[m]\vec{a_{1}}=-2\vec{e_{1}}+\vec{e_{2}}+0\vec{e_{3}}[/m]
[m]\vec{a_{2}}=\vec{e_{1}}+2\vec{e_{2}}+\vec{e_{3}}[/m]
[m]\vec{a_{3}}=-3\vec{e_{1}}-\vec{e_{2}}+2\vec{e_{3}}[/m]
[m]\vec{d}=1\vec{e_{1}}-2\vec{e_{2}}+3\vec{e_{3}}[/m]
Найти
[m]\vec{d}=x\vec{a_{1}}+y\vec{a_{2}}+z\vec{a_{3}}[/m]
Решение:
[m]1\vec{e_{1}}-2\vec{e_{2}}+3\vec{e_{3}}=x\cdot (-2\vec{e_{1}}+\vec{e_{2}}+0\cdot \vec{e_{3}})+y\cdot (\vec{e_{1}}+2\vec{e_{2}}+\vec{e_{3}})+z\cdot (-3\vec{e_{1}}-\vec{e_{2}}+2\vec{e_{3}})[/m]
[m]1\vec{e_{1}}-2\vec{e_{2}}+3\vec{e_{3}}=(-2x+y-3z)\cdot \vec{e_{1}}+(x+2y-z)\cdot \vec{e_{2}}+(0\cdot x+y+2z)\cdot \vec{e_{3}}[/m]
Приравниваем и получаем систему трех уравнений:
[m]\left\{\begin {matrix}1=-2x+y-3z\\-2=x+2y-z\\3=y+2z\end {matrix}\right.[/m]
Решаем систему[i] методом[/i] Крамера
О т в е т.:
[m]\vec{d}=-2\vec{a_{1}}+0,6\vec{a_{2}}+1,2\vec{a_{3}}[/m]