### Systems of linear equations and matrices: Systems and matrices

### Solving 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 using 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

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 solutions*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 \(n-r\) freely selectable unknowns; 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)\]

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