Трећи разред средње школе

Крамерово правило - примери 5

Дискусија решења хомогеног система једначина, решени задаци.

Задаци

Текст задатака објашњених у видео лекцији.

Пр.6)   Решити систем једначина са параметрима.

           $x + y + z = 0$

           $ax + 4y + z = 0$

           $6x + \left( {a + 2} \right)y + 2z = 0$

Пр.6) 

\[\begin{gathered}
{D_S} = \left| {\begin{array}{*{20}{c}}
1&1&1 \\
a&4&1 \\
6&{a + 2}&2
\end{array}} \right| = \left| {\begin{array}{*{20}{c}}
1&0&0 \\
a&{4 - a}&{1 - a} \\
6&{4 - a}&{ - 4}
\end{array}} \right| = \left| {\begin{array}{*{20}{c}}
{4 - a}&{1 - a} \\
{a - 4}&{ - 4}
\end{array}} \right| = \hfill \\
\hfill \\
= \left| {\begin{array}{*{20}{c}}
{4 - a}&{1 - a} \\
{ - \left( {4 - a} \right)}&{ - 4}
\end{array}} \right| = \left( {4 - a} \right)\left| {\begin{array}{*{20}{c}}
1&{1 - a} \\
{ - 1}&{ - 4}
\end{array}} \right| = \left( {4 - a} \right)\left( { - 4 + 1 - a} \right) = \hfill \\
\hfill \\
= \left( {4 - a} \right)\left( { - 3 - a} \right) = - \left( {4 - a} \right)\left( {a + 3} \right) = \left( {a - 4} \right)\left( {a + 3} \right) \hfill \\
\end{gathered} \]

Знамо да ${D_x} = {D_y} = {D_z} = 0$ 

1) ${D_s} \ne 0$

\[\begin{gathered}
\left( {a - 4} \right)\left( {a + 4} \right) \ne 0 \hfill \\
\begin{array}{*{20}{c}}
{a - 4 \ne 0}& \cap &{a + 3 \ne 0} \\
{a \ne 4}&{}&{a \ne - 3} \\
{}&{}&{}
\end{array} \hfill \\
\left( {x,y,z} \right) = \left( {0,0,0} \right) \hfill \\
\end{gathered} \]

2) ${D_s}= 0$

Ако је $a = 4$ 

\[\begin{gathered}
x + y + z = 0 \hfill \\
ax + 4y + z = 0 \hfill \\
\underline {6x + \left( {a + 2} \right)y + 2z = 0} \hfill \\
x + y + z = 0 \hfill \\
4x + 4y + z = 0 \hfill \\
\underline {6x + 6y + 2z = 0} \hfill \\
x + y + z = 0 \hfill \\
3x + 3y = 0 \hfill \\
\underline {4x + 4y = 0} \hfill \\
x + y + z = 0 \hfill \\
\underline {x + y = 0} \hfill \\
x = - y \hfill \\
z = 0 \hfill \\
\hfill \\
\left( {x,y,z} \right) = \left( { - y,y,0} \right) \hfill \\
\end{gathered} \]

Ако је $a = -3$

\[\begin{gathered}
x + y + z = 0 \hfill \\
ax + 4y + z = 0 \hfill \\
\underline {6x + \left( {a + 2} \right)y + 2z = 0} \hfill \\
x + y + z = 0 \hfill \\
- 3x + 4y + z = 0 \hfill \\
\underline {6x - y + 2z = 0} \hfill \\
x + y + z = 0 \hfill \\
- 4x + 3y = 0 \hfill \\
\underline {4x - 3y = 0} \hfill \\
x + y + z = 0 \hfill \\
\underline { - 4x + 3y = 0} \hfill \\
3y = 4x;y = \frac{{4x}}{3} \hfill \\
x + \frac{{4x}}{3} + z = 0 \hfill \\
z = - x - \frac{{4x}}{3};z = - \frac{7}{3}x \hfill \\
\hfill \\
\left( {x,y,z} \right) = \left( {x,\frac{{4x}}{3}, - \frac{7}{3}x} \right) \hfill \\
\end{gathered} \]