Mister Exam
4x-5(y+1)=1; 5/12y-1/2z=-1; 5/6+1/3y-3/2z=-1
The solution
You have entered
[src]
4*x - 5*(y + 1) = 1
$$4 x - 5 \left(y + 1\right) = 1$$
5*y z --- - - = -1 12 2
$$\frac{5 y}{12} - \frac{z}{2} = -1$$
5 y 3*z - + - - --- = -1 6 3 2
$$- \frac{3 z}{2} + \left(\frac{y}{3} + \frac{5}{6}\right) = -1$$
-3*z/2 + y/3 + 5/6 = -1
Rapid solution
$$x_{1} = - \frac{1}{11}$$
=
$$- \frac{1}{11}$$
=
$$y_{1} = - \frac{14}{11}$$
=
$$- \frac{14}{11}$$
=
$$z_{1} = \frac{31}{33}$$
=
$$\frac{31}{33}$$
=
=
$$- \frac{1}{11}$$
=
-0.0909090909090909
$$y_{1} = - \frac{14}{11}$$
=
$$- \frac{14}{11}$$
=
-1.27272727272727
$$z_{1} = \frac{31}{33}$$
=
$$\frac{31}{33}$$
=
0.939393939393939
Cramer's rule
$$4 x - 5 \left(y + 1\right) = 1$$
$$\frac{5 y}{12} - \frac{z}{2} = -1$$
$$- \frac{3 z}{2} + \left(\frac{y}{3} + \frac{5}{6}\right) = -1$$
We give the system of equations to the canonical form
$$4 x - 5 y = 6$$
$$\frac{5 y}{12} - \frac{z}{2} = -1$$
$$\frac{y}{3} - \frac{3 z}{2} = - \frac{11}{6}$$
Rewrite the system of linear equations as the matrix form
$$\left[\begin{matrix}4 x_{1} - 5 x_{2} + 0 x_{3}\\0 x_{1} + \frac{5 x_{2}}{12} - \frac{x_{3}}{2}\\0 x_{1} + \frac{x_{2}}{3} - \frac{3 x_{3}}{2}\end{matrix}\right] = \left[\begin{matrix}6\\-1\\- \frac{11}{6}\end{matrix}\right]$$
- 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:
$$A = \operatorname{det}{\left(\left[\begin{matrix}4 & -5 & 0\\0 & \frac{5}{12} & - \frac{1}{2}\\0 & \frac{1}{3} & - \frac{3}{2}\end{matrix}\right] \right)} = - \frac{11}{6}$$
, 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 )
$$x_{1} = - \frac{6 \operatorname{det}{\left(\left[\begin{matrix}6 & -5 & 0\\-1 & \frac{5}{12} & - \frac{1}{2}\\- \frac{11}{6} & \frac{1}{3} & - \frac{3}{2}\end{matrix}\right] \right)}}{11} = - \frac{1}{11}$$
$$x_{2} = - \frac{6 \operatorname{det}{\left(\left[\begin{matrix}4 & 6 & 0\\0 & -1 & - \frac{1}{2}\\0 & - \frac{11}{6} & - \frac{3}{2}\end{matrix}\right] \right)}}{11} = - \frac{14}{11}$$
$$x_{3} = - \frac{6 \operatorname{det}{\left(\left[\begin{matrix}4 & -5 & 6\\0 & \frac{5}{12} & -1\\0 & \frac{1}{3} & - \frac{11}{6}\end{matrix}\right] \right)}}{11} = \frac{31}{33}$$
$$\frac{5 y}{12} - \frac{z}{2} = -1$$
$$- \frac{3 z}{2} + \left(\frac{y}{3} + \frac{5}{6}\right) = -1$$
We give the system of equations to the canonical form
$$4 x - 5 y = 6$$
$$\frac{5 y}{12} - \frac{z}{2} = -1$$
$$\frac{y}{3} - \frac{3 z}{2} = - \frac{11}{6}$$
Rewrite the system of linear equations as the matrix form
$$\left[\begin{matrix}4 x_{1} - 5 x_{2} + 0 x_{3}\\0 x_{1} + \frac{5 x_{2}}{12} - \frac{x_{3}}{2}\\0 x_{1} + \frac{x_{2}}{3} - \frac{3 x_{3}}{2}\end{matrix}\right] = \left[\begin{matrix}6\\-1\\- \frac{11}{6}\end{matrix}\right]$$
- 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:
$$A = \operatorname{det}{\left(\left[\begin{matrix}4 & -5 & 0\\0 & \frac{5}{12} & - \frac{1}{2}\\0 & \frac{1}{3} & - \frac{3}{2}\end{matrix}\right] \right)} = - \frac{11}{6}$$
, 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 )
$$x_{1} = - \frac{6 \operatorname{det}{\left(\left[\begin{matrix}6 & -5 & 0\\-1 & \frac{5}{12} & - \frac{1}{2}\\- \frac{11}{6} & \frac{1}{3} & - \frac{3}{2}\end{matrix}\right] \right)}}{11} = - \frac{1}{11}$$
$$x_{2} = - \frac{6 \operatorname{det}{\left(\left[\begin{matrix}4 & 6 & 0\\0 & -1 & - \frac{1}{2}\\0 & - \frac{11}{6} & - \frac{3}{2}\end{matrix}\right] \right)}}{11} = - \frac{14}{11}$$
$$x_{3} = - \frac{6 \operatorname{det}{\left(\left[\begin{matrix}4 & -5 & 6\\0 & \frac{5}{12} & -1\\0 & \frac{1}{3} & - \frac{11}{6}\end{matrix}\right] \right)}}{11} = \frac{31}{33}$$
Gaussian elimination
Given the system of equations
$$4 x - 5 \left(y + 1\right) = 1$$
$$\frac{5 y}{12} - \frac{z}{2} = -1$$
$$- \frac{3 z}{2} + \left(\frac{y}{3} + \frac{5}{6}\right) = -1$$
We give the system of equations to the canonical form
$$4 x - 5 y = 6$$
$$\frac{5 y}{12} - \frac{z}{2} = -1$$
$$\frac{y}{3} - \frac{3 z}{2} = - \frac{11}{6}$$
Rewrite the system of linear equations as the matrix form
$$\left[\begin{matrix}4 & -5 & 0 & 6\\0 & \frac{5}{12} & - \frac{1}{2} & -1\\0 & \frac{1}{3} & - \frac{3}{2} & - \frac{11}{6}\end{matrix}\right]$$
In 1 -th column
$$\left[\begin{matrix}4\\0\\0\end{matrix}\right]$$
let’s convert all the elements, except
1 -th element into zero.
- To do this, let’s take 1 -th line
$$\left[\begin{matrix}4 & -5 & 0 & 6\end{matrix}\right]$$
,
and subtract it from other lines:
In 2 -th column
$$\left[\begin{matrix}-5\\\frac{5}{12}\\\frac{1}{3}\end{matrix}\right]$$
let’s convert all the elements, except
1 -th element into zero.
- To do this, let’s take 1 -th line
$$\left[\begin{matrix}4 & -5 & 0 & 6\end{matrix}\right]$$
,
and subtract it from other lines:
From 2 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}- \frac{\left(-1\right) 4}{12} & \frac{5}{12} - - \frac{-5}{12} & - \frac{1}{2} - \frac{\left(-1\right) 0}{12} & -1 - \frac{\left(-1\right) 6}{12}\end{matrix}\right] = \left[\begin{matrix}\frac{1}{3} & 0 & - \frac{1}{2} & - \frac{1}{2}\end{matrix}\right]$$
you get
$$\left[\begin{matrix}4 & -5 & 0 & 6\\\frac{1}{3} & 0 & - \frac{1}{2} & - \frac{1}{2}\\0 & \frac{1}{3} & - \frac{3}{2} & - \frac{11}{6}\end{matrix}\right]$$
From 3 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}- \frac{\left(-1\right) 4}{15} & \frac{1}{3} - - \frac{-1}{3} & - \frac{3}{2} - \frac{\left(-1\right) 0}{15} & - \frac{11}{6} - \frac{\left(-1\right) 6}{15}\end{matrix}\right] = \left[\begin{matrix}\frac{4}{15} & 0 & - \frac{3}{2} & - \frac{43}{30}\end{matrix}\right]$$
you get
$$\left[\begin{matrix}4 & -5 & 0 & 6\\\frac{1}{3} & 0 & - \frac{1}{2} & - \frac{1}{2}\\\frac{4}{15} & 0 & - \frac{3}{2} & - \frac{43}{30}\end{matrix}\right]$$
In 3 -th column
$$\left[\begin{matrix}0\\- \frac{1}{2}\\- \frac{3}{2}\end{matrix}\right]$$
let’s convert all the elements, except
2 -th element into zero.
- To do this, let’s take 2 -th line
$$\left[\begin{matrix}\frac{1}{3} & 0 & - \frac{1}{2} & - \frac{1}{2}\end{matrix}\right]$$
,
and subtract it from other lines:
From 3 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}\frac{4}{15} - \frac{3}{3} & - 0 \cdot 3 & - \frac{3}{2} - \frac{\left(-1\right) 3}{2} & - \frac{43}{30} - \frac{\left(-1\right) 3}{2}\end{matrix}\right] = \left[\begin{matrix}- \frac{11}{15} & 0 & 0 & \frac{1}{15}\end{matrix}\right]$$
you get
$$\left[\begin{matrix}4 & -5 & 0 & 6\\\frac{1}{3} & 0 & - \frac{1}{2} & - \frac{1}{2}\\- \frac{11}{15} & 0 & 0 & \frac{1}{15}\end{matrix}\right]$$
In 1 -th column
$$\left[\begin{matrix}4\\\frac{1}{3}\\- \frac{11}{15}\end{matrix}\right]$$
let’s convert all the elements, except
3 -th element into zero.
- To do this, let’s take 3 -th line
$$\left[\begin{matrix}- \frac{11}{15} & 0 & 0 & \frac{1}{15}\end{matrix}\right]$$
,
and subtract it from other lines:
From 1 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}4 - - -4 & -5 - \frac{\left(-60\right) 0}{11} & - \frac{\left(-60\right) 0}{11} & 6 - \frac{-60}{11 \cdot 15}\end{matrix}\right] = \left[\begin{matrix}0 & -5 & 0 & \frac{70}{11}\end{matrix}\right]$$
you get
$$\left[\begin{matrix}0 & -5 & 0 & \frac{70}{11}\\\frac{1}{3} & 0 & - \frac{1}{2} & - \frac{1}{2}\\- \frac{11}{15} & 0 & 0 & \frac{1}{15}\end{matrix}\right]$$
From 2 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}\frac{1}{3} - - \frac{-1}{3} & - \frac{\left(-5\right) 0}{11} & - \frac{1}{2} - \frac{\left(-5\right) 0}{11} & - \frac{1}{2} - \frac{-5}{11 \cdot 15}\end{matrix}\right] = \left[\begin{matrix}0 & 0 & - \frac{1}{2} & - \frac{31}{66}\end{matrix}\right]$$
you get
$$\left[\begin{matrix}0 & -5 & 0 & \frac{70}{11}\\0 & 0 & - \frac{1}{2} & - \frac{31}{66}\\- \frac{11}{15} & 0 & 0 & \frac{1}{15}\end{matrix}\right]$$
It is almost ready, all we have to do is to find variables, solving the elementary equations:
$$- 5 x_{2} - \frac{70}{11} = 0$$
$$\frac{31}{66} - \frac{x_{3}}{2} = 0$$
$$- \frac{11 x_{1}}{15} - \frac{1}{15} = 0$$
We get the answer:
$$x_{2} = - \frac{14}{11}$$
$$x_{3} = \frac{31}{33}$$
$$x_{1} = - \frac{1}{11}$$
$$4 x - 5 \left(y + 1\right) = 1$$
$$\frac{5 y}{12} - \frac{z}{2} = -1$$
$$- \frac{3 z}{2} + \left(\frac{y}{3} + \frac{5}{6}\right) = -1$$
We give the system of equations to the canonical form
$$4 x - 5 y = 6$$
$$\frac{5 y}{12} - \frac{z}{2} = -1$$
$$\frac{y}{3} - \frac{3 z}{2} = - \frac{11}{6}$$
Rewrite the system of linear equations as the matrix form
$$\left[\begin{matrix}4 & -5 & 0 & 6\\0 & \frac{5}{12} & - \frac{1}{2} & -1\\0 & \frac{1}{3} & - \frac{3}{2} & - \frac{11}{6}\end{matrix}\right]$$
In 1 -th column
$$\left[\begin{matrix}4\\0\\0\end{matrix}\right]$$
let’s convert all the elements, except
1 -th element into zero.
- To do this, let’s take 1 -th line
$$\left[\begin{matrix}4 & -5 & 0 & 6\end{matrix}\right]$$
,
and subtract it from other lines:
In 2 -th column
$$\left[\begin{matrix}-5\\\frac{5}{12}\\\frac{1}{3}\end{matrix}\right]$$
let’s convert all the elements, except
1 -th element into zero.
- To do this, let’s take 1 -th line
$$\left[\begin{matrix}4 & -5 & 0 & 6\end{matrix}\right]$$
,
and subtract it from other lines:
From 2 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}- \frac{\left(-1\right) 4}{12} & \frac{5}{12} - - \frac{-5}{12} & - \frac{1}{2} - \frac{\left(-1\right) 0}{12} & -1 - \frac{\left(-1\right) 6}{12}\end{matrix}\right] = \left[\begin{matrix}\frac{1}{3} & 0 & - \frac{1}{2} & - \frac{1}{2}\end{matrix}\right]$$
you get
$$\left[\begin{matrix}4 & -5 & 0 & 6\\\frac{1}{3} & 0 & - \frac{1}{2} & - \frac{1}{2}\\0 & \frac{1}{3} & - \frac{3}{2} & - \frac{11}{6}\end{matrix}\right]$$
From 3 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}- \frac{\left(-1\right) 4}{15} & \frac{1}{3} - - \frac{-1}{3} & - \frac{3}{2} - \frac{\left(-1\right) 0}{15} & - \frac{11}{6} - \frac{\left(-1\right) 6}{15}\end{matrix}\right] = \left[\begin{matrix}\frac{4}{15} & 0 & - \frac{3}{2} & - \frac{43}{30}\end{matrix}\right]$$
you get
$$\left[\begin{matrix}4 & -5 & 0 & 6\\\frac{1}{3} & 0 & - \frac{1}{2} & - \frac{1}{2}\\\frac{4}{15} & 0 & - \frac{3}{2} & - \frac{43}{30}\end{matrix}\right]$$
In 3 -th column
$$\left[\begin{matrix}0\\- \frac{1}{2}\\- \frac{3}{2}\end{matrix}\right]$$
let’s convert all the elements, except
2 -th element into zero.
- To do this, let’s take 2 -th line
$$\left[\begin{matrix}\frac{1}{3} & 0 & - \frac{1}{2} & - \frac{1}{2}\end{matrix}\right]$$
,
and subtract it from other lines:
From 3 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}\frac{4}{15} - \frac{3}{3} & - 0 \cdot 3 & - \frac{3}{2} - \frac{\left(-1\right) 3}{2} & - \frac{43}{30} - \frac{\left(-1\right) 3}{2}\end{matrix}\right] = \left[\begin{matrix}- \frac{11}{15} & 0 & 0 & \frac{1}{15}\end{matrix}\right]$$
you get
$$\left[\begin{matrix}4 & -5 & 0 & 6\\\frac{1}{3} & 0 & - \frac{1}{2} & - \frac{1}{2}\\- \frac{11}{15} & 0 & 0 & \frac{1}{15}\end{matrix}\right]$$
In 1 -th column
$$\left[\begin{matrix}4\\\frac{1}{3}\\- \frac{11}{15}\end{matrix}\right]$$
let’s convert all the elements, except
3 -th element into zero.
- To do this, let’s take 3 -th line
$$\left[\begin{matrix}- \frac{11}{15} & 0 & 0 & \frac{1}{15}\end{matrix}\right]$$
,
and subtract it from other lines:
From 1 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}4 - - -4 & -5 - \frac{\left(-60\right) 0}{11} & - \frac{\left(-60\right) 0}{11} & 6 - \frac{-60}{11 \cdot 15}\end{matrix}\right] = \left[\begin{matrix}0 & -5 & 0 & \frac{70}{11}\end{matrix}\right]$$
you get
$$\left[\begin{matrix}0 & -5 & 0 & \frac{70}{11}\\\frac{1}{3} & 0 & - \frac{1}{2} & - \frac{1}{2}\\- \frac{11}{15} & 0 & 0 & \frac{1}{15}\end{matrix}\right]$$
From 2 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}\frac{1}{3} - - \frac{-1}{3} & - \frac{\left(-5\right) 0}{11} & - \frac{1}{2} - \frac{\left(-5\right) 0}{11} & - \frac{1}{2} - \frac{-5}{11 \cdot 15}\end{matrix}\right] = \left[\begin{matrix}0 & 0 & - \frac{1}{2} & - \frac{31}{66}\end{matrix}\right]$$
you get
$$\left[\begin{matrix}0 & -5 & 0 & \frac{70}{11}\\0 & 0 & - \frac{1}{2} & - \frac{31}{66}\\- \frac{11}{15} & 0 & 0 & \frac{1}{15}\end{matrix}\right]$$
It is almost ready, all we have to do is to find variables, solving the elementary equations:
$$- 5 x_{2} - \frac{70}{11} = 0$$
$$\frac{31}{66} - \frac{x_{3}}{2} = 0$$
$$- \frac{11 x_{1}}{15} - \frac{1}{15} = 0$$
We get the answer:
$$x_{2} = - \frac{14}{11}$$
$$x_{3} = \frac{31}{33}$$
$$x_{1} = - \frac{1}{11}$$
Numerical answer
[src]
x1 = -0.09090909090909091 y1 = -1.272727272727273 z1 = 0.9393939393939394
x1 = -0.09090909090909091 y1 = -1.272727272727273 z1 = 0.9393939393939394