Mister Exam
2x1−3x2−x3=4; x1+x2−3x3=−3; 3x1+2x2+x3=2
The solution
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2*x1 - 3*x2 - x3 = 4
$$2 x_{1} - 3 x_{2} - x_{3} = 4$$
x1 + x2 - 3*x3 = -3
$$x_{1} + x_{2} - 3 x_{3} = -3$$
3*x1 + 2*x2 + x3 = 2
$$3 x_{1} + 2 x_{2} + x_{3} = 2$$
or
$$\begin{cases}2 x_{1} - 3 x_{2} - x_{3} = 4\\x_{1} + x_{2} - 3 x_{3} = -3\\3 x_{1} + 2 x_{2} + x_{3} = 2\end{cases}$$
Rapid solution
Gaussian elimination
Given the system of equations
$$2 x_{1} - 3 x_{2} - x_{3} = 4$$
$$x_{1} + x_{2} - 3 x_{3} = -3$$
$$3 x_{1} + 2 x_{2} + x_{3} = 2$$
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:
$$\left[\begin{matrix}0 & \frac{5}{2} & - \frac{5}{2} & -5\end{matrix}\right] = \left[\begin{matrix}0 & \frac{5}{2} & - \frac{5}{2} & -5\end{matrix}\right]$$
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:
$$\left[\begin{matrix}0 & \frac{13}{2} & \frac{5}{2} & -4\end{matrix}\right] = \left[\begin{matrix}0 & \frac{13}{2} & \frac{5}{2} & -4\end{matrix}\right]$$
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:
$$\left[\begin{matrix}2 & 0 & -4 & -2\end{matrix}\right] = \left[\begin{matrix}2 & 0 & -4 & -2\end{matrix}\right]$$
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:
$$\left[\begin{matrix}0 & 0 & 9 & 9\end{matrix}\right] = \left[\begin{matrix}0 & 0 & 9 & 9\end{matrix}\right]$$
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:
$$\left[\begin{matrix}2 & 0 & 0 & 2\end{matrix}\right] = \left[\begin{matrix}2 & 0 & 0 & 2\end{matrix}\right]$$
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:
$$\left[\begin{matrix}0 & \frac{5}{2} & 0 & - \frac{5}{2}\end{matrix}\right] = \left[\begin{matrix}0 & \frac{5}{2} & 0 & - \frac{5}{2}\end{matrix}\right]$$
you get
$$\left[\begin{matrix}2 & 0 & 0 & 2\\0 & \frac{5}{2} & 0 & - \frac{5}{2}\\0 & 0 & 9 & 9\end{matrix}\right]$$
Everything is almost ready - it remains only to find the unknowns by solving 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:
$$x_{1} = 1$$
$$x_{2} = -1$$
$$x_{3} = 1$$
$$2 x_{1} - 3 x_{2} - x_{3} = 4$$
$$x_{1} + x_{2} - 3 x_{3} = -3$$
$$3 x_{1} + 2 x_{2} + x_{3} = 2$$
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:
$$\left[\begin{matrix}0 & \frac{5}{2} & - \frac{5}{2} & -5\end{matrix}\right] = \left[\begin{matrix}0 & \frac{5}{2} & - \frac{5}{2} & -5\end{matrix}\right]$$
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:
$$\left[\begin{matrix}0 & \frac{13}{2} & \frac{5}{2} & -4\end{matrix}\right] = \left[\begin{matrix}0 & \frac{13}{2} & \frac{5}{2} & -4\end{matrix}\right]$$
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:
$$\left[\begin{matrix}2 & 0 & -4 & -2\end{matrix}\right] = \left[\begin{matrix}2 & 0 & -4 & -2\end{matrix}\right]$$
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:
$$\left[\begin{matrix}0 & 0 & 9 & 9\end{matrix}\right] = \left[\begin{matrix}0 & 0 & 9 & 9\end{matrix}\right]$$
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:
$$\left[\begin{matrix}2 & 0 & 0 & 2\end{matrix}\right] = \left[\begin{matrix}2 & 0 & 0 & 2\end{matrix}\right]$$
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:
$$\left[\begin{matrix}0 & \frac{5}{2} & 0 & - \frac{5}{2}\end{matrix}\right] = \left[\begin{matrix}0 & \frac{5}{2} & 0 & - \frac{5}{2}\end{matrix}\right]$$
you get
$$\left[\begin{matrix}2 & 0 & 0 & 2\\0 & \frac{5}{2} & 0 & - \frac{5}{2}\\0 & 0 & 9 & 9\end{matrix}\right]$$
Everything is almost ready - it remains only to find the unknowns by solving 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:
$$x_{1} = 1$$
$$x_{2} = -1$$
$$x_{3} = 1$$
Cramer's rule
$$2 x_{1} - 3 x_{2} - x_{3} = 4$$
$$x_{1} + x_{2} - 3 x_{3} = -3$$
$$3 x_{1} + 2 x_{2} + x_{3} = 2$$
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
$$\left[\begin{matrix}2 x_{1} - 3 x_{2} - x_{3}\\1 x_{1} + 1 x_{2} - 3 x_{3}\\3 x_{1} + 2 x_{2} + 1 x_{3}\end{matrix}\right] = \left[\begin{matrix}4\\-3\\2\end{matrix}\right]$$
- 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 )
$$x_{1} = \frac{\operatorname{det}{\left(\left[\begin{matrix}4 & -3 & -1\\-3 & 1 & -3\\2 & 2 & 1\end{matrix}\right] \right)}}{45} = 1$$
$$x_{2} = \frac{\operatorname{det}{\left(\left[\begin{matrix}2 & 4 & -1\\1 & -3 & -3\\3 & 2 & 1\end{matrix}\right] \right)}}{45} = -1$$
$$x_{3} = \frac{\operatorname{det}{\left(\left[\begin{matrix}2 & -3 & 4\\1 & 1 & -3\\3 & 2 & 2\end{matrix}\right] \right)}}{45} = 1$$
$$x_{1} + x_{2} - 3 x_{3} = -3$$
$$3 x_{1} + 2 x_{2} + x_{3} = 2$$
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
$$\left[\begin{matrix}2 x_{1} - 3 x_{2} - x_{3}\\1 x_{1} + 1 x_{2} - 3 x_{3}\\3 x_{1} + 2 x_{2} + 1 x_{3}\end{matrix}\right] = \left[\begin{matrix}4\\-3\\2\end{matrix}\right]$$
- 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 )
$$x_{1} = \frac{\operatorname{det}{\left(\left[\begin{matrix}4 & -3 & -1\\-3 & 1 & -3\\2 & 2 & 1\end{matrix}\right] \right)}}{45} = 1$$
$$x_{2} = \frac{\operatorname{det}{\left(\left[\begin{matrix}2 & 4 & -1\\1 & -3 & -3\\3 & 2 & 1\end{matrix}\right] \right)}}{45} = -1$$
$$x_{3} = \frac{\operatorname{det}{\left(\left[\begin{matrix}2 & -3 & 4\\1 & 1 & -3\\3 & 2 & 2\end{matrix}\right] \right)}}{45} = 1$$