x - 3 y + z = 17$$
Rewrite the system of linear equations as the matrix form
$$\left[\begin{matrix}5 & 1 & -3 & -2\\4 & 3 & 2 & 16\\2 & -3 & 1 & 17\end{matrix}\right]$$
In 1 -th column
$$\left[\begin{matrix}5\\4\\2\end{matrix}\right]$$
let’s convert all the elements, except
1 -th element into zero.
- To do this, let’s take 1 -th line
$$\left[\begin{matrix}5 & 1 & -3 & -2\end{matrix}\right]$$
,
and subtract it from other lines:
From 2 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}4 - \frac{4 \cdot 5}{5} & \frac{\left(-1\right) 4}{5} + 3 & 2 - - \frac{12}{5} & 16 - - \frac{8}{5}\end{matrix}\right] = \left[\begin{matrix}0 & \frac{11}{5} & \frac{22}{5} & \frac{88}{5}\end{matrix}\right]$$
you get
$$\left[\begin{matrix}5 & 1 & -3 & -2\\0 & \frac{11}{5} & \frac{22}{5} & \frac{88}{5}\\2 & -3 & 1 & 17\end{matrix}\right]$$
From 3 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}2 - \frac{2 \cdot 5}{5} & -3 + \frac{\left(-1\right) 2}{5} & 1 - - \frac{6}{5} & 17 - - \frac{4}{5}\end{matrix}\right] = \left[\begin{matrix}0 & - \frac{17}{5} & \frac{11}{5} & \frac{89}{5}\end{matrix}\right]$$
you get
$$\left[\begin{matrix}5 & 1 & -3 & -2\\0 & \frac{11}{5} & \frac{22}{5} & \frac{88}{5}\\0 & - \frac{17}{5} & \frac{11}{5} & \frac{89}{5}\end{matrix}\right]$$
In 2 -th column
$$\left[\begin{matrix}1\\\frac{11}{5}\\- \frac{17}{5}\end{matrix}\right]$$
let’s convert all the elements, except
2 -th element into zero.
- To do this, let’s take 2 -th line
$$\left[\begin{matrix}0 & \frac{11}{5} & \frac{22}{5} & \frac{88}{5}\end{matrix}\right]$$
,
and subtract it from other lines:
From 1 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}5 - \frac{0 \cdot 5}{11} & 1 - \frac{5 \cdot 11}{5 \cdot 11} & -3 - \frac{5 \cdot 22}{5 \cdot 11} & - \frac{5 \cdot 88}{5 \cdot 11} - 2\end{matrix}\right] = \left[\begin{matrix}5 & 0 & -5 & -10\end{matrix}\right]$$
you get
$$\left[\begin{matrix}5 & 0 & -5 & -10\\0 & \frac{11}{5} & \frac{22}{5} & \frac{88}{5}\\0 & - \frac{17}{5} & \frac{11}{5} & \frac{89}{5}\end{matrix}\right]$$
From 3 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}- \frac{\left(-17\right) 0}{11} & - \frac{17}{5} - \frac{\left(-17\right) 11}{5 \cdot 11} & \frac{11}{5} - \frac{\left(-17\right) 22}{5 \cdot 11} & \frac{89}{5} - \frac{\left(-17\right) 88}{5 \cdot 11}\end{matrix}\right] = \left[\begin{matrix}0 & 0 & 9 & 45\end{matrix}\right]$$
you get
$$\left[\begin{matrix}5 & 0 & -5 & -10\\0 & \frac{11}{5} & \frac{22}{5} & \frac{88}{5}\\0 & 0 & 9 & 45\end{matrix}\right]$$
In 3 -th column
$$\left[\begin{matrix}-5\\\frac{22}{5}\\9\end{matrix}\right]$$
let’s convert all the elements, except
3 -th element into zero.
- To do this, let’s take 3 -th line
$$\left[\begin{matrix}0 & 0 & 9 & 45\end{matrix}\right]$$
,
and subtract it from other lines:
From 1 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}5 - \frac{\left(-5\right) 0}{9} & - \frac{\left(-5\right) 0}{9} & -5 - \frac{\left(-5\right) 9}{9} & -10 - \frac{\left(-5\right) 45}{9}\end{matrix}\right] = \left[\begin{matrix}5 & 0 & 0 & 15\end{matrix}\right]$$
you get
$$\left[\begin{matrix}5 & 0 & 0 & 15\\0 & \frac{11}{5} & \frac{22}{5} & \frac{88}{5}\\0 & 0 & 9 & 45\end{matrix}\right]$$
From 2 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}- \frac{0 \cdot 22}{45} & \frac{11}{5} - \frac{0 \cdot 22}{45} & \frac{22}{5} - \frac{9 \cdot 22}{45} & \frac{88}{5} - \frac{22 \cdot 45}{45}\end{matrix}\right] = \left[\begin{matrix}0 & \frac{11}{5} & 0 & - \frac{22}{5}\end{matrix}\right]$$
you get
$$\left[\begin{matrix}5 & 0 & 0 & 15\\0 & \frac{11}{5} & 0 & - \frac{22}{5}\\0 & 0 & 9 & 45\end{matrix}\right]$$
It is almost ready, all we have to do is to find variables, solving the elementary equations:
$$5 x_{1} - 15 = 0$$
$$\frac{11 x_{2}}{5} + \frac{22}{5} = 0$$
$$9 x_{3} - 45 = 0$$
We get the answer:
$$x_{1} = 3$$
$$x_{2} = -2$$
$$x_{3} = 5$$
Cramer's rule
$$- 3 z + \left(5 x + y\right) = -2$$
$$2 z + \left(4 x + 3 y\right) = 16$$
$$z + \left(2 x - 3 y\right) = 17$$
We give the system of equations to the canonical form
$$5 x + y - 3 z = -2$$
$$4 x + 3 y + 2 z = 16$$
$$2 x - 3 y + z = 17$$
Rewrite the system of linear equations as the matrix form
$$\left[\begin{matrix}5 x_{1} + x_{2} - 3 x_{3}\\4 x_{1} + 3 x_{2} + 2 x_{3}\\2 x_{1} - 3 x_{2} + x_{3}\end{matrix}\right] = \left[\begin{matrix}-2\\16\\17\end{matrix}\right]$$
- 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:
$$A = \operatorname{det}{\left(\left[\begin{matrix}5 & 1 & -3\\4 & 3 & 2\\2 & -3 & 1\end{matrix}\right] \right)} = 99$$
, 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 )
$$x_{1} = \frac{\operatorname{det}{\left(\left[\begin{matrix}-2 & 1 & -3\\16 & 3 & 2\\17 & -3 & 1\end{matrix}\right] \right)}}{99} = 3$$
$$x_{2} = \frac{\operatorname{det}{\left(\left[\begin{matrix}5 & -2 & -3\\4 & 16 & 2\\2 & 17 & 1\end{matrix}\right] \right)}}{99} = -2$$
$$x_{3} = \frac{\operatorname{det}{\left(\left[\begin{matrix}5 & 1 & -2\\4 & 3 & 16\\2 & -3 & 17\end{matrix}\right] \right)}}{99} = 5$$
$$2 z + \left(4 x + 3 y\right) = 16$$
$$z + \left(2 x - 3 y\right) = 17$$
We give the system of equations to the canonical form
$$5 x + y - 3 z = -2$$
$$4 x + 3 y + 2 z = 16$$
$$2 x - 3 y + z = 17$$
Rewrite the system of linear equations as the matrix form
$$\left[\begin{matrix}5 x_{1} + x_{2} - 3 x_{3}\\4 x_{1} + 3 x_{2} + 2 x_{3}\\2 x_{1} - 3 x_{2} + x_{3}\end{matrix}\right] = \left[\begin{matrix}-2\\16\\17\end{matrix}\right]$$
- 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:
$$A = \operatorname{det}{\left(\left[\begin{matrix}5 & 1 & -3\\4 & 3 & 2\\2 & -3 & 1\end{matrix}\right] \right)} = 99$$
, 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 )
$$x_{1} = \frac{\operatorname{det}{\left(\left[\begin{matrix}-2 & 1 & -3\\16 & 3 & 2\\17 & -3 & 1\end{matrix}\right] \right)}}{99} = 3$$
$$x_{2} = \frac{\operatorname{det}{\left(\left[\begin{matrix}5 & -2 & -3\\4 & 16 & 2\\2 & 17 & 1\end{matrix}\right] \right)}}{99} = -2$$
$$x_{3} = \frac{\operatorname{det}{\left(\left[\begin{matrix}5 & 1 & -2\\4 & 3 & 16\\2 & -3 & 17\end{matrix}\right] \right)}}{99} = 5$$