Mister Exam
3x-2y+4z=11; 2x-y-z=4; 3x+4y-2z=11
The solution
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3*x - 2*y + 4*z = 11
$$4 z + \left(3 x - 2 y\right) = 11$$
2*x - y - z = 4
$$- z + \left(2 x - y\right) = 4$$
3*x + 4*y - 2*z = 11
$$- 2 z + \left(3 x + 4 y\right) = 11$$
-2*z + 3*x + 4*y = 11
Rapid solution
$$x_{1} = 3$$
=
$$3$$
=
$$y_{1} = 1$$
=
$$1$$
=
$$z_{1} = 1$$
=
$$1$$
=
=
$$3$$
=
3
$$y_{1} = 1$$
=
$$1$$
=
1
$$z_{1} = 1$$
=
$$1$$
=
1
Cramer's rule
$$4 z + \left(3 x - 2 y\right) = 11$$
$$- z + \left(2 x - y\right) = 4$$
$$- 2 z + \left(3 x + 4 y\right) = 11$$
We give the system of equations to the canonical form
$$3 x - 2 y + 4 z = 11$$
$$2 x - y - z = 4$$
$$3 x + 4 y - 2 z = 11$$
Rewrite the system of linear equations as the matrix form
$$\left[\begin{matrix}3 x_{1} - 2 x_{2} + 4 x_{3}\\2 x_{1} - x_{2} - x_{3}\\3 x_{1} + 4 x_{2} - 2 x_{3}\end{matrix}\right] = \left[\begin{matrix}11\\4\\11\end{matrix}\right]$$
- 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:
$$A = \operatorname{det}{\left(\left[\begin{matrix}3 & -2 & 4\\2 & -1 & -1\\3 & 4 & -2\end{matrix}\right] \right)} = 60$$
, 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 )
$$x_{1} = \frac{\operatorname{det}{\left(\left[\begin{matrix}11 & -2 & 4\\4 & -1 & -1\\11 & 4 & -2\end{matrix}\right] \right)}}{60} = 3$$
$$x_{2} = \frac{\operatorname{det}{\left(\left[\begin{matrix}3 & 11 & 4\\2 & 4 & -1\\3 & 11 & -2\end{matrix}\right] \right)}}{60} = 1$$
$$x_{3} = \frac{\operatorname{det}{\left(\left[\begin{matrix}3 & -2 & 11\\2 & -1 & 4\\3 & 4 & 11\end{matrix}\right] \right)}}{60} = 1$$
$$- z + \left(2 x - y\right) = 4$$
$$- 2 z + \left(3 x + 4 y\right) = 11$$
We give the system of equations to the canonical form
$$3 x - 2 y + 4 z = 11$$
$$2 x - y - z = 4$$
$$3 x + 4 y - 2 z = 11$$
Rewrite the system of linear equations as the matrix form
$$\left[\begin{matrix}3 x_{1} - 2 x_{2} + 4 x_{3}\\2 x_{1} - x_{2} - x_{3}\\3 x_{1} + 4 x_{2} - 2 x_{3}\end{matrix}\right] = \left[\begin{matrix}11\\4\\11\end{matrix}\right]$$
- 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:
$$A = \operatorname{det}{\left(\left[\begin{matrix}3 & -2 & 4\\2 & -1 & -1\\3 & 4 & -2\end{matrix}\right] \right)} = 60$$
, 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 )
$$x_{1} = \frac{\operatorname{det}{\left(\left[\begin{matrix}11 & -2 & 4\\4 & -1 & -1\\11 & 4 & -2\end{matrix}\right] \right)}}{60} = 3$$
$$x_{2} = \frac{\operatorname{det}{\left(\left[\begin{matrix}3 & 11 & 4\\2 & 4 & -1\\3 & 11 & -2\end{matrix}\right] \right)}}{60} = 1$$
$$x_{3} = \frac{\operatorname{det}{\left(\left[\begin{matrix}3 & -2 & 11\\2 & -1 & 4\\3 & 4 & 11\end{matrix}\right] \right)}}{60} = 1$$
Gaussian elimination
Given the system of equations
$$4 z + \left(3 x - 2 y\right) = 11$$
$$- z + \left(2 x - y\right) = 4$$
$$- 2 z + \left(3 x + 4 y\right) = 11$$
We give the system of equations to the canonical form
$$3 x - 2 y + 4 z = 11$$
$$2 x - y - z = 4$$
$$3 x + 4 y - 2 z = 11$$
Rewrite the system of linear equations as the matrix form
$$\left[\begin{matrix}3 & -2 & 4 & 11\\2 & -1 & -1 & 4\\3 & 4 & -2 & 11\end{matrix}\right]$$
In 1 -th column
$$\left[\begin{matrix}3\\2\\3\end{matrix}\right]$$
let’s convert all the elements, except
1 -th element into zero.
- To do this, let’s take 1 -th line
$$\left[\begin{matrix}3 & -2 & 4 & 11\end{matrix}\right]$$
,
and subtract it from other lines:
From 2 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}2 - \frac{2 \cdot 3}{3} & -1 - - \frac{4}{3} & - \frac{2 \cdot 4}{3} - 1 & 4 - \frac{2 \cdot 11}{3}\end{matrix}\right] = \left[\begin{matrix}0 & \frac{1}{3} & - \frac{11}{3} & - \frac{10}{3}\end{matrix}\right]$$
you get
$$\left[\begin{matrix}3 & -2 & 4 & 11\\0 & \frac{1}{3} & - \frac{11}{3} & - \frac{10}{3}\\3 & 4 & -2 & 11\end{matrix}\right]$$
From 3 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}\left(-1\right) 3 + 3 & 4 - -2 & \left(-1\right) 4 - 2 & \left(-1\right) 11 + 11\end{matrix}\right] = \left[\begin{matrix}0 & 6 & -6 & 0\end{matrix}\right]$$
you get
$$\left[\begin{matrix}3 & -2 & 4 & 11\\0 & \frac{1}{3} & - \frac{11}{3} & - \frac{10}{3}\\0 & 6 & -6 & 0\end{matrix}\right]$$
In 2 -th column
$$\left[\begin{matrix}-2\\\frac{1}{3}\\6\end{matrix}\right]$$
let’s convert all the elements, except
2 -th element into zero.
- To do this, let’s take 2 -th line
$$\left[\begin{matrix}0 & \frac{1}{3} & - \frac{11}{3} & - \frac{10}{3}\end{matrix}\right]$$
,
and subtract it from other lines:
From 1 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}3 - \left(-6\right) 0 & -2 - - 2 & 4 - - -22 & 11 - - -20\end{matrix}\right] = \left[\begin{matrix}3 & 0 & -18 & -9\end{matrix}\right]$$
you get
$$\left[\begin{matrix}3 & 0 & -18 & -9\\0 & \frac{1}{3} & - \frac{11}{3} & - \frac{10}{3}\\0 & 6 & -6 & 0\end{matrix}\right]$$
From 3 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}- 0 \cdot 18 & 6 - \frac{18}{3} & -6 - \frac{\left(-11\right) 18}{3} & - \frac{\left(-10\right) 18}{3}\end{matrix}\right] = \left[\begin{matrix}0 & 0 & 60 & 60\end{matrix}\right]$$
you get
$$\left[\begin{matrix}3 & 0 & -18 & -9\\0 & \frac{1}{3} & - \frac{11}{3} & - \frac{10}{3}\\0 & 0 & 60 & 60\end{matrix}\right]$$
In 3 -th column
$$\left[\begin{matrix}-18\\- \frac{11}{3}\\60\end{matrix}\right]$$
let’s convert all the elements, except
3 -th element into zero.
- To do this, let’s take 3 -th line
$$\left[\begin{matrix}0 & 0 & 60 & 60\end{matrix}\right]$$
,
and subtract it from other lines:
From 1 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}3 - \frac{\left(-3\right) 0}{10} & - \frac{\left(-3\right) 0}{10} & -18 - \frac{\left(-3\right) 60}{10} & -9 - \frac{\left(-3\right) 60}{10}\end{matrix}\right] = \left[\begin{matrix}3 & 0 & 0 & 9\end{matrix}\right]$$
you get
$$\left[\begin{matrix}3 & 0 & 0 & 9\\0 & \frac{1}{3} & - \frac{11}{3} & - \frac{10}{3}\\0 & 0 & 60 & 60\end{matrix}\right]$$
From 2 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}- \frac{\left(-11\right) 0}{180} & \frac{1}{3} - \frac{\left(-11\right) 0}{180} & - \frac{11}{3} - \frac{\left(-11\right) 60}{180} & - \frac{10}{3} - \frac{\left(-11\right) 60}{180}\end{matrix}\right] = \left[\begin{matrix}0 & \frac{1}{3} & 0 & \frac{1}{3}\end{matrix}\right]$$
you get
$$\left[\begin{matrix}3 & 0 & 0 & 9\\0 & \frac{1}{3} & 0 & \frac{1}{3}\\0 & 0 & 60 & 60\end{matrix}\right]$$
It is almost ready, all we have to do is to find variables, solving the elementary equations:
$$3 x_{1} - 9 = 0$$
$$\frac{x_{2}}{3} - \frac{1}{3} = 0$$
$$60 x_{3} - 60 = 0$$
We get the answer:
$$x_{1} = 3$$
$$x_{2} = 1$$
$$x_{3} = 1$$
$$4 z + \left(3 x - 2 y\right) = 11$$
$$- z + \left(2 x - y\right) = 4$$
$$- 2 z + \left(3 x + 4 y\right) = 11$$
We give the system of equations to the canonical form
$$3 x - 2 y + 4 z = 11$$
$$2 x - y - z = 4$$
$$3 x + 4 y - 2 z = 11$$
Rewrite the system of linear equations as the matrix form
$$\left[\begin{matrix}3 & -2 & 4 & 11\\2 & -1 & -1 & 4\\3 & 4 & -2 & 11\end{matrix}\right]$$
In 1 -th column
$$\left[\begin{matrix}3\\2\\3\end{matrix}\right]$$
let’s convert all the elements, except
1 -th element into zero.
- To do this, let’s take 1 -th line
$$\left[\begin{matrix}3 & -2 & 4 & 11\end{matrix}\right]$$
,
and subtract it from other lines:
From 2 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}2 - \frac{2 \cdot 3}{3} & -1 - - \frac{4}{3} & - \frac{2 \cdot 4}{3} - 1 & 4 - \frac{2 \cdot 11}{3}\end{matrix}\right] = \left[\begin{matrix}0 & \frac{1}{3} & - \frac{11}{3} & - \frac{10}{3}\end{matrix}\right]$$
you get
$$\left[\begin{matrix}3 & -2 & 4 & 11\\0 & \frac{1}{3} & - \frac{11}{3} & - \frac{10}{3}\\3 & 4 & -2 & 11\end{matrix}\right]$$
From 3 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}\left(-1\right) 3 + 3 & 4 - -2 & \left(-1\right) 4 - 2 & \left(-1\right) 11 + 11\end{matrix}\right] = \left[\begin{matrix}0 & 6 & -6 & 0\end{matrix}\right]$$
you get
$$\left[\begin{matrix}3 & -2 & 4 & 11\\0 & \frac{1}{3} & - \frac{11}{3} & - \frac{10}{3}\\0 & 6 & -6 & 0\end{matrix}\right]$$
In 2 -th column
$$\left[\begin{matrix}-2\\\frac{1}{3}\\6\end{matrix}\right]$$
let’s convert all the elements, except
2 -th element into zero.
- To do this, let’s take 2 -th line
$$\left[\begin{matrix}0 & \frac{1}{3} & - \frac{11}{3} & - \frac{10}{3}\end{matrix}\right]$$
,
and subtract it from other lines:
From 1 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}3 - \left(-6\right) 0 & -2 - - 2 & 4 - - -22 & 11 - - -20\end{matrix}\right] = \left[\begin{matrix}3 & 0 & -18 & -9\end{matrix}\right]$$
you get
$$\left[\begin{matrix}3 & 0 & -18 & -9\\0 & \frac{1}{3} & - \frac{11}{3} & - \frac{10}{3}\\0 & 6 & -6 & 0\end{matrix}\right]$$
From 3 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}- 0 \cdot 18 & 6 - \frac{18}{3} & -6 - \frac{\left(-11\right) 18}{3} & - \frac{\left(-10\right) 18}{3}\end{matrix}\right] = \left[\begin{matrix}0 & 0 & 60 & 60\end{matrix}\right]$$
you get
$$\left[\begin{matrix}3 & 0 & -18 & -9\\0 & \frac{1}{3} & - \frac{11}{3} & - \frac{10}{3}\\0 & 0 & 60 & 60\end{matrix}\right]$$
In 3 -th column
$$\left[\begin{matrix}-18\\- \frac{11}{3}\\60\end{matrix}\right]$$
let’s convert all the elements, except
3 -th element into zero.
- To do this, let’s take 3 -th line
$$\left[\begin{matrix}0 & 0 & 60 & 60\end{matrix}\right]$$
,
and subtract it from other lines:
From 1 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}3 - \frac{\left(-3\right) 0}{10} & - \frac{\left(-3\right) 0}{10} & -18 - \frac{\left(-3\right) 60}{10} & -9 - \frac{\left(-3\right) 60}{10}\end{matrix}\right] = \left[\begin{matrix}3 & 0 & 0 & 9\end{matrix}\right]$$
you get
$$\left[\begin{matrix}3 & 0 & 0 & 9\\0 & \frac{1}{3} & - \frac{11}{3} & - \frac{10}{3}\\0 & 0 & 60 & 60\end{matrix}\right]$$
From 2 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}- \frac{\left(-11\right) 0}{180} & \frac{1}{3} - \frac{\left(-11\right) 0}{180} & - \frac{11}{3} - \frac{\left(-11\right) 60}{180} & - \frac{10}{3} - \frac{\left(-11\right) 60}{180}\end{matrix}\right] = \left[\begin{matrix}0 & \frac{1}{3} & 0 & \frac{1}{3}\end{matrix}\right]$$
you get
$$\left[\begin{matrix}3 & 0 & 0 & 9\\0 & \frac{1}{3} & 0 & \frac{1}{3}\\0 & 0 & 60 & 60\end{matrix}\right]$$
It is almost ready, all we have to do is to find variables, solving the elementary equations:
$$3 x_{1} - 9 = 0$$
$$\frac{x_{2}}{3} - \frac{1}{3} = 0$$
$$60 x_{3} - 60 = 0$$
We get the answer:
$$x_{1} = 3$$
$$x_{2} = 1$$
$$x_{3} = 1$$