Крамерово правило – примери 4

Текст задатака објашњених у видео лекцији.

Пр.4)   Решити систем једначина са параметрима.

            $x + y + z = a$

            $x + \left( {1 + a} \right)y + z = 2a$

            $x + y - \left( {1 + a} \right)z = 0$

4)

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

1) ${D_s} \ne 0$ јединствено решење

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

2) ${D_s} = 0$

Ако је $a = 1$

${D_s} = {D_x} = {D_y} = {D_z} = 0$ - неодређен систем

\[\begin{gathered}
x + y + z = 0 \hfill \\
x + y + z = 0 \hfill \\
x + y - z = 0 \hfill \\
\end{gathered} \] једноструко неодређен систем

\[\begin{gathered}
x + y + z = 0 \hfill \\
x + y + z = 0 \hfill \\
\underline {x + y - z = 0} \hfill \\
x + y + z = 0 \hfill \\
\underline {2x + 2y = 0} \hfill \\
x + y + z = 0 \hfill \\
x + y = 0 \Rightarrow y = - x \hfill \\
x + \left( { - x} \right) + z = 0 \hfill \\
z = 0 \hfill \\
\left( {x,y,z} \right) = \left( {x, - x,0} \right) \hfill \\
\end{gathered} \]

Ако је $a = -2$

\[\begin{gathered}
{D_s} = 0 \hfill \\
{D_x} = a\left( {2 - {a^2}} \right) = - 2\left( {2 - 4} \right) = 4 \ne 0 \hfill \\
\end{gathered} \] систем нема решења