Several methods for solving linear systems of equations.
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Solving systems of two
linear equations
Method of equalization: In the method of
equalization, we isolate the same variable in 2 equations and equal the
obtained expressions. By this, we obtain an equation which does not use
one of the given variables.
Method of substitution: In the method of
substitution, we isolate some variable in an equation and replace it in
the other equations. By this, we obtain equations which do not use one
of the given variables.
Solving by graphing (2 variables): Consider a
system of linear equations with 2 variables. Then each equation can be
represented by a line in a \(2\)-dimensional coordinate system. The set
of solutions then is represented by the intersection of all
lines.
Method of elimination: In the method of
elimination two equations (or multiples of them) are added/subtracted
such that at least one variable is eliminated.
Example 1. The solution of the system of linear
equations \[\begin{array}{rrrrr}
4x_1 & - & 2x_2 & = & 6\\
2x_1 & + & x_2 & = & 5
\end{array}\] with the two variables \(x_1\) and \(x_2\) is \[x_1 =
-1 \qquad \text{and} \qquad x_2=3.\]
Solvability:
number of solutions
For any lines there exist three kinds of intersections:
In general the following is true: Every system of linear equations
has either (i) exactly one solution or (ii) no solution or (iii)
infinitely many solutions.
Solving bigger systems
of linear equations
Observations:
Recipe for calculations:
Bring the system to a "triangle form" or "row echelon
form".
Afterwards, determine the set of solutions.
Example 2. We determine the unique solution of the
following system of linear equations:
\[\begin{array}{lrrrrrll}
I: & 1x_1 & + & 3x_2 & - & 3x_3 & = ~ 3 & \\
II: & 2x_1 & + & 7x_2 & - & 5x_3 & = ~ 4 &
-2\cdot I\\
III: & -1x_1 & + & 1x_2 & + & 9x_3 & = ~ -13
& +1 \cdot I
\\ \\
I: & 1x_1 & + & 3x_2 & - & 3x_3 & = ~ 3 &\\
II: & 0x_1 & + & 1x_2 & + & 1x_3 & = ~ -2
&\\
III: & 0x_1 & + & 4x_2 & + & 6x_3 & = ~ -10
& -4\cdot II
\\ \\
I: & 1x_1 & + & 3x_2 & - & 3x_3 & = ~ 3\\
II: & 0x_1 & + & 1x_2 & + & 1x_3 & = ~ -2
&\\
III: & 0x_1 & + & 0x_2 & + & 2x_3 & = ~ -2
\end{array}\] Using back substitution we obtain the solution:
\[x_3 = -1, \qquad x_2 = -1, \qquad x_1 =
3.\]
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