next up previous contents
Next: A special case: homogeneous Up: Solution of a system Previous: Solution of a system

The general case.

    

The algorithm:

% latex2html id marker 1379
\fbox{
\begin{minipage}{11cm}
\begin{enumerate}
\ite...
...the solutions of the system ~\ref{general system}
\end{enumerate}\end{minipage}}

Proposition 3.3.1  
1.
The system  1 has at least a solution if, and only if, there is no row leading coefficient in the last column.
2.
Suppose that the system  1 has at least a solution. If there exists at least one column without row leading coefficient, then the system has an infinity of solutions.

Example 3.3.2   We solve the system: $\begin{cases}x-y+z=2 \\ 2x-4y+z=-5 \\ x-y+2z=5 \end{cases}$.


\begin{align*}\begin{pmatrix}1&-1&1&2\\ 2&-4&1&-5\\ 1&-1&2&5 \end{pmatrix} \over...
...grightarrow}\begin{pmatrix}1&0&0&2\\ 0&1&0&3\\ 0&0&1&3 \end{pmatrix}\end{align*}
Therefore the system has an unique solution given by (x,y,z)=(2,3,3).

Example 3.3.3   We solve the system: $\begin{cases}x-2y+z=2 \\ 2x+4y+z=1 \\ 5x+6y+3z=0 \end{cases}$.
\begin{align*}\begin{pmatrix}1&-2&1&2\\ 2&4&1&1\\ 5&6&3&0 \end{pmatrix} \overset...
...htarrow}\begin{pmatrix}1&-2&1&2\\ 0&8&-1&-3\\ 0&0&0&-4 \end{pmatrix}\end{align*}
The solution set of the given system is empty.

Example 3.3.4   We solve the system: $\begin{cases}x-2y+z=3 \\ 2x+y+3z=4 \\ 5x+7z=11 \end{cases}$.
\begin{align*}\begin{pmatrix}1&-2&1&3\\ 2&1&3&4\\ 5&0&7&11 \end{pmatrix} \overse...
...ightarrow}\begin{pmatrix}1&-2&1&3\\ 0&5&1&-2\\ 0&0&0&0 \end{pmatrix}\end{align*}
We see already that the system has an infinity of solutions.
\begin{align*}\overset{R_2 \leftarrow \frac 15 R_2}{\longrightarrow}\begin{pmatr...
...&0&0 \end{pmatrix}\overset{R_1 \leftarrow R_1+2R_2}{\longrightarrow}\end{align*}
This matrix is associated to the system of equations:

\begin{displaymath}\begin{cases}x + \frac 75 z = \frac {11}{5} \\ y - \frac 15 z = -\frac 25 \end{cases}\end{displaymath}

The solution set of the given system is: $S= \left\{ (x,y,z)= \left( \frac {11}{5} - \frac 75 z, -\frac 25 + \frac 15 z , z \right); z \in \mathbb{R}\right\}$.

Suppose that A is in row-echelon form. An unknown in the column of whom there is a leading coefficient is called a principal unknown; otherwise, it is called a free unknown. In example  3.4, the unknowns x and y are principal, and the unknown z is free. \fbox{
\begin{minipage}{11cm}
If the system has solutions and if there are free ...
...expressed as a function of the free unknowns on his right side.
\end{minipage} }

Example 3.3.5   We solve the system: $\begin{cases}
x + 2y -z+t=1 \\ x + 2y +3z \qquad =1 \\ x + 2y + 3z -2t =-1
\end{cases}$

The augmented matrix A of this system is the matrix given in Example  2.6, i.e.

\begin{displaymath}A= \begin{pmatrix}
1 & 2 & -1 & 1 & 1 \\ 1 & 2 & 3 & 0 & 1 \\ 1 & 2 & 3 & -2 & -1
\end{pmatrix}\end{displaymath}

The row-reduced matrix B equivalent to A has been computed in Example  2.6:

\begin{displaymath}B=\begin{pmatrix}
\boldsymbol{1} & 2 & -1 & 0 & 0 \\ 0 & 0 & ...
...bol{1} & 0 & 1 \\
0 & 0 & 0 & \boldsymbol{1} & 1
\end{pmatrix}\end{displaymath}

Thus, the given system of equations is equivalent to the following system:

\begin{displaymath}\begin{cases}
x + 2y -z \qquad =1 \\ \qquad \qquad z \qquad =1 \\ \qquad \qquad \qquad t =-1
\end{cases}\end{displaymath}

The unknowns z and t are principal, as there are pivots in their respective colums. As there is no free unknown on their right, we find constant values for z and t from the two last equations. The y unknown is free, as there is no pivot in its column. By substitution of the values we found for z and t into the first equation, we find an expression for the principal unknown x as a function of y, namely x=2-2y. The set of solutions of the given system of equation is thus:

\begin{displaymath}\mathcal{S}= \{ (2-2y, y, 1 -1)\; ; \; y \in \mathbb{R}\}
\end{displaymath}


next up previous contents
Next: A special case: homogeneous Up: Solution of a system Previous: Solution of a system
Noah Dana-Picard
2001-02-26