Linear Equations: Difference between revisions

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=Introduction=
=Introduction=
This is the page on Linear Equation
This is the page on Linear Equation
=What is a System=
A system of linear equations is a collection of two or more linear equations, and a solution to a system of linear equations consists of values of each of the unknown variables in the system that satisfies all of its equations, or makes them true
=Getting Started=
=Getting Started=
Already fine with this but did pick up a new technique. In the video we had the following to solve.
I actually spent more time understanding how to do the matrix in wiki. An important thing to remember is you can add two equation together in a system because if you thing about 2=2, 4=4 then adding them together does not break the equality. The first time through I did not use matrices but wrote them out. Not sure which I prefer
<syntaxhighlight lang="txt">
<syntaxhighlight lang="txt">
  x + 4y -4z = 0
  x + 4y -4z = 0
Line 14: Line 16:
2 & -3 & -1 & 3
2 & -3 & -1 & 3
\end{bmatrix}
\end{bmatrix}
'''Multiply row 1 by -1'''<br>
'''Multiply row 1 by -1'''<br>
<math>
<math>
Line 19: Line 22:
</math><br>
</math><br>
\begin{bmatrix}
\begin{bmatrix}
-1 & -4 & 4 & 0 \\
-1 & 4 & -4 & 0 \\
1 & 4 & 2 & 12 \\
1 & 4 & 2 & 12 \\
2 & -3 & -1 & 3
2 & -3 & -1 & 3
\end{bmatrix}
\end{bmatrix}
Now add the original top row to the middle row
 
'''Now add the original Row 1 to Row 2'''
<math>
R_1 = R_1 + R_2 \implies
\left( -1 + 1, \; -4 + 4, \; 4 + 2, \; 0 + 12 \right) = \left( 0, \; 0, \; 6, \; 12 \right)
</math>
 
\begin{bmatrix}
\begin{bmatrix}
-1 & 4 & -4 & 0 \\
0 & 0 & 6 & 12 \\
0 & 0 & 6 & 12 \\
2 & -3 & -1 & 3
\end{bmatrix}
\end{bmatrix}


<math>
Therefore 6z = 12
R_2 \to R_2 - R_1:
<br>
</math>
 
He said for the middle row by times the top row by -1  and add it to the middle to transform. I struggled to understand as my approach is to move the values around to get to one  e.g for the top one I would make it
<syntaxhighlight lang="txt">
x = 4z -4y
</syntaxhighlight>
And replace all the instances of x but his approach is to make the x amount 0 by multiplying by whatever would make it 0 when added to the next row. So
<syntaxhighlight lang="txt">
-x - 4y + 4z = 0  // top row multiplied by -1
x + 4y + 2z = 12 // now add middle row to it
        + 6z = 12 // Removed the x
</syntaxhighlight>
Need for me some explanation given I have never seen this approach. For the 3rd line we multiply by -2 and add to row 3
<syntaxhighlight lang="txt">
-2x - 8y + 8z = 0 // top row multiplied by -2
2x  -3y - 1z = 3 // now add 3rd row to top row
    -11y + 7z = 3 // Removed the x
</syntaxhighlight>
Now we have
<syntaxhighlight lang="txt">
x + 4y - 4z = 0
  -11y + 7z = 3
        + 6z = 12  
</syntaxhighlight>
This is called '''Gaussian elimination method'''. Where you are left with 3 equations with 3, 2 and one variables and is known as '''the echelon form'''
This is called '''Gaussian elimination method'''. Where you are left with 3 equations with 3, 2 and one variables and is known as '''the echelon form'''


=Infinite Solutions=
=Infinite Solutions=
Sometimes the equations are not solvable when put in echlon form. You can see this when you see maybe x = x. This means the are infinite answers and example might be
Sometimes the equations are not solvable when put in echlon form. You can see this when you see maybe x = x. This means the are infinite answers and example might be
<syntaxhighlight lang="txt">
𝑥 + 2𝑦 = 4
-x -y + 3z = 3
2𝑥 + 4𝑦 = 8
x   + 1z =  3
 
3x -y + 7z = 15
\begin{bmatrix}
</syntaxhighlight>
1 & 2 & 4 \\
Apply Gaussian
2 & 4 & 8
<syntaxhighlight lang="txt">
\end{bmatrix}
-x -y + 3z = 3
Eliminate x from the second row
  -y + 4z = 6
<br>
        0 = 0
    <math>
</syntaxhighlight>
R_2 \to R_2 -2R_1
    </math>
Gives
(2 -2.1, 4 -2.2, 8 -2.4) = (0,0,0)
Now we have<br>
\begin{bmatrix}
1 & 2 & 4 \\
0 & 0 & 0
\end{bmatrix}
 
=Parameterizing=
=Parameterizing=
We can, when we have infinite solution express one or more variable in terms of the other e.g y = -6 + 4z. This is called parameterizing. There are  
We can, when we have infinite solution express one or more variable in terms of the other e.g y = -6 + 4z. This is called parameterizing. There are  
*leading variables
*leading variables
*free variables
*free variables
=What is a System=
A system of linear equations is a collection of two or more linear equations, and a solution to a system of linear equations consists of values of each of the unknown variables in the system that satisfies all of its equations, or makes them true

Latest revision as of 23:57, 22 April 2025

Introduction

This is the page on Linear Equation

What is a System

A system of linear equations is a collection of two or more linear equations, and a solution to a system of linear equations consists of values of each of the unknown variables in the system that satisfies all of its equations, or makes them true

Getting Started

I actually spent more time understanding how to do the matrix in wiki. An important thing to remember is you can add two equation together in a system because if you thing about 2=2, 4=4 then adding them together does not break the equality. The first time through I did not use matrices but wrote them out. Not sure which I prefer

 x + 4y -4z = 0
 x + 4y + 2z = 12
 2x -3y -z = 3

In new video the put the value in a matrix so we now have \begin{bmatrix} 1 & 4 & -4 & 0 \\ 1 & 4 & 2 & 12 \\ 2 & -3 & -1 & 3 \end{bmatrix}

Multiply row 1 by -1
R2R2+(1).R1
\begin{bmatrix} -1 & 4 & -4 & 0 \\ 1 & 4 & 2 & 12 \\ 2 & -3 & -1 & 3 \end{bmatrix}

Now add the original Row 1 to Row 2 R1=R1+R2(1+1,4+4,4+2,0+12)=(0,0,6,12)

\begin{bmatrix} -1 & 4 & -4 & 0 \\ 0 & 0 & 6 & 12 \\ 2 & -3 & -1 & 3 \end{bmatrix}

Therefore 6z = 12
This is called Gaussian elimination method. Where you are left with 3 equations with 3, 2 and one variables and is known as the echelon form

Infinite Solutions

Sometimes the equations are not solvable when put in echlon form. You can see this when you see maybe x = x. This means the are infinite answers and example might be

𝑥 + 2𝑦 = 4
2𝑥 + 4𝑦 = 8

\begin{bmatrix} 1 & 2 & 4 \\ 2 & 4 & 8 \end{bmatrix} Eliminate x from the second row

   R2R22R1

Gives

(2 -2.1, 4 -2.2, 8 -2.4) = (0,0,0)

Now we have
\begin{bmatrix} 1 & 2 & 4 \\ 0 & 0 & 0 \end{bmatrix}

Parameterizing

We can, when we have infinite solution express one or more variable in terms of the other e.g y = -6 + 4z. This is called parameterizing. There are

  • leading variables
  • free variables