Mister Exam
5x+3y-z=-7; 3x-4y+2z=10; x+y-3z=-5
The solution
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5*x + 3*y - z = -7
$$- z + \left(5 x + 3 y\right) = -7$$
3*x - 4*y + 2*z = 10
$$2 z + \left(3 x - 4 y\right) = 10$$
x + y - 3*z = -5
$$- 3 z + \left(x + y\right) = -5$$
-3*z + x + y = -5
Rapid solution
$$x_{1} = 0$$
=
$$0$$
=
$$y_{1} = -2$$
=
$$-2$$
=
$$z_{1} = 1$$
=
$$1$$
=
=
$$0$$
=
0
$$y_{1} = -2$$
=
$$-2$$
=
-2
$$z_{1} = 1$$
=
$$1$$
=
1
Gaussian elimination
Given the system of equations
$$- z + \left(5 x + 3 y\right) = -7$$
$$2 z + \left(3 x - 4 y\right) = 10$$
$$- 3 z + \left(x + y\right) = -5$$
We give the system of equations to the canonical form
$$5 x + 3 y - z = -7$$
$$3 x - 4 y + 2 z = 10$$
$$x + y - 3 z = -5$$
Rewrite the system of linear equations as the matrix form
$$\left[\begin{matrix}5 & 3 & -1 & -7\\3 & -4 & 2 & 10\\1 & 1 & -3 & -5\end{matrix}\right]$$
In 1 -th column
$$\left[\begin{matrix}5\\3\\1\end{matrix}\right]$$
let’s convert all the elements, except
1 -th element into zero.
- To do this, let’s take 1 -th line
$$\left[\begin{matrix}5 & 3 & -1 & -7\end{matrix}\right]$$
,
and subtract it from other lines:
From 2 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}3 - \frac{3 \cdot 5}{5} & -4 - \frac{3 \cdot 3}{5} & 2 - - \frac{3}{5} & 10 - - \frac{21}{5}\end{matrix}\right] = \left[\begin{matrix}0 & - \frac{29}{5} & \frac{13}{5} & \frac{71}{5}\end{matrix}\right]$$
you get
$$\left[\begin{matrix}5 & 3 & -1 & -7\\0 & - \frac{29}{5} & \frac{13}{5} & \frac{71}{5}\\1 & 1 & -3 & -5\end{matrix}\right]$$
From 3 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}1 - \frac{5}{5} & 1 - \frac{3}{5} & -3 - - \frac{1}{5} & -5 - - \frac{7}{5}\end{matrix}\right] = \left[\begin{matrix}0 & \frac{2}{5} & - \frac{14}{5} & - \frac{18}{5}\end{matrix}\right]$$
you get
$$\left[\begin{matrix}5 & 3 & -1 & -7\\0 & - \frac{29}{5} & \frac{13}{5} & \frac{71}{5}\\0 & \frac{2}{5} & - \frac{14}{5} & - \frac{18}{5}\end{matrix}\right]$$
In 2 -th column
$$\left[\begin{matrix}3\\- \frac{29}{5}\\\frac{2}{5}\end{matrix}\right]$$
let’s convert all the elements, except
2 -th element into zero.
- To do this, let’s take 2 -th line
$$\left[\begin{matrix}0 & - \frac{29}{5} & \frac{13}{5} & \frac{71}{5}\end{matrix}\right]$$
,
and subtract it from other lines:
From 1 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}5 - \frac{\left(-15\right) 0}{29} & 3 - - -3 & -1 - \frac{\left(-15\right) 13}{5 \cdot 29} & -7 - \frac{\left(-15\right) 71}{5 \cdot 29}\end{matrix}\right] = \left[\begin{matrix}5 & 0 & \frac{10}{29} & \frac{10}{29}\end{matrix}\right]$$
you get
$$\left[\begin{matrix}5 & 0 & \frac{10}{29} & \frac{10}{29}\\0 & - \frac{29}{5} & \frac{13}{5} & \frac{71}{5}\\0 & \frac{2}{5} & - \frac{14}{5} & - \frac{18}{5}\end{matrix}\right]$$
From 3 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}- \frac{\left(-2\right) 0}{29} & \frac{2}{5} - - \frac{-2}{5} & - \frac{14}{5} - \frac{\left(-2\right) 13}{5 \cdot 29} & - \frac{18}{5} - \frac{\left(-2\right) 71}{5 \cdot 29}\end{matrix}\right] = \left[\begin{matrix}0 & 0 & - \frac{76}{29} & - \frac{76}{29}\end{matrix}\right]$$
you get
$$\left[\begin{matrix}5 & 0 & \frac{10}{29} & \frac{10}{29}\\0 & - \frac{29}{5} & \frac{13}{5} & \frac{71}{5}\\0 & 0 & - \frac{76}{29} & - \frac{76}{29}\end{matrix}\right]$$
In 3 -th column
$$\left[\begin{matrix}\frac{10}{29}\\\frac{13}{5}\\- \frac{76}{29}\end{matrix}\right]$$
let’s convert all the elements, except
3 -th element into zero.
- To do this, let’s take 3 -th line
$$\left[\begin{matrix}0 & 0 & - \frac{76}{29} & - \frac{76}{29}\end{matrix}\right]$$
,
and subtract it from other lines:
From 1 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}5 - \frac{\left(-5\right) 0}{38} & - \frac{\left(-5\right) 0}{38} & \frac{10}{29} - - \frac{-10}{29} & \frac{10}{29} - - \frac{-10}{29}\end{matrix}\right] = \left[\begin{matrix}5 & 0 & 0 & 0\end{matrix}\right]$$
you get
$$\left[\begin{matrix}5 & 0 & 0 & 0\\0 & - \frac{29}{5} & \frac{13}{5} & \frac{71}{5}\\0 & 0 & - \frac{76}{29} & - \frac{76}{29}\end{matrix}\right]$$
From 2 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}- \frac{\left(-377\right) 0}{380} & - \frac{29}{5} - \frac{\left(-377\right) 0}{380} & \frac{13}{5} - - \frac{-13}{5} & \frac{71}{5} - - \frac{-13}{5}\end{matrix}\right] = \left[\begin{matrix}0 & - \frac{29}{5} & 0 & \frac{58}{5}\end{matrix}\right]$$
you get
$$\left[\begin{matrix}5 & 0 & 0 & 0\\0 & - \frac{29}{5} & 0 & \frac{58}{5}\\0 & 0 & - \frac{76}{29} & - \frac{76}{29}\end{matrix}\right]$$
It is almost ready, all we have to do is to find variables, solving the elementary equations:
$$5 x_{1} = 0$$
$$- \frac{29 x_{2}}{5} - \frac{58}{5} = 0$$
$$\frac{76}{29} - \frac{76 x_{3}}{29} = 0$$
We get the answer:
$$x_{1} = 0$$
$$x_{2} = -2$$
$$x_{3} = 1$$
$$- z + \left(5 x + 3 y\right) = -7$$
$$2 z + \left(3 x - 4 y\right) = 10$$
$$- 3 z + \left(x + y\right) = -5$$
We give the system of equations to the canonical form
$$5 x + 3 y - z = -7$$
$$3 x - 4 y + 2 z = 10$$
$$x + y - 3 z = -5$$
Rewrite the system of linear equations as the matrix form
$$\left[\begin{matrix}5 & 3 & -1 & -7\\3 & -4 & 2 & 10\\1 & 1 & -3 & -5\end{matrix}\right]$$
In 1 -th column
$$\left[\begin{matrix}5\\3\\1\end{matrix}\right]$$
let’s convert all the elements, except
1 -th element into zero.
- To do this, let’s take 1 -th line
$$\left[\begin{matrix}5 & 3 & -1 & -7\end{matrix}\right]$$
,
and subtract it from other lines:
From 2 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}3 - \frac{3 \cdot 5}{5} & -4 - \frac{3 \cdot 3}{5} & 2 - - \frac{3}{5} & 10 - - \frac{21}{5}\end{matrix}\right] = \left[\begin{matrix}0 & - \frac{29}{5} & \frac{13}{5} & \frac{71}{5}\end{matrix}\right]$$
you get
$$\left[\begin{matrix}5 & 3 & -1 & -7\\0 & - \frac{29}{5} & \frac{13}{5} & \frac{71}{5}\\1 & 1 & -3 & -5\end{matrix}\right]$$
From 3 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}1 - \frac{5}{5} & 1 - \frac{3}{5} & -3 - - \frac{1}{5} & -5 - - \frac{7}{5}\end{matrix}\right] = \left[\begin{matrix}0 & \frac{2}{5} & - \frac{14}{5} & - \frac{18}{5}\end{matrix}\right]$$
you get
$$\left[\begin{matrix}5 & 3 & -1 & -7\\0 & - \frac{29}{5} & \frac{13}{5} & \frac{71}{5}\\0 & \frac{2}{5} & - \frac{14}{5} & - \frac{18}{5}\end{matrix}\right]$$
In 2 -th column
$$\left[\begin{matrix}3\\- \frac{29}{5}\\\frac{2}{5}\end{matrix}\right]$$
let’s convert all the elements, except
2 -th element into zero.
- To do this, let’s take 2 -th line
$$\left[\begin{matrix}0 & - \frac{29}{5} & \frac{13}{5} & \frac{71}{5}\end{matrix}\right]$$
,
and subtract it from other lines:
From 1 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}5 - \frac{\left(-15\right) 0}{29} & 3 - - -3 & -1 - \frac{\left(-15\right) 13}{5 \cdot 29} & -7 - \frac{\left(-15\right) 71}{5 \cdot 29}\end{matrix}\right] = \left[\begin{matrix}5 & 0 & \frac{10}{29} & \frac{10}{29}\end{matrix}\right]$$
you get
$$\left[\begin{matrix}5 & 0 & \frac{10}{29} & \frac{10}{29}\\0 & - \frac{29}{5} & \frac{13}{5} & \frac{71}{5}\\0 & \frac{2}{5} & - \frac{14}{5} & - \frac{18}{5}\end{matrix}\right]$$
From 3 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}- \frac{\left(-2\right) 0}{29} & \frac{2}{5} - - \frac{-2}{5} & - \frac{14}{5} - \frac{\left(-2\right) 13}{5 \cdot 29} & - \frac{18}{5} - \frac{\left(-2\right) 71}{5 \cdot 29}\end{matrix}\right] = \left[\begin{matrix}0 & 0 & - \frac{76}{29} & - \frac{76}{29}\end{matrix}\right]$$
you get
$$\left[\begin{matrix}5 & 0 & \frac{10}{29} & \frac{10}{29}\\0 & - \frac{29}{5} & \frac{13}{5} & \frac{71}{5}\\0 & 0 & - \frac{76}{29} & - \frac{76}{29}\end{matrix}\right]$$
In 3 -th column
$$\left[\begin{matrix}\frac{10}{29}\\\frac{13}{5}\\- \frac{76}{29}\end{matrix}\right]$$
let’s convert all the elements, except
3 -th element into zero.
- To do this, let’s take 3 -th line
$$\left[\begin{matrix}0 & 0 & - \frac{76}{29} & - \frac{76}{29}\end{matrix}\right]$$
,
and subtract it from other lines:
From 1 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}5 - \frac{\left(-5\right) 0}{38} & - \frac{\left(-5\right) 0}{38} & \frac{10}{29} - - \frac{-10}{29} & \frac{10}{29} - - \frac{-10}{29}\end{matrix}\right] = \left[\begin{matrix}5 & 0 & 0 & 0\end{matrix}\right]$$
you get
$$\left[\begin{matrix}5 & 0 & 0 & 0\\0 & - \frac{29}{5} & \frac{13}{5} & \frac{71}{5}\\0 & 0 & - \frac{76}{29} & - \frac{76}{29}\end{matrix}\right]$$
From 2 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}- \frac{\left(-377\right) 0}{380} & - \frac{29}{5} - \frac{\left(-377\right) 0}{380} & \frac{13}{5} - - \frac{-13}{5} & \frac{71}{5} - - \frac{-13}{5}\end{matrix}\right] = \left[\begin{matrix}0 & - \frac{29}{5} & 0 & \frac{58}{5}\end{matrix}\right]$$
you get
$$\left[\begin{matrix}5 & 0 & 0 & 0\\0 & - \frac{29}{5} & 0 & \frac{58}{5}\\0 & 0 & - \frac{76}{29} & - \frac{76}{29}\end{matrix}\right]$$
It is almost ready, all we have to do is to find variables, solving the elementary equations:
$$5 x_{1} = 0$$
$$- \frac{29 x_{2}}{5} - \frac{58}{5} = 0$$
$$\frac{76}{29} - \frac{76 x_{3}}{29} = 0$$
We get the answer:
$$x_{1} = 0$$
$$x_{2} = -2$$
$$x_{3} = 1$$
Cramer's rule
$$- z + \left(5 x + 3 y\right) = -7$$
$$2 z + \left(3 x - 4 y\right) = 10$$
$$- 3 z + \left(x + y\right) = -5$$
We give the system of equations to the canonical form
$$5 x + 3 y - z = -7$$
$$3 x - 4 y + 2 z = 10$$
$$x + y - 3 z = -5$$
Rewrite the system of linear equations as the matrix form
$$\left[\begin{matrix}5 x_{1} + 3 x_{2} - x_{3}\\3 x_{1} - 4 x_{2} + 2 x_{3}\\x_{1} + x_{2} - 3 x_{3}\end{matrix}\right] = \left[\begin{matrix}-7\\10\\-5\end{matrix}\right]$$
- 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:
$$A = \operatorname{det}{\left(\left[\begin{matrix}5 & 3 & -1\\3 & -4 & 2\\1 & 1 & -3\end{matrix}\right] \right)} = 76$$
, 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 )
$$x_{1} = \frac{\operatorname{det}{\left(\left[\begin{matrix}-7 & 3 & -1\\10 & -4 & 2\\-5 & 1 & -3\end{matrix}\right] \right)}}{76} = 0$$
$$x_{2} = \frac{\operatorname{det}{\left(\left[\begin{matrix}5 & -7 & -1\\3 & 10 & 2\\1 & -5 & -3\end{matrix}\right] \right)}}{76} = -2$$
$$x_{3} = \frac{\operatorname{det}{\left(\left[\begin{matrix}5 & 3 & -7\\3 & -4 & 10\\1 & 1 & -5\end{matrix}\right] \right)}}{76} = 1$$
$$2 z + \left(3 x - 4 y\right) = 10$$
$$- 3 z + \left(x + y\right) = -5$$
We give the system of equations to the canonical form
$$5 x + 3 y - z = -7$$
$$3 x - 4 y + 2 z = 10$$
$$x + y - 3 z = -5$$
Rewrite the system of linear equations as the matrix form
$$\left[\begin{matrix}5 x_{1} + 3 x_{2} - x_{3}\\3 x_{1} - 4 x_{2} + 2 x_{3}\\x_{1} + x_{2} - 3 x_{3}\end{matrix}\right] = \left[\begin{matrix}-7\\10\\-5\end{matrix}\right]$$
- 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:
$$A = \operatorname{det}{\left(\left[\begin{matrix}5 & 3 & -1\\3 & -4 & 2\\1 & 1 & -3\end{matrix}\right] \right)} = 76$$
, 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 )
$$x_{1} = \frac{\operatorname{det}{\left(\left[\begin{matrix}-7 & 3 & -1\\10 & -4 & 2\\-5 & 1 & -3\end{matrix}\right] \right)}}{76} = 0$$
$$x_{2} = \frac{\operatorname{det}{\left(\left[\begin{matrix}5 & -7 & -1\\3 & 10 & 2\\1 & -5 & -3\end{matrix}\right] \right)}}{76} = -2$$
$$x_{3} = \frac{\operatorname{det}{\left(\left[\begin{matrix}5 & 3 & -7\\3 & -4 & 10\\1 & 1 & -5\end{matrix}\right] \right)}}{76} = 1$$
Numerical answer
[src]
x1 = -2.067951531382569e-25 y1 = -2.0 z1 = 1.0
x2 = 2.067951531382569e-25 y2 = -2.0 z2 = 1.0
x3 = 1.033975765691285e-25 y3 = -2.0 z3 = 1.0
x4 = 1.550963648536927e-25 y4 = -2.0 z4 = 1.0
x5 = -1.033975765691285e-25 y5 = -2.0 z5 = 1.0
x6 = 0 y6 = -2.0 z6 = 1.0
x7 = 5.169878828456423e-26 y7 = -2.0 z7 = 1.0
x7 = 5.169878828456423e-26 y7 = -2.0 z7 = 1.0