\(a)\;\;\left\{ \begin{array}{l}2x - 3y + z = -7\;\;\;\left( 1 \right)\\ - 4x + 5y + 3z = 6\;\;\;\left( 2 \right)\\x + 2y - 2z = 5\;\;\;\;\;\left( 3 \right)\end{array} \right.\)
Ta có: \(\left( 1 \right) \Leftrightarrow z = - 7 - 2x + 3y.\) Thế \(z\) vào \((2), \, \, (3)\) ta được hệ phương trình:
\(\begin{array}{l}\left\{ \begin{array}{l} - 4x + 5y + 3\left( { - 7 - 2x + 3y} \right) = 6\\x + 2y - 2\left( { - 7 - 2x + 3y} \right) = 5\end{array} \right.\\ \Leftrightarrow \left\{ \begin{array}{l}10x - 14y = - 27\\5x - 4y = - 9\end{array} \right.\\ \Leftrightarrow \left\{ \begin{array}{l}10x - 14y = - 27\\10x - 8y = - 18\end{array} \right.\\ \Leftrightarrow \left\{ \begin{array}{l}6y = 9\\5x - 4y = - 9\end{array} \right.\\ \Leftrightarrow \left\{ \begin{array}{l}x = - \frac{3}{5}\\y = \frac{3}{2}\end{array} \right.\\ \Rightarrow z = - 7 - 2.\left( { - \frac{3}{5}} \right) + 3.\frac{3}{2} = - \frac{{13}}{{10}}.\end{array}\)
Vậy hệ phương trình có nghiệm: \(\left( {x;\;y;\;z} \right) = \left( { - \frac{3}{5};\;\frac{3}{2};\; - \frac{{13}}{{10}}} \right).\)
\(\begin{array}{l}
b)\;\;\left\{ \begin{array}{l}
x + 4y - 2z = 1\\
- 2x + 3y + z = - 6\\
3x + 8y - z = 12
\end{array} \right.\\ \Leftrightarrow \left\{ \begin{array}{l}
x + 4y - 2z = 1\\
3x + 8y - z = 12\\
x + 11y = 6
\end{array} \right. \\ \Leftrightarrow \left\{ \begin{array}{l}
x + 4y - 2z = 1\\
6x + 16y - 2z = 24\\
x + 11y = 6
\end{array} \right.
\\ \Leftrightarrow \left\{ \begin{array}{l}
3x + 8y - z = 12\\
5x + 12y = 23\\
x + 11y = 6
\end{array} \right.\\ \Leftrightarrow \left\{ \begin{array}{l}
3x + 8y - z = 12\\
5x + 12y = 23\\
5x + 55y = 30
\end{array} \right.\\ \Leftrightarrow \left\{ \begin{array}{l}
3x + 8y - z = 12\\
x + 11y = 6\\
43y = 7
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
z = 3x + 8y - 12\\
y = \frac{7}{{43}}\\
x = 6 - 11.\frac{7}{{43}}
\end{array} \right.\\ \Leftrightarrow \left\{ \begin{array}{l}
x = \frac{{181}}{{43}}\\
y = \frac{7}{{43}}\\
z = \frac{{83}}{{43}}
\end{array} \right..
\end{array}\)
Vậy phương trình có nghiệm \(\left( {x;\;y;\;z} \right) = \left( {\frac{{181}}{{43}};\frac{7}{{43}};\;\frac{{83}}{{43}}} \right).\)
