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Детерминанте - примери 3

Решавање једначина задатих у облику детерминанте

Задаци

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Пр.3)   Решити једначину задату преко детерминанте.

            $\left| {\begin{array}{*{20}{c}}
  a&1&1 \\
  1&a&1 \\
  1&1&a
\end{array}} \right| = 0$

Пр.4)   $\left| {\begin{array}{*{20}{c}}
  {x - 3}&{x + 2}&{x - 1} \\
  {x + 2}&{x - 4}&x \\
  {x - 1}&{x + 4}&{x - 5}
\end{array}} \right| = 0$

Пр.3)

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

Пр.4)

\[\begin{gathered}
\left| {\begin{array}{*{20}{c}}
{x - 3}&{x + 2}&{x - 1} \\
{x + 2}&{x - 4}&x \\
{x - 1}&{x + 4}&{x - 5}
\end{array}} \right| = \left| {\begin{array}{*{20}{c}}
{x - 3}&{x + 2}&{x - 1} \\
{x + 2 - x + 3}&{x - 4 - x - 2}&{x - x + 1} \\
{x - 1 - x + 3}&{x + 4 - x - 2}&{x - 5 - x + 1}
\end{array}} \right| = \hfill \\
\hfill \\
= \left| {\begin{array}{*{20}{c}}
{x - 3}&{x + 2}&{x - 1} \\
5&{ - 6}&1 \\
2&2&{ - 4}
\end{array}} \right| = 2 \cdot \left| {\begin{array}{*{20}{c}}
{x - 3}&{x + 2}&{x - 1} \\
5&{ - 6}&1 \\
1&1&{ - 1}
\end{array}} \right| = \hfill \\
\hfill \\
= 2 \cdot \left| {\begin{array}{*{20}{c}}
{x - 3}&{x + 2 - x + 3}&{x - 1 + 2x - 6} \\
5&{ - 11}&{11} \\
1&0&0
\end{array}} \right| = 2 \cdot \left| {\begin{array}{*{20}{c}}
5&{3x - 7} \\
{ - 11}&{11}
\end{array}} \right| = \hfill \\
\hfill \\
= 22 \cdot \left| {\begin{array}{*{20}{c}}
5&{3x - 7} \\
{ - 1}&1
\end{array}} \right| = 22\left( {5 + 3x - 7} \right) = 22\left( {3x - 2} \right) \hfill \\
\hfill \\
22\left( {3x - 2} \right) = 0 \hfill \\
3x - 2 = 0 \hfill \\
3x = 2 \hfill \\
x = \frac{2}{3} \hfill \\
\end{gathered} \]