Kapitel 12: Lineare Algebra und Analytische Geometrie (S. 199)

Aufgabe 1

(a) Es ist

\(\begin{align*}\vec{x}+\vec{y}=\left(\begin{array}{c}1\\-2\\2\end{array}\right)+\left(\begin{array}{c}0\\1\\1\end{array}\right)=\left(\begin{array}{c}1+0\\-2+1\\2+1\end{array}\right)=\left(\begin{array}{c}1\\-1\\3\end{array}\right)\,.\end{align*}\)

(b) Es ist

\(\begin{align*}\vec{y}-\vec{z}=\left(\begin{array}{c}0\\1\\1\end{array}\right)-\left(\begin{array}{c}3\\0\\2\end{array}\right)=\left(\begin{array}{c}0-3\\1-0\\1-2\end{array}\right)=\left(\begin{array}{c}-3\\1\\-1\end{array}\right)\,.\end{align*}\)

(c) Es ist

\(\begin{align*}2\cdot\vec{x}-3\cdot(\vec{y}+\vec{z})&=2\cdot\left(\begin{array}{c}1\\-2\\2\end{array}\right)-3\cdot\left(\left(\begin{array}{c}0\\1\\1\end{array}\right)+\left(\begin{array}{c}3\\0\\2\end{array}\right)\right)\\&=\left(\begin{array}{c}2\cdot 1\\2\cdot(-2)\\2\cdot2\end{array}\right)-3\cdot\left(\begin{array}{c}0+3\\1+0\\1+2\end{array}\right)=\left(\begin{array}{c}2\\-4\\4\end{array}\right)-3\cdot\left(\begin{array}{c}3\\1\\3\end{array}\right)\\&=\left(\begin{array}{c}2\\-4\\4\end{array}\right)-\left(\begin{array}{c}3\cdot 3\\3\cdot 1\\3\cdot 3\end{array}\right)=\left(\begin{array}{c}2\\-4\\4\end{array}\right)-\left(\begin{array}{c}9\\3\\9\end{array}\right)=\left(\begin{array}{c}2-9\\-4-3\\4-9\end{array}\right)=\left(\begin{array}{c}-7\\-7\\-5\end{array}\right)\,.\end{align*}\)

(d) Es ist

\(\begin{align*}\vec{z}+\frac{1}{2}\cdot(\vec{z}-\vec{y})&=\left(\begin{array}{c}3\\0\\2\end{array}\right)+\frac{1}{2}\cdot\left(\left(\begin{array}{c}3\\0\\2\end{array}\right)-\left(\begin{array}{c}0\\1\\1\end{array}\right)\right)\\&=\left(\begin{array}{c}3\\0\\2\end{array}\right)+\frac{1}{2}\cdot\left(\begin{array}{c}3-0\\0-1\\2-1\end{array}\right)=\left(\begin{array}{c}3\\0\\2\end{array}\right)+\frac{1}{2}\cdot\left(\begin{array}{c}3\\-1\\1\end{array}\right)\\&=\left(\begin{array}{c}3\\0\\2\end{array}\right)+\left(\begin{array}{c}\frac{1}{2}\cdot3\\\frac{1}{2}\cdot (-1)\\\frac{1}{2}\cdot 1\end{array}\right)=\left(\begin{array}{c}3\\0\\2\end{array}\right)+\left(\begin{array}{c}\frac{3}{2}\\-\frac{1}{2}\\\frac{1}{2}\end{array}\right)=\left(\begin{array}{c}3+\frac{3}{2}\\0-\frac{1}{2}\\2+\frac{1}{2}\end{array}\right)=\left(\begin{array}{c}\frac{9}{2}\\-\frac{1}{2}\\\frac{5}{2}\end{array}\right)\,.\end{align*}\)

Aufgabe 2

(a) Es ist

\(\begin{align*}0,5\cdot\left(\begin{array}{c}-1\\0\\1\end{array}\right)+\left[\vec{x}-\left(\begin{array}{c}2\\1\\1\end{array}\right)\right]=\left(\begin{array}{c}0\\0\\0\end{array}\right)&\;\Longleftrightarrow\;0,5\cdot\left(\begin{array}{c}-1\\0\\1\end{array}\right)+\left[\vec{x}-\left(\begin{array}{c}2\\1\\1\end{array}\right)\right]=\left(\begin{array}{c}0\\0\\0\end{array}\right)\\&\;\Longleftrightarrow\;\left(\begin{array}{c}-\frac{1}{2}\\0\\\frac{1}{2}\end{array}\right)-\left(\begin{array}{c}2\\1\\1\end{array}\right)+\vec{x}=\left(\begin{array}{c}0\\0\\0\end{array}\right)\\&\;\Longleftrightarrow\;\left(\begin{array}{c}-\frac{5}{2}\\-1\\-\frac{1}{2}\end{array}\right)+\vec{x}=\left(\begin{array}{c}0\\0\\0\end{array}\right)\\&\;\Longleftrightarrow\;\vec{x}=\left(\begin{array}{c}0\\0\\0\end{array}\right)-\left(\begin{array}{c}-\frac{5}{2}\\-1\\-\frac{1}{2}\end{array}\right)=\left(\begin{array}{c}\frac{5}{2}\\1\\\frac{1}{2}\end{array}\right)\,.\end{align*}\)

(b) Es ist

\(\begin{align*}\frac{3}{2}\cdot\left[\vec{x}-\left(\begin{array}{c}-3\\2\\-1\end{array}\right)\right]=\vec{x}+\left(\begin{array}{c}-1\\1\\1\end{array}\right)&\;\Longleftrightarrow\;\vec{x}-\left(\begin{array}{c}-3\\2\\-1\end{array}\right)=\frac{2}{3}\cdot\left[\vec{x}+\left(\begin{array}{c}-1\\1\\1\end{array}\right)\right]\\&\;\Longleftrightarrow\;\vec{x}=\frac{2}{3}\cdot\vec{x}+\frac{2}{3}\cdot\left(\begin{array}{c}-1\\1\\1\end{array}\right)+\left(\begin{array}{c}-3\\2\\-1\end{array}\right)\\&\;\Longleftrightarrow\;\vec{x}-\frac{2}{3}\cdot\vec{x}=\left(\begin{array}{c}-\frac{2}{3}\\\frac{2}{3}\\\frac{2}{3}\end{array}\right)+\left(\begin{array}{c}-3\\2\\-1\end{array}\right)\\&\;\Longleftrightarrow\;\frac{1}{3}\cdot\vec{x}=\left(\begin{array}{c}-\frac{11}{3}\\\frac{8}{3}\\-\frac{1}{3}\end{array}\right)\;\Longleftrightarrow\;\vec{x}=3\cdot\left(\begin{array}{c}-\frac{11}{3}\\\frac{8}{3}\\-\frac{1}{3}\end{array}\right)=\left(\begin{array}{c}-11\\8\\-1\end{array}\right)\,.\end{align*}\)