Разложить вектор по векторам (по базису) онлайн

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Задание:

Разложить вектор $$\overline{a}$$ по базису $$\overline{x}$$ , $$\overline{y}$$ , $$\overline{z}$$
$$\overline{a}\lbrace{3}, {2}, {5}\rbrace$$ ;      $$\overline{x}\lbrace{\frac{1}{2}}, {\frac{1}{3}}, {1}\rbrace$$ ;      $$\overline{y}\lbrace{-1}, {-4}, {2}\rbrace$$ ;      $$\overline{z}\lbrace{1}, {-1}, {3}\rbrace$$ ;

Решение:


$${{\overline{a}}={{{{{\alpha}\cdot {\overline{x}}}}+{{{{{\beta}\cdot {\overline{y}}}}+{{{\gamma}\cdot {\overline{z}}}}}}}}}$$

$$\left\{\begin{array}{l}{{{-\beta+\gamma+\frac{1}{2} \alpha}={3}}}\\{{{-4 \beta-\gamma+\frac{1}{3} \alpha}={2}}}\\{{{2 \beta+3 \gamma+\alpha}={5}}}\end{array}\right.$$

$${\left(\begin{array}{ccc|c}{\frac{1}{2}}&{-1}&{1}&{3}\\{\frac{1}{3}}&{-4}&{-1}&{2}\\{1}&{2}&{3}&{5}\end{array}\right)}$$ ~ $${\left(\begin{array}{ccc|c}{1}&{-2}&{2}&{6}\\{\frac{1}{3}}&{-4}&{-1}&{2}\\{1}&{2}&{3}&{5}\end{array}\right)}$$ ~ $${\left(\begin{array}{ccc|c}{1}&{-2}&{2}&{6}\\{0}&{-\frac{10}{3}}&{-\frac{5}{3}}&{0}\\{1}&{2}&{3}&{5}\end{array}\right)}$$ ~ $${\left(\begin{array}{ccc|c}{1}&{-2}&{2}&{6}\\{0}&{-\frac{10}{3}}&{-\frac{5}{3}}&{0}\\{0}&{4}&{1}&{-1}\end{array}\right)}$$ ~ $${\left(\begin{array}{ccc|c}{1}&{-2}&{2}&{6}\\{0}&{1}&{\frac{1}{2}}&{0}\\{0}&{4}&{1}&{-1}\end{array}\right)}$$ ~ $${\left(\begin{array}{ccc|c}{1}&{0}&{3}&{6}\\{0}&{1}&{\frac{1}{2}}&{0}\\{0}&{4}&{1}&{-1}\end{array}\right)}$$ ~ $${\left(\begin{array}{ccc|c}{1}&{0}&{3}&{6}\\{0}&{1}&{\frac{1}{2}}&{0}\\{0}&{0}&{-1}&{-1}\end{array}\right)}$$ ~ $${\left(\begin{array}{ccc|c}{1}&{0}&{3}&{6}\\{0}&{1}&{\frac{1}{2}}&{0}\\{0}&{0}&{1}&{1}\end{array}\right)}$$ ~ $${\left(\begin{array}{ccc|c}{1}&{0}&{0}&{3}\\{0}&{1}&{\frac{1}{2}}&{0}\\{0}&{0}&{1}&{1}\end{array}\right)}$$ ~ $${\left(\begin{array}{ccc|c}{1}&{0}&{0}&{3}\\{0}&{1}&{0}&{-\frac{1}{2}}\\{0}&{0}&{1}&{1}\end{array}\right)}$$

$$\left\{\begin{array}{l}{{{\alpha}={3}}}\\{{{\beta}={-\frac{1}{2}}}}\\{{{\gamma}={1}}}\end{array}\right.$$

$${{\overline{a}}={{{3 \overline{x}}+\left({{{-\frac{1}{2} \overline{y}}+{\overline{z}}}}\right)}}}$$

Ответ:

$${{\overline{a}}={\overline{z}+3 \overline{x}-\frac{1}{2} \overline{y}}}$$