Трећи разред средње школе

Крамерово правило - примери 1

Решавање једначина задатих у облику детерминанте.

Задаци

Текст задатака објашњених у видео лекцији.

Пр.1)   Решити дати ситем једначина.

            $x + y + 2 = 6$

            $2x - y + z = 3$

            $x - 2y + 2x = 3$

Пр.3)   Решити дати систем једначина.

            $x + y + z = 3$

            $2x - 3y + 2z = 1$

            $3x - 2y + 3z = 7$

Пр.1)

\[\begin{gathered}
{D_s} = \left| {\begin{array}{*{20}{c}}
1&1&1 \\
2&{ - 1}&1 \\
1&{ - 2}&2
\end{array}} \right|\begin{array}{*{20}{c}}
1 \\
2 \\
1
\end{array}\begin{array}{*{20}{c}}
{} \\
{} \\
{}
\end{array}\begin{array}{*{20}{c}}
1 \\
{ - 1} \\
{ - 2}
\end{array} = - 2 + 1 - 4 + 1 + 2 - 4 = - 6 \ne 0 \hfill \\
\hfill \\
{D_x} = \left| {\begin{array}{*{20}{c}}
6&1&1 \\
3&{ - 1}&1 \\
3&{ - 2}&2
\end{array}} \right|\begin{array}{*{20}{c}}
6 \\
3 \\
3
\end{array}\begin{array}{*{20}{c}}
{} \\
{} \\
{}
\end{array}\begin{array}{*{20}{c}}
1 \\
{ - 1} \\
{ - 2}
\end{array} = - 12 + 3 - 6 + 3 + 12 - 6 = - 6\begin{array}{*{20}{c}}
{}&{}
\end{array} \hfill \\
\hfill \\
{D_y} = \left| {\begin{array}{*{20}{c}}
1&6&1 \\
2&3&1 \\
1&3&2
\end{array}} \right|\begin{array}{*{20}{c}}
1 \\
2 \\
1
\end{array}\begin{array}{*{20}{c}}
{} \\
{} \\
{}
\end{array}\begin{array}{*{20}{c}}
6 \\
3 \\
3
\end{array} = 6 + 6 + 6 - 3 - 3 - 24 = - 12\begin{array}{*{20}{c}}
{}&{}&{}&{}
\end{array} \hfill \\
\hfill \\
{D_z} = \left| {\begin{array}{*{20}{c}}
1&1&6 \\
2&{ - 1}&3 \\
1&{ - 2}&3
\end{array}} \right|\begin{array}{*{20}{c}}
1 \\
2 \\
1
\end{array}\begin{array}{*{20}{c}}
{} \\
{} \\
{}
\end{array}\begin{array}{*{20}{c}}
1 \\
{ - 1} \\
{ - 2}
\end{array} = - 3 + 3 - 24 + 6 + 6 - 6 = - 18\begin{array}{*{20}{c}}
{}&{}
\end{array} \hfill \\
\hfill \\
x = \frac{{{D_x}}}{{{D_s}}} = \frac{{ - 6}}{{ - 6}} = 1 \hfill \\
\hfill \\
y = \frac{{{D_y}}}{{{D_s}}} = \frac{{ - 12}}{{ - 6}} = 2 \hfill \\
\hfill \\
z = \frac{{{D_z}}}{{{D_s}}} = \frac{{ - 18}}{{ - 6}} = 3 \hfill \\
\hfill \\
\left( {x,y,z} \right) = \left( {1,2,3} \right) \hfill \\
\end{gathered} \]

Пр.3)

\[\begin{gathered}
{D_s} = \left| {\begin{array}{*{20}{c}}
1&1&1 \\
2&{ - 3}&2 \\
3&{ - 2}&3
\end{array}} \right|\begin{array}{*{20}{c}}
1 \\
2 \\
3
\end{array}\begin{array}{*{20}{c}}
{} \\
{} \\
{}
\end{array}\begin{array}{*{20}{c}}
1 \\
{ - 3} \\
{ - 2}
\end{array} = - 9 + 6 - 4 + 9 + 4 - 6 = 0\begin{array}{*{20}{c}}
{}&{}&{}
\end{array} \hfill \\
\hfill \\
{D_x} = \left| {\begin{array}{*{20}{c}}
3&1&1 \\
1&{ - 3}&2 \\
7&{ - 2}&3
\end{array}} \right|\begin{array}{*{20}{c}}
3 \\
1 \\
7
\end{array}\begin{array}{*{20}{c}}
{} \\
{} \\
{}
\end{array}\begin{array}{*{20}{c}}
1 \\
{ - 3} \\
{ - 2}
\end{array} = - 27 + 14 - 2 + 21 + 12 - 3 = 15 \ne 0 \hfill \\
\end{gathered} \]

Систем немогућ.