Mister Exam
x+3y+4z=17; 2x-3y+5z=16; 3x+4y-z=7
The solution
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x + 3*y + 4*z = 17
$$4 z + \left(x + 3 y\right) = 17$$
2*x - 3*y + 5*z = 16
$$5 z + \left(2 x - 3 y\right) = 16$$
3*x + 4*y - z = 7
$$- z + \left(3 x + 4 y\right) = 7$$
-z + 3*x + 4*y = 7
Rapid solution
$$x_{1} = 2$$
=
$$2$$
=
$$y_{1} = 1$$
=
$$1$$
=
$$z_{1} = 3$$
=
$$3$$
=
=
$$2$$
=
2
$$y_{1} = 1$$
=
$$1$$
=
1
$$z_{1} = 3$$
=
$$3$$
=
3
Cramer's rule
$$4 z + \left(x + 3 y\right) = 17$$
$$5 z + \left(2 x - 3 y\right) = 16$$
$$- z + \left(3 x + 4 y\right) = 7$$
We give the system of equations to the canonical form
$$x + 3 y + 4 z = 17$$
$$2 x - 3 y + 5 z = 16$$
$$3 x + 4 y - z = 7$$
Rewrite the system of linear equations as the matrix form
$$\left[\begin{matrix}x_{1} + 3 x_{2} + 4 x_{3}\\2 x_{1} - 3 x_{2} + 5 x_{3}\\3 x_{1} + 4 x_{2} - x_{3}\end{matrix}\right] = \left[\begin{matrix}17\\16\\7\end{matrix}\right]$$
- 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:
$$A = \operatorname{det}{\left(\left[\begin{matrix}1 & 3 & 4\\2 & -3 & 5\\3 & 4 & -1\end{matrix}\right] \right)} = 102$$
, 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 )
$$x_{1} = \frac{\operatorname{det}{\left(\left[\begin{matrix}17 & 3 & 4\\16 & -3 & 5\\7 & 4 & -1\end{matrix}\right] \right)}}{102} = 2$$
$$x_{2} = \frac{\operatorname{det}{\left(\left[\begin{matrix}1 & 17 & 4\\2 & 16 & 5\\3 & 7 & -1\end{matrix}\right] \right)}}{102} = 1$$
$$x_{3} = \frac{\operatorname{det}{\left(\left[\begin{matrix}1 & 3 & 17\\2 & -3 & 16\\3 & 4 & 7\end{matrix}\right] \right)}}{102} = 3$$
$$5 z + \left(2 x - 3 y\right) = 16$$
$$- z + \left(3 x + 4 y\right) = 7$$
We give the system of equations to the canonical form
$$x + 3 y + 4 z = 17$$
$$2 x - 3 y + 5 z = 16$$
$$3 x + 4 y - z = 7$$
Rewrite the system of linear equations as the matrix form
$$\left[\begin{matrix}x_{1} + 3 x_{2} + 4 x_{3}\\2 x_{1} - 3 x_{2} + 5 x_{3}\\3 x_{1} + 4 x_{2} - x_{3}\end{matrix}\right] = \left[\begin{matrix}17\\16\\7\end{matrix}\right]$$
- 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:
$$A = \operatorname{det}{\left(\left[\begin{matrix}1 & 3 & 4\\2 & -3 & 5\\3 & 4 & -1\end{matrix}\right] \right)} = 102$$
, 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 )
$$x_{1} = \frac{\operatorname{det}{\left(\left[\begin{matrix}17 & 3 & 4\\16 & -3 & 5\\7 & 4 & -1\end{matrix}\right] \right)}}{102} = 2$$
$$x_{2} = \frac{\operatorname{det}{\left(\left[\begin{matrix}1 & 17 & 4\\2 & 16 & 5\\3 & 7 & -1\end{matrix}\right] \right)}}{102} = 1$$
$$x_{3} = \frac{\operatorname{det}{\left(\left[\begin{matrix}1 & 3 & 17\\2 & -3 & 16\\3 & 4 & 7\end{matrix}\right] \right)}}{102} = 3$$
Gaussian elimination
Given the system of equations
$$4 z + \left(x + 3 y\right) = 17$$
$$5 z + \left(2 x - 3 y\right) = 16$$
$$- z + \left(3 x + 4 y\right) = 7$$
We give the system of equations to the canonical form
$$x + 3 y + 4 z = 17$$
$$2 x - 3 y + 5 z = 16$$
$$3 x + 4 y - z = 7$$
Rewrite the system of linear equations as the matrix form
$$\left[\begin{matrix}1 & 3 & 4 & 17\\2 & -3 & 5 & 16\\3 & 4 & -1 & 7\end{matrix}\right]$$
In 1 -th column
$$\left[\begin{matrix}1\\2\\3\end{matrix}\right]$$
let’s convert all the elements, except
1 -th element into zero.
- To do this, let’s take 1 -th line
$$\left[\begin{matrix}1 & 3 & 4 & 17\end{matrix}\right]$$
,
and subtract it from other lines:
From 2 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}\left(-1\right) 2 + 2 & - 2 \cdot 3 - 3 & 5 - 2 \cdot 4 & 16 - 2 \cdot 17\end{matrix}\right] = \left[\begin{matrix}0 & -9 & -3 & -18\end{matrix}\right]$$
you get
$$\left[\begin{matrix}1 & 3 & 4 & 17\\0 & -9 & -3 & -18\\3 & 4 & -1 & 7\end{matrix}\right]$$
From 3 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}\left(-1\right) 3 + 3 & 4 - 3 \cdot 3 & - 3 \cdot 4 - 1 & 7 - 3 \cdot 17\end{matrix}\right] = \left[\begin{matrix}0 & -5 & -13 & -44\end{matrix}\right]$$
you get
$$\left[\begin{matrix}1 & 3 & 4 & 17\\0 & -9 & -3 & -18\\0 & -5 & -13 & -44\end{matrix}\right]$$
In 2 -th column
$$\left[\begin{matrix}3\\-9\\-5\end{matrix}\right]$$
let’s convert all the elements, except
2 -th element into zero.
- To do this, let’s take 2 -th line
$$\left[\begin{matrix}0 & -9 & -3 & -18\end{matrix}\right]$$
,
and subtract it from other lines:
From 1 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}1 - \frac{\left(-1\right) 0}{3} & 3 - - -3 & 4 - - -1 & 17 - - -6\end{matrix}\right] = \left[\begin{matrix}1 & 0 & 3 & 11\end{matrix}\right]$$
you get
$$\left[\begin{matrix}1 & 0 & 3 & 11\\0 & -9 & -3 & -18\\0 & -5 & -13 & -44\end{matrix}\right]$$
From 3 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}- \frac{0 \cdot 5}{9} & -5 - - 5 & -13 - - \frac{5}{3} & -44 - - 10\end{matrix}\right] = \left[\begin{matrix}0 & 0 & - \frac{34}{3} & -34\end{matrix}\right]$$
you get
$$\left[\begin{matrix}1 & 0 & 3 & 11\\0 & -9 & -3 & -18\\0 & 0 & - \frac{34}{3} & -34\end{matrix}\right]$$
In 3 -th column
$$\left[\begin{matrix}3\\-3\\- \frac{34}{3}\end{matrix}\right]$$
let’s convert all the elements, except
3 -th element into zero.
- To do this, let’s take 3 -th line
$$\left[\begin{matrix}0 & 0 & - \frac{34}{3} & -34\end{matrix}\right]$$
,
and subtract it from other lines:
From 1 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}1 - \frac{\left(-9\right) 0}{34} & - \frac{\left(-9\right) 0}{34} & 3 - - -3 & 11 - - -9\end{matrix}\right] = \left[\begin{matrix}1 & 0 & 0 & 2\end{matrix}\right]$$
you get
$$\left[\begin{matrix}1 & 0 & 0 & 2\\0 & -9 & -3 & -18\\0 & 0 & - \frac{34}{3} & -34\end{matrix}\right]$$
From 2 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}- \frac{0 \cdot 9}{34} & -9 - \frac{0 \cdot 9}{34} & -3 - - 3 & -18 - - 9\end{matrix}\right] = \left[\begin{matrix}0 & -9 & 0 & -9\end{matrix}\right]$$
you get
$$\left[\begin{matrix}1 & 0 & 0 & 2\\0 & -9 & 0 & -9\\0 & 0 & - \frac{34}{3} & -34\end{matrix}\right]$$
It is almost ready, all we have to do is to find variables, solving the elementary equations:
$$x_{1} - 2 = 0$$
$$9 - 9 x_{2} = 0$$
$$34 - \frac{34 x_{3}}{3} = 0$$
We get the answer:
$$x_{1} = 2$$
$$x_{2} = 1$$
$$x_{3} = 3$$
$$4 z + \left(x + 3 y\right) = 17$$
$$5 z + \left(2 x - 3 y\right) = 16$$
$$- z + \left(3 x + 4 y\right) = 7$$
We give the system of equations to the canonical form
$$x + 3 y + 4 z = 17$$
$$2 x - 3 y + 5 z = 16$$
$$3 x + 4 y - z = 7$$
Rewrite the system of linear equations as the matrix form
$$\left[\begin{matrix}1 & 3 & 4 & 17\\2 & -3 & 5 & 16\\3 & 4 & -1 & 7\end{matrix}\right]$$
In 1 -th column
$$\left[\begin{matrix}1\\2\\3\end{matrix}\right]$$
let’s convert all the elements, except
1 -th element into zero.
- To do this, let’s take 1 -th line
$$\left[\begin{matrix}1 & 3 & 4 & 17\end{matrix}\right]$$
,
and subtract it from other lines:
From 2 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}\left(-1\right) 2 + 2 & - 2 \cdot 3 - 3 & 5 - 2 \cdot 4 & 16 - 2 \cdot 17\end{matrix}\right] = \left[\begin{matrix}0 & -9 & -3 & -18\end{matrix}\right]$$
you get
$$\left[\begin{matrix}1 & 3 & 4 & 17\\0 & -9 & -3 & -18\\3 & 4 & -1 & 7\end{matrix}\right]$$
From 3 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}\left(-1\right) 3 + 3 & 4 - 3 \cdot 3 & - 3 \cdot 4 - 1 & 7 - 3 \cdot 17\end{matrix}\right] = \left[\begin{matrix}0 & -5 & -13 & -44\end{matrix}\right]$$
you get
$$\left[\begin{matrix}1 & 3 & 4 & 17\\0 & -9 & -3 & -18\\0 & -5 & -13 & -44\end{matrix}\right]$$
In 2 -th column
$$\left[\begin{matrix}3\\-9\\-5\end{matrix}\right]$$
let’s convert all the elements, except
2 -th element into zero.
- To do this, let’s take 2 -th line
$$\left[\begin{matrix}0 & -9 & -3 & -18\end{matrix}\right]$$
,
and subtract it from other lines:
From 1 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}1 - \frac{\left(-1\right) 0}{3} & 3 - - -3 & 4 - - -1 & 17 - - -6\end{matrix}\right] = \left[\begin{matrix}1 & 0 & 3 & 11\end{matrix}\right]$$
you get
$$\left[\begin{matrix}1 & 0 & 3 & 11\\0 & -9 & -3 & -18\\0 & -5 & -13 & -44\end{matrix}\right]$$
From 3 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}- \frac{0 \cdot 5}{9} & -5 - - 5 & -13 - - \frac{5}{3} & -44 - - 10\end{matrix}\right] = \left[\begin{matrix}0 & 0 & - \frac{34}{3} & -34\end{matrix}\right]$$
you get
$$\left[\begin{matrix}1 & 0 & 3 & 11\\0 & -9 & -3 & -18\\0 & 0 & - \frac{34}{3} & -34\end{matrix}\right]$$
In 3 -th column
$$\left[\begin{matrix}3\\-3\\- \frac{34}{3}\end{matrix}\right]$$
let’s convert all the elements, except
3 -th element into zero.
- To do this, let’s take 3 -th line
$$\left[\begin{matrix}0 & 0 & - \frac{34}{3} & -34\end{matrix}\right]$$
,
and subtract it from other lines:
From 1 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}1 - \frac{\left(-9\right) 0}{34} & - \frac{\left(-9\right) 0}{34} & 3 - - -3 & 11 - - -9\end{matrix}\right] = \left[\begin{matrix}1 & 0 & 0 & 2\end{matrix}\right]$$
you get
$$\left[\begin{matrix}1 & 0 & 0 & 2\\0 & -9 & -3 & -18\\0 & 0 & - \frac{34}{3} & -34\end{matrix}\right]$$
From 2 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}- \frac{0 \cdot 9}{34} & -9 - \frac{0 \cdot 9}{34} & -3 - - 3 & -18 - - 9\end{matrix}\right] = \left[\begin{matrix}0 & -9 & 0 & -9\end{matrix}\right]$$
you get
$$\left[\begin{matrix}1 & 0 & 0 & 2\\0 & -9 & 0 & -9\\0 & 0 & - \frac{34}{3} & -34\end{matrix}\right]$$
It is almost ready, all we have to do is to find variables, solving the elementary equations:
$$x_{1} - 2 = 0$$
$$9 - 9 x_{2} = 0$$
$$34 - \frac{34 x_{3}}{3} = 0$$
We get the answer:
$$x_{1} = 2$$
$$x_{2} = 1$$
$$x_{3} = 3$$