Systems of linear equations: System of linear equations and matrices
Solving systems of linear equations by Gaussian elimination
Thanks to the foregoing we have a procedure for solving a system of linear equations.
Solving systems of linear equations by Gaussian elimination A system of linear equations can be solved as follows:
- write down its augmented matrix
- reduce the matrix to reduced echelon form by means of Gaussian elimination
- read off the solution from the reduced echelon form
- write down its matching the augmented matrix: The augmented matrix contains all the information about the system, except for the names of the variables. We remember those names and the order in which they are chosen for the columns.
- reduce the matrix to reduced row echelon form: The reduction of the given matrix turns the matrix into an augmented matrix of a system of linear equations which is equivalent to the original system. This also applies to the matrix in reduced echelon form.
- read the solution off from the reduced echelon form: The solution can be read off directly from the augmented matrix in the reduced echelon form, as will be described below.
In order to describe the solution of a system of which the augmented matrix is in reduced echelon form, we distinguish three cases:
Solutions of systems of linear equations Suppose that the augmented matrix of a system of \(m\) linear equations in \(n\) unknowns has the reduced echelon form shown below with \(m-r\) null rows at the bottom. \[ \left(\,\begin{array}{ccccccccccccccc|c}
0 & \cdots & 0 & 1 & \ast & \cdots & \ast & 0 & \ast & \cdots & \ast & 0 & \ast & \cdots & \ast & b_1\\
0 & \cdots & 0 & 0 & 0 & \cdots & 0 & 1 & \ast & \cdots & \ast & 0 & \ast & \cdots & \ast & b_2\\
0 & \vdots & 0 & 0 & 0 & \vdots & 0 & 0 & \ddots & \ddots & \ddots & 0 & \ast & \vdots & \ast & \vdots \\
0 & \cdots & 0 & 0 & 0 & \cdots & 0 & 0 & 0 & \cdots & 0 & 1 & \ast & \cdots & \ast & b_r\\
0 & \cdots & 0 & 0 & 0 & \cdots & 0 & 0 & 0 & \cdots & 0 & 0 & 0 & \cdots & 0 & b_{r+1}\\
\vdots & & & & & & & & & & & & & & \vdots & \vdots \\
0 & \cdots & & & & & & & & & & & & \cdots & 0 & b_m\end{array}\,\right)
\]
Then this system has
- no solution if and only if at least one of the numbers \(b_{r+1}, \ldots, b_m\) is distinct from zero;
- exactly one solution if and only if no column with an #\ast# occurs in the reduced echelon form, that is, if and only if \(b_{r+1}= \cdots =b_m=0\) and \(r=n\); in this case, the unique solution \(\rv{b_1,\ldots,b_n}\);
- infinitely many solutions if and only if \(b_{r+1}= \cdots =b_m=0\) and \(r\lt n\); then there are \(n-r\) unknowns free to choose; choose a parameter for each unknown that corresponds to a column with #\ast# in the reduced echelon form. Subtract the scalar product with the parameter of the corresponding column vector from the last column vector and remove the column used. The result is an augmented matrix in a reduced echelon form that belongs to a system of equations in the unknowns that correspond to a column with a leading element #1# in the original reduced echelon form. If we interpret the parameters as constants, then the remaining matrix satisfies the conditions for a unique solution. Therefore, for each choice of the \(n-r\) free parameters, we find a unique solution.
A system that has a solution, is also called consistent. An insolvable system is also known as inconsistent.
Here are two examples illustrating this solution procedure.
1 & 2 & 3 & 2\\
0 & 1 & 2 & 1\\
3 & 1 & 1 & 3
\end{array}\,\right)
\] Row reduction leads to the reduced form
\[
\left(\,\begin{array}{rrr|r}
1 & 0 & 0 & 1\\
0 & 1 & 0 & -\,1\\
0 & 0 & 1 & 1
\end{array}\,\right)
\]
Therefore, the system has exactly one solution, namely \[ \cv{x_1\\x_2\\x_3}=
\left(\!\!\begin{array}{r} 1\\-1\\1\end{array}\right)\]
