System of linear equation is
a collection of linear equations involving the same set of variables. So let’s
say we have system of m equations with n unknowns and they are written in the
following form:
$$\begin{align}
& {{f}_{1}}\left( {{x}_{1}},{{x}_{2}},...,{{x}_{n}} \right)={{b}_{1}} \\
& {{f}_{2}}\left( {{x}_{1}},{{x}_{2}},...,{{x}_{n}} \right)={{b}_{2}} \\
& ......................\\
& {{f}_{m}}\left( {{x}_{1}},{{x}_{2}},...,{{x}_{n}} \right)={{b}_{m}} \\
\end{align}$$
Solving system means that
you determine a value of every unknown starting from
to
so when they are inserted in the previous system of m
equations on left side they give exactly the same value which is on the right
side of the equation. Every n-tuple with that feature we call solution of the
linear equation.
Equivalent system
There are three basic
rules which give us the way we can switch the existing system with other
system and that other system is identical to initial system. For two systems
we can say that are equivalent if every solution of the first system is also
the solution of the second system and if every solution of second system is
also equal to the solution of the first system.
Two systems are equivalent if we replace the existing system with
other system and doing this we neither loose solution nor gain a new solution
of the system.
|
The three rules are:
1) Replacement of any two equations of the system,
2) Multiplying the equations of the system by number that is not zero and 3) Adding one equation a second equation which is multiplied by some number |
If we think a little bit, we
can see that the first rule gives us the equivalent system because in linear
system 1.1 if we change the order of equations we still get the same solution.
So the order of equations doesn’t affect the solution of the linear system. For
solution it is very important that in the end; or when you insert the solution
of the equation into linear system left hand side of the equation is equivalent
to the right hand side of the equation. By solution I mean the values of all
unknowns.
Example 1
Let’s consider the following
system of equations.
$$\begin{align}
& x+y-z=-1 \\
& 2x+4y\text{ }=-4 \\
& -x+3y+2z=-6 \\
\end{align}$$
Now to solve this linear
system we will use the previous three rules. First we will simplify the system
by replacing the third equation with the sum of the first and the third. Then
you will get:
$$\begin{align}
& \text{ }x+\text{ }y-z=-1 \\
& 2x+4y\text{ }=-4 \\
& \text{ 4}y+\text{ }z\text{ }=-7 \\
\end{align}$$
Notice that any solution of
the original system will be a solution of the new system. Indeed, the first two
equations are exactly the same, and the third holds since it is the result of
adding the equal quantities in the first equation to the equal quantities in
the third equation of the original system. Also note that any solution of the
new system is a solution of the original system. This is because we can recover
the third equation of the original system by subtracting the first equation
from the new third equation. It follows, then, that the new system has exactly
the same solution set as the old system. The new system has the advantage of
being slightly simpler. To make further progress toward simplification, we can
eliminate the variable x in the second equation by adding -2 times the first
equation to the second equation. This produces the system
$$\begin{align}
& \text{ }x+\text{ }y-z=-1 \\
& \text{ 2}y+2z=-2 \\
& \text{ 4}y+\text{ }z\text{ }=-7 \\
\end{align}$$
Notice that we can recover
the previous system by adding 2 times the first equation to the second. Thus,
we can argue as above that no solutions have been lost and no new solutions
have appeared. Hence, we still have exactly the same solution set. Next we would like to use the y in the second
equation to eliminate the occurrences of y in the other equations. First we
will multiply the second equation by ½. The system is changed again:
$$\begin{align}
& \text{ }x+\text{ }y-z=-1 \\
& \text{ }y+z=-1 \\
& \text{ 4}y+\text{ }z\text{ }=-7 \\
\end{align}$$
Take a minute to convince
yourself that this did not affect the solution set. Now add -4 times the second
equation to the third to give the following system (again with the same
solution set):
$$\begin{align}
& \text{ }x+\text{ }y-z=-1 \\
& \text{ }y+z=-1 \\
& \text{ }-3z\text{ }=-3 \\
\end{align}$$
Multiply the third equation
by -½ to obtain z = 1. This gives the following system, where it is clear what value z must have in
order for the equations to hold:
$$\begin{align}
& \text{ }x+\text{ }y-z=-1 \\
& \text{ }y+z=-1 \\
& \text{ }z\text{ }=\text{ }1 \\
\end{align}$$
Subtracting the third equation from the second and then adding the third equation to the first will further simplify things to
Finally, subtract the second
equation from the first to product the simplest possible system:
$$\begin{align}
& \text{ }x\text{ }=-1 \\
& \text{ }y\text{ }=-2 \\
& \text{ }z\text{ }=\text{ 1} \\
\end{align}$$
Now let’s take a look at another example:
Example
2
Show that, for arbitrary solutions values
of s and t,
$$\begin{align}
& {{x}_{1}}=t-s+1, \\
& {{x}_{2}}=t+s+2, \\
& {{x}_{3}}=s, \\
& {{x}_{4}}=t, \\
\end{align}$$
is a solution to the system:
$$\underline{\begin{align}
& {{x}_{1}}-2{{x}_{2}}+3{{x}_{3}}+{{x}_{4}}=-3, \\
& 2{{x}_{1}}-{{x}_{2}}+3{{x}_{3}}-{{x}_{4}}=0 \\
\end{align}}$$
Solution: Simply substitute these values
for $${{x}_{1}},{{x}_{2}},{{x}_{3}}\text{ and }{{x}_{4}}$$in each equation:
$$\begin{align}
& t-s+1-2\left( t+s+2 \right)+3\left( s \right)+t=-3, \\
& 2\left( t-s+1 \right)-\left( t+s+2 \right)+3\left( s \right)-t=0, \\
\end{align}$$
Because both equations are satisfied, it’s
a solution for all s and t. The quantities s and t in the previous example are
called parameters, and the set of solutions, described in this way, is said to
be given in parametric form and is called the general solution to the system.
It turns out that the solutions to every system of equations can be given in
parametric form.
Example
3
Describe all solutions to $$3x-y=4$$in parametric form.
Solution: The previous equation can be
written in the following form.
$$y=3x-4$$
Thus, if t denotes any number at all, we
can set x = t and then obtain y = 3t - . This is clearly a solution for any
value of t. On the other hand, every solution to 3x-y=4 arises in this way (t
is just the value of x). Hence the set of all solutions can be described
parametrically as:
$$\underline{\begin{align}
& x=t, \\
& y=3t-4, \\
\end{align}}\text{ }t\text{ arbitrary}$$
Note that there are infinitely many
distinct solutions, one for each choice of the parameter t. It is important to
realize that the solutions to 3x-y=4 can be given in a parametric form in
sveral ways. Instead of writing y=3x-4 as before, we could have found x in
terms of y
$$x=\frac{1}{3}\left( y+4 \right)$$
and then chosen y=s where s is a parameter.
Hence the solutions are:
$$\underline{\begin{align}
& x=\frac{1}{3}\left( s+4 \right) \\
& y=s \\
\end{align}}\text{ }s\text{ arbitrary}$$
This is also a correct parametric
representation of the solution to 3x-y=4. In fact, the parameter are related by
s=3t-4 or t=(1/3)(s+4).
Let’s take a look to a more complicated
example:
Example
4
Describe all solution’s to 3x-y+2z=6 in
parametric form.
Solution: Solving the equation for y in
terms of x and z, we get y=3x+2z-6. If s and t are arbitrary, then setting x=s
and z =t we get a following solution.
$$\underline{\begin{align}
& x=s, \\
& y=3s+2t-6, \\
& z=t \\
\end{align}}\text{ }s\text{ and }t\text{ arbitrary}$$
When we have only two variables, the
solution to the system of linear equations can be described geometrically
because the graph of a linear equation ax+by=c is a straight line. Moreover, a
point P(s,t) with coordinates s and t lies on the line if and only if as+bt=c;
that is x=s and y=t is a solution to the equation.
In particular, if the system consists of
just one equation there must be infinitely many solutions because there are
infinitely many points on a line. If the system has two equations there are
three possibilities for the corresponding straight lines:
1) The lines intersect in a single
point. Then the system has a unique solution corresponding to that point.
2) The lines intersect in a single
point. Then the system has a unique solution.
3) The lines are identical. Then
the system has infinitely many solutions one for eaxh point on the (common)
line.
However the graphical method has its
limitations: When more than three variables are involved, no physical image of
the graphs is possible. It is necessary to turn to a more algebraic method of
solution.
Now we will introduce a concept that
simplifies the computations so let’s consider the following system:
$$\underline{\begin{align}
& 3{{x}_{1}}+2{{x}_{2}}-{{x}_{3}}+{{x}_{4}}=-1 \\
& 2{{x}_{1}}\text{ }-{{x}_{3}}+2{{x}_{4}}=0 \\
&
3{{x}_{1}}+{{x}_{2}}+2{{x}_{3}}+5{{x}_{4}}=2 \\
\end{align}}$$
The previous system consists of three equations
with four variables. We can also write the previous system in the following
form:
$$\left[ \left. \begin{matrix}
$$\left[ \left. \begin{matrix}
3
& 2 & -1 & 1 \\
2
& 0 & -1 & 2 \\
3
& 1 & 2 & 5 \\
\end{matrix} \right|\begin{matrix}
-1 \\
0 \\
2 \\
\end{matrix}
\right]$$
occurring in the system is called the
augmented matrix of the system. Each row of the matrix consists of the
coefficient of the variables from the corresponding equation, together with the
constant term. For clarity, the constants are separated by a vertical line. The
augmented matrix is just a different way of describing the system of equations.
The array of coefficients of the variables
$$\left[ \begin{matrix}
3
& 2 & -1 & 1 \\
2
& 0 & -1 & 2 \\
3
& 1 & 2 & 5 \\
\end{matrix}
\right]$$
is called coefficient matrix of the system
and
$$\left[ \begin{matrix}
$$\left[ \begin{matrix}
-1 \\
0 \\
2 \\
\end{matrix} \right]$$
is called the constant matrix of the
system. 
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