Mister Exam
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- 2x1−3x2−x3=4; x1+x2−3x3=−3; 3x1+2x2+x3=2
- Х^2+у^2=13; Ху=-6
- X+84248=105000; 1-y^2=2+x
- (nx+2x)*22.4=3.136; nx*44+2x*36.5=5.86
-
Identical expressions
- two x1− three x2−x3= four ; x1+x2−3x3=−3; 3x1+2x2+x3=2
- 2x1−3x2−x3 equally 4; x1 plus x2−3x3 equally −3; 3x1 plus 2x2 plus x3 equally 2
- two x1− three x2−x3 equally four ; x1 plus x2−3x3 equally −3; 3x1 plus 2x2 plus x3 equally 2
-
Similar expressions
- 2x1−3x2−x3=4; x1-x2−3x3=−3; 3x1+2x2+x3=2
- 2x1−3x2−x3=4; x1+x2−3x3=−3; 3x1-2x2+x3=2
- 2x1−3x2−x3=4; x1+x2−3x3=−3; 3x1+2x2-x3=2
2x1−3x2−x3=4; x1+x2−3x3=−3; 3x1+2x2+x3=2
The solution
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2*x1 - 3*x2 - x3 = 4
$$- x_{3} + \left(2 x_{1} - 3 x_{2}\right) = 4$$
x1 + x2 - 3*x3 = -3
$$- 3 x_{3} + \left(x_{1} + x_{2}\right) = -3$$
3*x1 + 2*x2 + x3 = 2
$$x_{3} + \left(3 x_{1} + 2 x_{2}\right) = 2$$
x3 + 3*x1 + 2*x2 = 2
Rapid solution
$$x_{11} = 1$$
=
$$1$$
=
$$x_{21} = -1$$
=
$$-1$$
=
$$x_{31} = 1$$
=
$$1$$
=
=
$$1$$
=
1
$$x_{21} = -1$$
=
$$-1$$
=
-1
$$x_{31} = 1$$
=
$$1$$
=
1
Cramer's rule
$$- x_{3} + \left(2 x_{1} - 3 x_{2}\right) = 4$$
$$- 3 x_{3} + \left(x_{1} + x_{2}\right) = -3$$
$$x_{3} + \left(3 x_{1} + 2 x_{2}\right) = 2$$
We give the system of equations to the canonical form
$$2 x_{1} - 3 x_{2} - x_{3} = 4$$
$$x_{1} + x_{2} - 3 x_{3} = -3$$
$$3 x_{1} + 2 x_{2} + x_{3} = 2$$
Rewrite the system of linear equations as the matrix form
$$\left[\begin{matrix}2 x_{1} - 3 x_{2} - x_{3}\\x_{1} + x_{2} - 3 x_{3}\\3 x_{1} + 2 x_{2} + x_{3}\end{matrix}\right] = \left[\begin{matrix}4\\-3\\2\end{matrix}\right]$$
- 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:
$$A = \operatorname{det}{\left(\left[\begin{matrix}2 & -3 & -1\\1 & 1 & -3\\3 & 2 & 1\end{matrix}\right] \right)} = 45$$
, 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 )
$$x_{1} = \frac{\operatorname{det}{\left(\left[\begin{matrix}4 & -3 & -1\\-3 & 1 & -3\\2 & 2 & 1\end{matrix}\right] \right)}}{45} = 1$$
$$x_{2} = \frac{\operatorname{det}{\left(\left[\begin{matrix}2 & 4 & -1\\1 & -3 & -3\\3 & 2 & 1\end{matrix}\right] \right)}}{45} = -1$$
$$x_{3} = \frac{\operatorname{det}{\left(\left[\begin{matrix}2 & -3 & 4\\1 & 1 & -3\\3 & 2 & 2\end{matrix}\right] \right)}}{45} = 1$$
$$- 3 x_{3} + \left(x_{1} + x_{2}\right) = -3$$
$$x_{3} + \left(3 x_{1} + 2 x_{2}\right) = 2$$
We give the system of equations to the canonical form
$$2 x_{1} - 3 x_{2} - x_{3} = 4$$
$$x_{1} + x_{2} - 3 x_{3} = -3$$
$$3 x_{1} + 2 x_{2} + x_{3} = 2$$
Rewrite the system of linear equations as the matrix form
$$\left[\begin{matrix}2 x_{1} - 3 x_{2} - x_{3}\\x_{1} + x_{2} - 3 x_{3}\\3 x_{1} + 2 x_{2} + x_{3}\end{matrix}\right] = \left[\begin{matrix}4\\-3\\2\end{matrix}\right]$$
- 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:
$$A = \operatorname{det}{\left(\left[\begin{matrix}2 & -3 & -1\\1 & 1 & -3\\3 & 2 & 1\end{matrix}\right] \right)} = 45$$
, 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 )
$$x_{1} = \frac{\operatorname{det}{\left(\left[\begin{matrix}4 & -3 & -1\\-3 & 1 & -3\\2 & 2 & 1\end{matrix}\right] \right)}}{45} = 1$$
$$x_{2} = \frac{\operatorname{det}{\left(\left[\begin{matrix}2 & 4 & -1\\1 & -3 & -3\\3 & 2 & 1\end{matrix}\right] \right)}}{45} = -1$$
$$x_{3} = \frac{\operatorname{det}{\left(\left[\begin{matrix}2 & -3 & 4\\1 & 1 & -3\\3 & 2 & 2\end{matrix}\right] \right)}}{45} = 1$$
Gaussian elimination
Given the system of equations
$$- x_{3} + \left(2 x_{1} - 3 x_{2}\right) = 4$$
$$- 3 x_{3} + \left(x_{1} + x_{2}\right) = -3$$
$$x_{3} + \left(3 x_{1} + 2 x_{2}\right) = 2$$
We give the system of equations to the canonical form
$$2 x_{1} - 3 x_{2} - x_{3} = 4$$
$$x_{1} + x_{2} - 3 x_{3} = -3$$
$$3 x_{1} + 2 x_{2} + x_{3} = 2$$
Rewrite the system of linear equations as the matrix form
$$\left[\begin{matrix}2 & -3 & -1 & 4\\1 & 1 & -3 & -3\\3 & 2 & 1 & 2\end{matrix}\right]$$
In 1 -th column
$$\left[\begin{matrix}2\\1\\3\end{matrix}\right]$$
let’s convert all the elements, except
1 -th element into zero.
- To do this, let’s take 1 -th line
$$\left[\begin{matrix}2 & -3 & -1 & 4\end{matrix}\right]$$
,
and subtract it from other lines:
From 2 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}1 - \frac{2}{2} & 1 - - \frac{3}{2} & -3 - - \frac{1}{2} & -3 - \frac{4}{2}\end{matrix}\right] = \left[\begin{matrix}0 & \frac{5}{2} & - \frac{5}{2} & -5\end{matrix}\right]$$
you get
$$\left[\begin{matrix}2 & -3 & -1 & 4\\0 & \frac{5}{2} & - \frac{5}{2} & -5\\3 & 2 & 1 & 2\end{matrix}\right]$$
From 3 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}3 - \frac{2 \cdot 3}{2} & 2 - - \frac{9}{2} & 1 - - \frac{3}{2} & 2 - \frac{3 \cdot 4}{2}\end{matrix}\right] = \left[\begin{matrix}0 & \frac{13}{2} & \frac{5}{2} & -4\end{matrix}\right]$$
you get
$$\left[\begin{matrix}2 & -3 & -1 & 4\\0 & \frac{5}{2} & - \frac{5}{2} & -5\\0 & \frac{13}{2} & \frac{5}{2} & -4\end{matrix}\right]$$
In 2 -th column
$$\left[\begin{matrix}-3\\\frac{5}{2}\\\frac{13}{2}\end{matrix}\right]$$
let’s convert all the elements, except
2 -th element into zero.
- To do this, let’s take 2 -th line
$$\left[\begin{matrix}0 & \frac{5}{2} & - \frac{5}{2} & -5\end{matrix}\right]$$
,
and subtract it from other lines:
From 1 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}2 - \frac{\left(-6\right) 0}{5} & -3 - \frac{\left(-6\right) 5}{2 \cdot 5} & - \frac{\left(-6\right) \left(-1\right) 5}{2 \cdot 5} - 1 & 4 - - -6\end{matrix}\right] = \left[\begin{matrix}2 & 0 & -4 & -2\end{matrix}\right]$$
you get
$$\left[\begin{matrix}2 & 0 & -4 & -2\\0 & \frac{5}{2} & - \frac{5}{2} & -5\\0 & \frac{13}{2} & \frac{5}{2} & -4\end{matrix}\right]$$
From 3 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}- \frac{0 \cdot 13}{5} & \frac{13}{2} - \frac{5 \cdot 13}{2 \cdot 5} & \frac{5}{2} - - \frac{13}{2} & -4 - - 13\end{matrix}\right] = \left[\begin{matrix}0 & 0 & 9 & 9\end{matrix}\right]$$
you get
$$\left[\begin{matrix}2 & 0 & -4 & -2\\0 & \frac{5}{2} & - \frac{5}{2} & -5\\0 & 0 & 9 & 9\end{matrix}\right]$$
In 3 -th column
$$\left[\begin{matrix}-4\\- \frac{5}{2}\\9\end{matrix}\right]$$
let’s convert all the elements, except
3 -th element into zero.
- To do this, let’s take 3 -th line
$$\left[\begin{matrix}0 & 0 & 9 & 9\end{matrix}\right]$$
,
and subtract it from other lines:
From 1 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}2 - \frac{\left(-4\right) 0}{9} & - \frac{\left(-4\right) 0}{9} & -4 - \frac{\left(-4\right) 9}{9} & -2 - \frac{\left(-4\right) 9}{9}\end{matrix}\right] = \left[\begin{matrix}2 & 0 & 0 & 2\end{matrix}\right]$$
you get
$$\left[\begin{matrix}2 & 0 & 0 & 2\\0 & \frac{5}{2} & - \frac{5}{2} & -5\\0 & 0 & 9 & 9\end{matrix}\right]$$
From 2 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}- \frac{\left(-5\right) 0}{18} & \frac{5}{2} - \frac{\left(-5\right) 0}{18} & - \frac{5}{2} - \frac{\left(-5\right) 9}{18} & -5 - \frac{\left(-5\right) 9}{18}\end{matrix}\right] = \left[\begin{matrix}0 & \frac{5}{2} & 0 & - \frac{5}{2}\end{matrix}\right]$$
you get
$$\left[\begin{matrix}2 & 0 & 0 & 2\\0 & \frac{5}{2} & 0 & - \frac{5}{2}\\0 & 0 & 9 & 9\end{matrix}\right]$$
It is almost ready, all we have to do is to find variables, solving the elementary equations:
$$2 x_{1} - 2 = 0$$
$$\frac{5 x_{2}}{2} + \frac{5}{2} = 0$$
$$9 x_{3} - 9 = 0$$
We get the answer:
$$x_{1} = 1$$
$$x_{2} = -1$$
$$x_{3} = 1$$
$$- x_{3} + \left(2 x_{1} - 3 x_{2}\right) = 4$$
$$- 3 x_{3} + \left(x_{1} + x_{2}\right) = -3$$
$$x_{3} + \left(3 x_{1} + 2 x_{2}\right) = 2$$
We give the system of equations to the canonical form
$$2 x_{1} - 3 x_{2} - x_{3} = 4$$
$$x_{1} + x_{2} - 3 x_{3} = -3$$
$$3 x_{1} + 2 x_{2} + x_{3} = 2$$
Rewrite the system of linear equations as the matrix form
$$\left[\begin{matrix}2 & -3 & -1 & 4\\1 & 1 & -3 & -3\\3 & 2 & 1 & 2\end{matrix}\right]$$
In 1 -th column
$$\left[\begin{matrix}2\\1\\3\end{matrix}\right]$$
let’s convert all the elements, except
1 -th element into zero.
- To do this, let’s take 1 -th line
$$\left[\begin{matrix}2 & -3 & -1 & 4\end{matrix}\right]$$
,
and subtract it from other lines:
From 2 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}1 - \frac{2}{2} & 1 - - \frac{3}{2} & -3 - - \frac{1}{2} & -3 - \frac{4}{2}\end{matrix}\right] = \left[\begin{matrix}0 & \frac{5}{2} & - \frac{5}{2} & -5\end{matrix}\right]$$
you get
$$\left[\begin{matrix}2 & -3 & -1 & 4\\0 & \frac{5}{2} & - \frac{5}{2} & -5\\3 & 2 & 1 & 2\end{matrix}\right]$$
From 3 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}3 - \frac{2 \cdot 3}{2} & 2 - - \frac{9}{2} & 1 - - \frac{3}{2} & 2 - \frac{3 \cdot 4}{2}\end{matrix}\right] = \left[\begin{matrix}0 & \frac{13}{2} & \frac{5}{2} & -4\end{matrix}\right]$$
you get
$$\left[\begin{matrix}2 & -3 & -1 & 4\\0 & \frac{5}{2} & - \frac{5}{2} & -5\\0 & \frac{13}{2} & \frac{5}{2} & -4\end{matrix}\right]$$
In 2 -th column
$$\left[\begin{matrix}-3\\\frac{5}{2}\\\frac{13}{2}\end{matrix}\right]$$
let’s convert all the elements, except
2 -th element into zero.
- To do this, let’s take 2 -th line
$$\left[\begin{matrix}0 & \frac{5}{2} & - \frac{5}{2} & -5\end{matrix}\right]$$
,
and subtract it from other lines:
From 1 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}2 - \frac{\left(-6\right) 0}{5} & -3 - \frac{\left(-6\right) 5}{2 \cdot 5} & - \frac{\left(-6\right) \left(-1\right) 5}{2 \cdot 5} - 1 & 4 - - -6\end{matrix}\right] = \left[\begin{matrix}2 & 0 & -4 & -2\end{matrix}\right]$$
you get
$$\left[\begin{matrix}2 & 0 & -4 & -2\\0 & \frac{5}{2} & - \frac{5}{2} & -5\\0 & \frac{13}{2} & \frac{5}{2} & -4\end{matrix}\right]$$
From 3 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}- \frac{0 \cdot 13}{5} & \frac{13}{2} - \frac{5 \cdot 13}{2 \cdot 5} & \frac{5}{2} - - \frac{13}{2} & -4 - - 13\end{matrix}\right] = \left[\begin{matrix}0 & 0 & 9 & 9\end{matrix}\right]$$
you get
$$\left[\begin{matrix}2 & 0 & -4 & -2\\0 & \frac{5}{2} & - \frac{5}{2} & -5\\0 & 0 & 9 & 9\end{matrix}\right]$$
In 3 -th column
$$\left[\begin{matrix}-4\\- \frac{5}{2}\\9\end{matrix}\right]$$
let’s convert all the elements, except
3 -th element into zero.
- To do this, let’s take 3 -th line
$$\left[\begin{matrix}0 & 0 & 9 & 9\end{matrix}\right]$$
,
and subtract it from other lines:
From 1 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}2 - \frac{\left(-4\right) 0}{9} & - \frac{\left(-4\right) 0}{9} & -4 - \frac{\left(-4\right) 9}{9} & -2 - \frac{\left(-4\right) 9}{9}\end{matrix}\right] = \left[\begin{matrix}2 & 0 & 0 & 2\end{matrix}\right]$$
you get
$$\left[\begin{matrix}2 & 0 & 0 & 2\\0 & \frac{5}{2} & - \frac{5}{2} & -5\\0 & 0 & 9 & 9\end{matrix}\right]$$
From 2 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}- \frac{\left(-5\right) 0}{18} & \frac{5}{2} - \frac{\left(-5\right) 0}{18} & - \frac{5}{2} - \frac{\left(-5\right) 9}{18} & -5 - \frac{\left(-5\right) 9}{18}\end{matrix}\right] = \left[\begin{matrix}0 & \frac{5}{2} & 0 & - \frac{5}{2}\end{matrix}\right]$$
you get
$$\left[\begin{matrix}2 & 0 & 0 & 2\\0 & \frac{5}{2} & 0 & - \frac{5}{2}\\0 & 0 & 9 & 9\end{matrix}\right]$$
It is almost ready, all we have to do is to find variables, solving the elementary equations:
$$2 x_{1} - 2 = 0$$
$$\frac{5 x_{2}}{2} + \frac{5}{2} = 0$$
$$9 x_{3} - 9 = 0$$
We get the answer:
$$x_{1} = 1$$
$$x_{2} = -1$$
$$x_{3} = 1$$