Трећи разред средње школе

Крамерово правило - примери 3

Решавање једначина задатих у облику детерминанте.

Задаци

Текст задатака објашњених у видео лекцији.

Пр.4)   Решити систем једначина са параметрима.

            $ax + y + z = 1$

            $x - ay + z = 1$

            $x + y + az = 1$

Пр.4)

\[\begin{gathered}
{D_s} = \left| {\begin{array}{*{20}{c}}
a&1&1 \\
1&a&1 \\
1&1&a
\end{array}} \right| = \left| {\begin{array}{*{20}{c}}
a&1&1 \\
{1 - a}&{a - 1}&0 \\
{1 - {a^2}}&{1 - a}&0
\end{array}} \right| = \left| {\begin{array}{*{20}{c}}
{1 - a}&{ - \left( {1 - a} \right)} \\
{\left( {1 - a} \right)\left( {1 + a} \right)}&{1 - a}
\end{array}} \right| = \hfill \\
\hfill \\
= {\left( {1 - a} \right)^2}\left| {\begin{array}{*{20}{c}}
1&{ - 1} \\
{1 + a}&1
\end{array}} \right| = {\left( {1 - a} \right)^2}\left( {2 + a} \right) \hfill \\
\hfill \\
{D_x} = \left| {\begin{array}{*{20}{c}}
1&1&1 \\
1&a&1 \\
1&1&a
\end{array}} \right| = \left| {\begin{array}{*{20}{c}}
1&0&0 \\
1&{a - 1}&0 \\
1&0&{a - 1}
\end{array}} \right| = \left| {\begin{array}{*{20}{c}}
{a - 1}&0 \\
0&{a - 1}
\end{array}} \right| = {\left( {a - 1} \right)^2} \hfill \\
\hfill \\
{D_y} = \left| {\begin{array}{*{20}{c}}
a&1&1 \\
1&1&1 \\
1&1&a
\end{array}} \right| = \left| {\begin{array}{*{20}{c}}
a&{ - 1}&1 \\
{1 - a}&0&0 \\
{1 - a}&0&{a - 1}
\end{array}} \right| = - 1 \cdot \left| {\begin{array}{*{20}{c}}
{1 - a}&0 \\
{1 - a}&{a - 1}
\end{array}} \right| = \hfill \\
\hfill \\
= - \left( {1 - a} \right)\left| {\begin{array}{*{20}{c}}
1&0 \\
1&{a - 1}
\end{array}} \right| = \left( {a - 1} \right)\left( {a - 1} \right) = {\left( {a - 1} \right)^2} \hfill \\
\hfill \\
{D_z} = \left| {\begin{array}{*{20}{c}}
a&1&1 \\
1&a&1 \\
1&1&1
\end{array}} \right| = \left| {\begin{array}{*{20}{c}}
a&{1 - a}&{1 - a} \\
1&{a - 1}&0 \\
1&0&0
\end{array}} \right| = \left( {1 - a} \right)\left| {\begin{array}{*{20}{c}}
1&{a - 1} \\
1&0
\end{array}} \right| = \hfill \\
\hfill \\
= \left( {1 - a} \right)\left( {0 - \left( {a - 1} \right)} \right) = {\left( {1 - a} \right)^2} \hfill \\
\end{gathered} \]

1) ${D_s} \ne 0$ јединствено решење

\[\begin{gathered}
{\left( {1 - a} \right)^2}\left( {2 + a} \right) \ne 0 \hfill \\
{\left( {1 - a} \right)^2} \ne 0 \cap \left( {2 + a} \right) \ne 0 \hfill \\
1 - a \ne 0 \hfill \\
a \ne 1 \hfill \\
\begin{array}{*{20}{c}}
{{{\left( {1 - a} \right)}^2} \ne 0}& \cap &{\left( {2 + a} \right) \ne 0} \\
{1 - a \ne 0}&{}&{a \ne - 2} \\
{a \ne 1}&{}&{}
\end{array} \hfill \\
x = \frac{{{D_x}}}{{{D_s}}} = \frac{{{{\left( {1 - a} \right)}^2}}}{{{{\left( {1 - a} \right)}^2}\left( {2 + a} \right)}} = \frac{1}{{2 + a}} \hfill \\
\hfill \\
x = y = z = \frac{1}{{2 + a}} \hfill \\
\left( {x,y,z} \right) = \left( {\frac{1}{{2 + a}},\frac{1}{{2 + a}},\frac{1}{{2 + a}}} \right) \hfill \\
\end{gathered} \]

2) ${D_s} = 0$

Ако је $a = 1$

\[\begin{gathered}
{D_s} = {\left( {1 - a} \right)^2}\left( {2 + a} \right) = 0 \hfill \\
{D_x} = {D_y} = {D_z} = 0 \hfill \\
x + y + z = 1 \hfill \\
x + y + z = 1 \hfill \\
x + y + z = 1 \hfill \\
\end{gathered} \]  двоструко неодређен систем

\[\begin{gathered}
x = 1 - y - z \hfill \\
\left( {x,y,z} \right) = \left( {1 - y - z,y,z} \right) \hfill \\
\end{gathered} \]

Ако је $a = -2$

\[\begin{gathered}
{D_s} = {\left( {1 - a} \right)^2}\left( {2 - 2} \right) = 0 \hfill \\
{D_x} = {\left( {1 + 2} \right)^2} = 9 \hfill \\
\end{gathered} \] систем нема решења