1. 다음 행렬 중 정칙행렬을 찾고 각 역행렬을 구하라.
$$ \small (1)\ A=\begin{pmatrix}3&-7\\-2&5\end{pmatrix} \qquad (2)\ B=\begin{pmatrix}-1&1\\1&-1\end{pmatrix}$$
$$ \small (3)\ C=\begin{pmatrix}2&3&1\\5&7&3\\1&1&1\end{pmatrix} \qquad (4)\ D=\begin{pmatrix}1&1&1\\3&4&5\\6&7&8\end{pmatrix}$$
$$ \small (5)\ E=\begin{pmatrix}1&2&3&1\\1&1&-2&-1\\-1&-1&2&-1\\-1&2&1&1\end{pmatrix} \qquad (6)\ F=\begin{pmatrix}0&2&1&-1\\1&-3&-2&1\\3&3&1&4\\3&2&0&1\end{pmatrix}$$
1. 정칙행렬
- n차 정방행렬 A에 행렬 B가 존재하여 AB=BA=I를 만족할 때 이다.
2. 역행렬
- 정칙행렬 A에 대해 역연산이 가능한 행렬
(1)
$$ \small (A|I_{2})=\left(\begin{array}{cc|cc}3&-7&1&0\\-2&5&0&1\end{array}\right) $$
$$ \small \underrightarrow{R_{1}(\frac{1}{3})} \left(\begin{array}{cc|cc}1&-\frac{7}{3}&\frac{1}{3}&0\\-2&5&0&1\end{array}\right) $$
$$ \small \underrightarrow{R_{1,\ 2}(2)} \left(\begin{array}{cc|cc}1&-\frac{7}{3}&\frac{1}{3}&0\\0&\frac{1}{3}&\frac{2}{3}&1\end{array}\right) $$
$$ \small \underrightarrow{R_{2,\ 1}(7)} \left(\begin{array}{cc|cc}1&0&5&7\\0&\frac{1}{3}&\frac{2}{3}&1\end{array}\right) $$
$$ \small \underrightarrow{R_{2}(3)} \left(\begin{array}{cc|cc}1&0&5&7\\0&1&2&1\end{array}\right) $$
소거행제형 행렬로 변환하여 영행이 없으므로, 정칙행렬이다.
그러므로 역행렬은
$$ \small \begin{pmatrix}5&7\\2&1\end{pmatrix}$$
(2)
$$ \small (A|I_{2})=\left(\begin{array}{cc|cc}-1&1&1&0\\1&-1&0&1\end{array}\right) $$
$$ \small \underrightarrow{R_{1,\ 2}} \left(\begin{array}{cc|cc}1&-1&0&1\\-1&1&1&0\end{array}\right) $$
$$ \small \underrightarrow{R_{1,\ 2}(1)} \left(\begin{array}{cc|cc}1&-1&0&1\\0&0&1&1\end{array}\right) $$
소거행제형으로 변환하였는데 영행이 존재하므로, 정칙행렬이 아니다.
(3)
$$ \small (A|I_{3})=\left(\begin{array}{ccc|ccc}2&3&1&1&0&0\\5&7&3&0&1&0\\1&1&1&0&0&1\end{array}\right) $$
$$ \small \underrightarrow{R_{3,\ 1}} \left(\begin{array}{ccc|ccc}1&1&1&0&0&1\\5&7&3&0&1&0\\2&3&1&1&0&0\end{array}\right) $$
$$ \small \underrightarrow{R_{1,\ 2}(-5)} \left(\begin{array}{ccc|ccc}1&1&1&0&0&1\\0&2&-2&0&1&-5\\2&3&1&1&0&0\end{array}\right) $$
$$ \small \underrightarrow{R_{1,\ 3}(-2)} \left(\begin{array}{ccc|ccc}1&1&1&0&0&1\\0&2&-2&0&1&-5\\0&1&-1&1&0&-2\end{array}\right) $$
$$ \small \underrightarrow{R_{3,\ 2}} \left(\begin{array}{ccc|ccc}1&1&1&0&0&1\\0&1&-1&1&0&-2\\0&2&-2&0&1&-5\end{array}\right) $$
$$ \small \underrightarrow{R_{2,\ 1}(-1)} \left(\begin{array}{ccc|ccc}1&0&2&-1&0&3\\0&1&-1&1&0&-2\\0&2&-2&0&1&-5\end{array}\right) $$
$$ \small \underrightarrow{R_{2,\ 3}(-2)} \left(\begin{array}{ccc|ccc}1&0&2&-1&0&3\\0&1&-1&1&0&-2\\0&0&0&-2&1&-1\end{array}\right) $$
세번째 행에서 영행이 존재하므로, 정칙행렬이 아니다.
(4)
$$ \small (A|I_{3})=\left(\begin{array}{ccc|ccc}1&1&1&1&0&0\\3&4&5&0&1&0\\6&7&8&0&0&1\end{array}\right) $$
$$ \small \underrightarrow{R_{1,\ 2}(-3)} \left(\begin{array}{ccc|ccc}1&1&1&1&0&0\\0&1&2&-3&1&0\\6&7&8&0&0&1\end{array}\right) $$
$$ \small \underrightarrow{R_{1,\ 3}(-6)} \left(\begin{array}{ccc|ccc}1&1&1&1&0&0\\0&1&2&-3&1&0\\0&1&2&-6&0&1\end{array}\right) $$
$$ \small \underrightarrow{R_{2,\ 3}(-1)} \left(\begin{array}{ccc|ccc}1&1&1&1&0&0\\0&1&2&-3&1&0\\0&0&0&-3&-1&1\end{array}\right) $$
세번째 행에서 영행이 존재하므로, 정칙행렬이 아니다.
(5)
$$ \small (A|I_{4})= \left(\begin{array}{cccc|cccc}1&2&3&1&1&0&0&0\\1&1&-2&-1&0&1&0&0\\-1&-1&2&-1&0&0&1&0\\-1&2&1&1&0&0&0&1\end{array}\right) $$
$$ \small \underrightarrow{R_{1,\ 2}(-1)} \left(\begin{array}{cccc|cccc}1&2&3&1&1&0&0&0\\0&-1&-5&-2&-1&1&0&0\\-1&-1&2&-1&0&0&1&0\\-1&2&1&1&0&0&0&1\end{array}\right) $$
$$ \small \underrightarrow{R_{1,\ 3}(1)} \left(\begin{array}{cccc|cccc}1&2&3&1&1&0&0&0\\0&-1&-5&-2&-1&1&0&0\\0&1&5&0&1&0&1&0\\-1&2&1&1&0&0&0&1\end{array}\right) $$
$$ \small \underrightarrow{R_{1,\ 4}(1)} \left(\begin{array}{cccc|cccc}1&2&3&1&1&0&0&0\\0&-1&-5&-2&-1&1&0&0\\0&1&5&0&1&0&1&0\\0&4&4&2&1&0&0&1\end{array}\right) $$
$$ \small \underrightarrow{R_{2,\ 3}} \left(\begin{array} {cccc|cccc}1&2&3&1&1&0&0&0\\0&1&5&0&1&0&1&0\\0&-1&-5&-2&-1&1&0&0\\0&4&4&2&1&0&0&1\end{array}\right) $$
$$ \small \underrightarrow{R_{2,\ 3}(1)} \left(\begin{array}{cccc|cccc}1&2&3&1&1&0&0&0\\0&1&5&0&1&0&1&0\\0&0&0&-2&0&1&1&0\\0&4&4&2&1&0&0&1\end{array}\right) $$
$$ \small \underrightarrow{R_{2,\ 1}(-2)} \left(\begin{array}{cccc|cccc}1&0&-7&1&-1&0&-2&0\\0&1&5&0&1&0&1&0\\0&0&0&-2&0&1&1&0\\0&4&4&2&1&0&0&1\end{array}\right) $$
$$ \small \underrightarrow{R_{2,\ 4}(-4)} \left(\begin{array}{cccc|cccc}1&0&-7&1&-1&0&-2&0\\0&1&5&0&1&0&1&0\\0&0&0&-2&0&1&1&0\\0&0&-16&2&-3&0&-4&1\end{array}\right) $$
$$ \small \underrightarrow{R_{4}(-\frac{1}{16})} \left(\begin{array}{cccc|cccc}1&0&-7&1&-1&0&-2&0\\0&1&5&0&1&0&1&0\\0&0&0&-2&0&1&1&0\\0&0&1&-\frac{1}{8}&\frac{3}{16}&0&\frac{1}{4}&-\frac{1}{16}\end{array}\right) $$
$$ \small \underrightarrow{R_{3,\ 4}} \left(\begin{array}{cccc|cccc}1&0&-7&1&-1&0&-2&0\\0&1&5&0&1&0&1&0\\0&0&1&-\frac{1}{8}&\frac{3}{16}&0&\frac{1}{4}&-\frac{1}{16}\\0&0&0&-2&0&1&1&0\end{array}\right) $$
$$ \small \underrightarrow{R_{3,\ 1}(7)} \left(\begin{array}{cccc|cccc}1&0&0&\frac{1}{8}&\frac{5}{16}&0&-\frac{1}{4}&-\frac{7}{16}\\0&1&5&0&1&0&1&0\\0&0&1&-\frac{1}{8}&\frac{3}{16}&0&\frac{1}{4}&-\frac{1}{16}\\0&0&0&-2&0&1&1&0\end{array}\right) $$
$$ \small \underrightarrow{R_{3,\ 2}(-5)} \left(\begin{array}{cccc|cccc}1&0&0&\frac{1}{8}&\frac{5}{16}&0&-\frac{1}{4}&-\frac{7}{16}\\0&1&0&\frac{5}{8}&\frac{1}{16}&0&-\frac{1}{4}&\frac{5}{16}\\0&0&1&-\frac{1}{8}&\frac{3}{16}&0&\frac{1}{4}&-\frac{1}{16}\\0&0&0&-2&0&1&1&0\end{array}\right) $$
$$ \small \underrightarrow{R_{4}(-\frac{1}{2})} \left(\begin{array}{cccc|cccc}1&0&0&\frac{1}{8}&\frac{5}{16}&0&-\frac{1}{4}&-\frac{7}{16}\\0&1&0&\frac{5}{8}&\frac{1}{16}&0&-\frac{1}{4}&\frac{5}{16}\\0&0&1&-\frac{1}{8}&\frac{3}{16}&0&\frac{1}{4}&-\frac{1}{16}\\0&0&0&1&0&-\frac{1}{2}&-\frac{1}{2}&0\end{array}\right) $$
$$ \small \underrightarrow{R_{4,\ 1}(-\frac{1}{8})} \left(\begin{array}{cccc|cccc}1&0&0&0&\frac{5}{16}&\frac{1}{16}&-\frac{3}{16}&-\frac{7}{16}\\0&1&0&\frac{5}{8}&\frac{1}{16}&0&-\frac{1}{4}&\frac{5}{16}\\0&0&1&-\frac{1}{8}&\frac{3}{16}&0&\frac{1}{4}&-\frac{1}{16}\\0&0&0&1&0&-\frac{1}{2}&-\frac{1}{2}&0\end{array}\right) $$
$$ \small \underrightarrow{R_{4,\ 2}(-\frac{5}{8})} \left(\begin{array}{cccc|cccc}1&0&0&0&\frac{5}{16}&\frac{1}{16}&-\frac{3}{16}&-\frac{7}{16}\\0&1&0&0&\frac{1}{16}&\frac{5}{16}&\frac{1}{16}&\frac{5}{16}\\0&0&1&-\frac{1}{8}&\frac{3}{16}&0&\frac{1}{4}&-\frac{1}{16}\\0&0&0&1&0&-\frac{1}{2}&-\frac{1}{2}&0\end{array}\right) $$
$$ \small \underrightarrow{R_{4,\ 3}(\frac{1}{8})} \left(\begin{array}{cccc|cccc}1&0&0&0&\frac{5}{16}&\frac{1}{16}&-\frac{3}{16}&-\frac{7}{16}\\0&1&0&0&\frac{1}{16}&\frac{5}{16}&\frac{1}{16}&\frac{5}{16}\\0&0&1&0&\frac{3}{16}&-\frac{1}{16}&\frac{3}{16}&-\frac{1}{16}\\0&0&0&1&0&-\frac{1}{2}&-\frac{1}{2}&0\end{array}\right) $$
소거행제형 행렬로 변환하여 영행이 없으므로, 정칙행렬이다.
이에 따른 역행렬은
$$ \small \begin{pmatrix}\frac{5}{16}&\frac{1}{16}&-\frac{3}{16}&-\frac{7}{16}\\\frac{1}{16}&\frac{5}{16}&\frac{1}{16}&\frac{5}{16}\\\frac{3}{16}&-\frac{1}{16}&\frac{3}{16}&-\frac{1}{16}\\0&-\frac{1}{2}&-\frac{1}{2}&0\end{pmatrix}$$
(6)
$$ \small (A|I_{4})= \left(\begin{array}{cccc|cccc}0&2&1&-1&1&0&0&0\\1&-3&-2&1&0&1&0&0\\3&3&1&4&0&0&1&0\\3&2&0&1&0&0&0&1\end{array}\right) $$
$$ \small \underrightarrow{R_{2,\ 1}}\left(\begin{array}{cccc|cccc}1&-3&-2&1&0&1&0&0\\0&2&1&-1&1&0&0&0\\3&3&1&4&0&0&1&0\\3&2&0&1&0&0&0&1\end{array}\right) $$
$$ \small \underrightarrow{R_{1,\ 3}(-3)}\left(\begin{array}{cccc|cccc}1&-3&-2&1&0&1&0&0\\0&2&1&-1&1&0&0&0\\0&12&7&1&0&-3&1&0\\3&2&0&1&0&0&0&1\end{array}\right) $$
$$ \small \underrightarrow{R_{1,\ 4}(-3)}\left(\begin{array}{cccc|cccc}1&-3&-2&1&0&1&0&0\\0&2&1&-1&1&0&0&0\\0&12&7&1&0&-3&1&0\\0&11&6&-2&0&-3&0&1\end{array}\right) $$
$$ \small \underrightarrow{R_{2}(\frac{1}{2})}\left(\begin{array}{cccc|cccc}1&-3&-2&1&0&1&0&0\\0&1&\frac{1}{2}&-\frac{1}{2}&\frac{1}{2}&0&0&0\\0&12&7&1&0&-3&1&0\\0&11&6&-2&0&-3&0&1\end{array}\right) $$
$$ \small \underrightarrow{R_{2,\ 1}(3)}\left(\begin{array}{cccc|cccc}1&0&-\frac{1}{2}&-\frac{1}{2}&\frac{3}{2}&1&0&0\\0&1&\frac{1}{2}&-\frac{1}{2}&\frac{1}{2}&0&0&0\\0&12&7&1&0&-3&1&0\\0&11&6&-2&0&-3&0&1\end{array}\right) $$
$$ \small \underrightarrow{R_{2,\ 3}(-12)}\left(\begin{array}{cccc|cccc}1&0&-\frac{1}{2}&-\frac{1}{2}&\frac{3}{2}&1&0&0\\0&1&\frac{1}{2}&-\frac{1}{2}&\frac{1}{2}&0&0&0\\0&0&1&7&-6&-3&1&0\\0&11&6&-2&0&-3&0&1\end{array}\right) $$
$$ \small \underrightarrow{R_{2,\ 4}(-11)}\left(\begin{array}{cccc|cccc}1&0&-\frac{1}{2}&-\frac{1}{2}&\frac{3}{2}&1&0&0\\0&1&\frac{1}{2}&-\frac{1}{2}&\frac{1}{2}&0&0&0\\0&0&1&7&-6&-3&1&0\\0&0&\frac{1}{2}&\frac{7}{2}&-\frac{11}{2}&-3&0&1\end{array}\right) $$
$$ \small \underrightarrow{R_{3,\ 1}(\frac{1}{2})}\left(\begin{array}{cccc|cccc}1&0&0&3&-\frac{3}{2}&-\frac{1}{2}&\frac{1}{2}&0\\0&1&\frac{1}{2}&-\frac{1}{2}&\frac{1}{2}&0&0&0\\0&0&1&7&-6&-3&1&0\\0&0&\frac{1}{2}&\frac{7}{2}&-\frac{11}{2}&-3&0&1\end{array}\right) $$
$$ \small \underrightarrow{R_{3,\ 2}(-\frac{1}{2})}\left(\begin{array}{cccc|cccc}1&0&0&3&-\frac{3}{2}&-\frac{1}{2}&\frac{1}{2}&0\\0&1&0&-4&\frac{7}{2}&\frac{3}{2}&-\frac{1}{2}&0\\0&0&1&7&-6&-3&1&0\\0&0&\frac{1}{2}&\frac{7}{2}&-\frac{11}{2}&-3&0&1\end{array}\right) $$
$$ \small \underrightarrow{R_{3,\ 4}(-\frac{1}{2})}\left(\begin{array}{cccc|cccc}1&0&0&3&-\frac{3}{2}&-\frac{1}{2}&\frac{1}{2}&0\\0&1&0&-4&\frac{7}{2}&\frac{3}{2}&-\frac{1}{2}&0\\0&0&1&7&-6&-3&1&0\\0&0&0&0&-\frac{5}{2}&-\frac{3}{2}&-\frac{1}{2}&1\end{array}\right) $$
네번째 행에서 영행이 존재하므로, 정칙행렬이 아니다.
2. 역행렬을 구하라.
$$ \small (1)\ \begin{pmatrix}\frac{2}{11}&-\frac{1}{11}\\\frac{5}{11}&\frac{3}{11}\end{pmatrix} \qquad (2)\ \begin{pmatrix}cos\theta&-sin\theta\\sin\theta&cos\theta \end{pmatrix}$$
$$ \small (3)\ \begin{pmatrix}a&0\\b&1\end{pmatrix} \qquad (4)\ \begin{pmatrix}a&0\\0&b\end{pmatrix} (단,\ a,\ b \ne 0)$$
(1)
$$ \small(A|I_{2})=\left( \begin{array}{cc|cc}\frac{2}{11}&-\frac{1}{11}&1&0\\\frac{5}{11}&\frac{3}{11}&0&1\end{array} \right)$$
$$ \small \underrightarrow{R_{1}(\frac{11}{2})} \left( \begin{array}{cc|cc}1&-\frac{1}{2}&\frac{11}{2}&0\\\frac{5}{11}&\frac{3}{11}&0&1\end{array} \right)$$
$$ \small \underrightarrow{R_{1,\ 2}(-\frac{5}{11})} \left( \begin{array}{cc|cc}1&-\frac{1}{2}&\frac{11}{2}&0\\0&\frac{1}{2}&-\frac{5}{2}&1\end{array} \right)$$
$$ \small \underrightarrow{R_{2,\ 1}(1)} \left( \begin{array}{cc|cc}1&0&3&1\\0&\frac{1}{2}&-\frac{5}{2}&1\end{array} \right)$$
$$ \small \underrightarrow{R_{2}(2)} \left( \begin{array}{cc|cc}1&0&3&1\\0&1&-5&2\end{array} \right)$$
$$ \small A^{-1}=\begin{pmatrix}3&1\\-5&2\end{pmatrix}$$
(2)
$$ \small (A|I_{2})=\left( \begin{array}{cc|cc} cos\theta&-sin\theta&1&0\\sin\theta&cos\theta&0&1\end{array} \right) $$
$$ \small \underrightarrow{R_{1,\ 2}(-\frac{sin\theta}{cos\theta})} \left( \begin{array}{cc|cc} cos\theta&-sin\theta&1&0\\0&\frac{sin\theta^{2}+cos\theta^{2}}{cos\theta}&-\frac{sin\theta}{cos\theta}&1 \end{array} \right) $$
$$ \small \underrightarrow{R_{2}(cos\theta)} \left( \begin{array}{cc|cc}cos\theta&-sin\theta&1&0\\0&sin\theta^{2}+cos\theta^{2}&-sin\theta&cos\theta \end{array} \right) $$
$$ \small 이때,\ cos\theta^{2}+sin\theta^{2} = 1 이므로 $$
$$ \small \left( \begin{array}{cc|cc}cos\theta&-sin\theta&1&0\\0&1&-sin\theta&cos\theta \end{array} \right) $$
$$ \small \underrightarrow{R_{2\, 1}(sin\theta)} \left( \begin{array}{cc|cc}cos\theta&0&1-sin\theta^{2}&cos\theta \cdot sin\theta\\0&1&-sin\theta&cos\theta \end{array} \right) $$
$$ \small 위에서\ sin\theta^{2}+cos\theta^{2}=1\ 이었으므로$$
$$ \small 1-sin\theta=cos\theta\ 이다.$$
$$ \small \left( \begin{array}{cc|cc}cos\theta&0&cos\theta^{2}&cos\theta \cdot sin\theta\\0&1&-sin\theta&cos\theta \end{array} \right) $$
$$ \small \underrightarrow{R_{1}(\frac{1}{cos\theta})} \left( \begin{array}{cc|cc}1&0&cos\theta&sin\theta\\0&1&-sin\theta&cos\theta \end{array} \right) $$
$$ \small A^{-1}=\begin{pmatrix}cos\theta&sin\theta\\-sin\theta&cos\theta \end{pmatrix} $$
(3)
$$ \small (A|I_{2})= \left( \begin{array}{cc|cc}a&0&1&0\\b&1&0&1\end{array} \right) $$
$$ \small \underrightarrow{R_{1}(\frac{1}{a})} \left( \begin{array}{cc|cc}1&0&\frac{1}{a}&0\\b&1&0&1 \end{array} \right) $$
$$ \small \underrightarrow{R_{1,\ 2}(-b)} \left( \begin{array}{cc|cc}1&0&\frac{1}{a}&0\\0&1&-\frac{b}{a}&1 \end{array} \right) $$
$$ \small A^{-1}=\begin{pmatrix}\frac{1}{a}&0\\-\frac{b}{a}&1 \end{pmatrix} $$
(4)
$$ \small (A|I_{2})=\left( \begin{array}{cc|cc}a&0&1&0\\0&b&0&1 \end{array} \right) $$
$$ \small \underrightarrow{R_{1}(\frac{1}{a})} \left( \begin{array}{cc|cc}1&0&\frac{1}{a}&0\\0&b&0&1\end{array} \right)$$
$$ \small \underrightarrow{R_{2}(\frac{1}{b})} \left( \begin{array}{cc|cc}1&0&\frac{1}{a}&0\\0&1&0&\frac{1}{b} \end{array} \right) $$
$$ \small A^{-1}=\begin{pmatrix}\frac{1}{a}&0\\0&\frac{1}{b}\end{pmatrix} $$
$$ \small 3.\ 행렬\ A가\ 역행렬\ A^{-1}을\ 갖도록\ 하는\ t의\ 값을\ 구하라. $$
$$ \small A=\begin{pmatrix}1&0&3\\0&0&1\\5&t&2\end{pmatrix}$$
$$ \small (A|I_{3})=\left( \begin{array}{ccc|ccc}1&0&3&1&0&0\\0&0&1&0&1&0\\5&t&2&0&0&1\end{array} \right) $$
$$ \small \underrightarrow{R_{2,\ 3}} \left( \begin{array}{ccc|ccc}1&0&3&1&0&0\\5&t&2&0&0&1\\0&0&1&0&1&0 \end{array} \right) $$
$$ \small \underrightarrow{R_{1,\ 2}(-5)} \left( \begin{array}{ccc|ccc}1&0&3&1&0&0\\0&t&-13&-5&0&1\\0&0&1&0&1&0 \end{array} \right) $$
$$ \small \underrightarrow{R_{3,\ 2}(13)} \left( \begin{array}{ccc|ccc}1&0&3&1&0&0\\0&t&0&-5&13&1\\0&0&1&0&1&0 \end{array} \right) $$
$$ \small \underrightarrow{R_{3,\ 1}(-3)} \left( \begin{array}{ccc|ccc}1&0&0&1&-3&0\\0&t&0&-5&13&1\\0&0&1&0&1&0 \end{array} \right) $$
이때, 역행렬이 존재하기 위해서는 영행이 존재하면 안되므로
t의 값은 0이 아닌 모든 실수이다.
4. 다음 정칙행렬에서 각 기본행렬의 곱으로 표현하라.
$$ \small (1)\ A=\begin{pmatrix}2&1\\0&3\end{pmatrix} \qquad (2)\ B=\begin{pmatrix}1&2\\3&4\end{pmatrix}$$
$$ \small (3)\ C=\begin{pmatrix}1&0&0\\2&3&0\\6&1&2\end{pmatrix} \qquad (4)\ D=\begin{pmatrix}1&3&2\\1&2&1\\2&4&1\end{pmatrix}$$
기본행렬에 대한 곱으로 나타내기 위해서는 단위행렬 I에
기본행연산 한 번만 사용하여 얻는 행렬 E를 A에 여러번 곱하여 나타낸다.
$$ \small 즉,\ E_{1}E_{2}E_{3}A=I\ 이고\ A=E_{1}^{-1}E_{2}^{-1}E_{3}^{-1}\ 이다.$$
(1)
$$ \small A=\begin{pmatrix}2&1\\0&3\end{pmatrix} $$
$$ \small \underrightarrow{R_{1}(\frac{1}{2})} \begin{pmatrix}1&\frac{1}{2}\\0&3\end{pmatrix}, \qquad E_{1}=\begin{pmatrix}\frac{1}{2}&0\\0&1\end{pmatrix} $$
$$ \small \underrightarrow{R_{1}(\frac{1}{3})} \begin{pmatrix}1&\frac{1}{2}\\0&1\end{pmatrix}, \qquad E_{2}=\begin{pmatrix}1&0\\0&\frac{1}{3}\end{pmatrix} $$
$$ \small \underrightarrow{R_{2,\ 1}(-\frac{1}{2})} \begin{pmatrix}1&-\frac{1}{2}\\0&1\end{pmatrix}, \qquad E_{3}=\begin{pmatrix}1&-\frac{1}{2}\\0&1\end{pmatrix} $$
위에서 구한 기본행렬연산의 역행렬을 구해주면 된다.
즉, 각 E에 대한 역행렬을 구하면 된다.
$$ \small E_{1}^{-1}=\begin{pmatrix}2&0\\0&1\end{pmatrix} $$
$$ \small E_{2}^{-1}=\begin{pmatrix}1&0\\0&3\end{pmatrix} $$
$$ \small E_{3}^{-1}=\begin{pmatrix}1&\frac{1}{2}\\0&1\end{pmatrix} $$
$$ \small A=\begin{pmatrix}2&0\\0&1\end{pmatrix}\begin{pmatrix}1&0\\0&3\end{pmatrix}\begin{pmatrix}1&\frac{1}{2}\\0&1\end{pmatrix} $$
(2)
$$ \small B=\begin{pmatrix}1&2\\3&4\end{pmatrix}$$
$$ \small \underrightarrow{R_{1,\ 2}(-3)}\begin{pmatrix}1&2\\0&-2\end{pmatrix},\ \qquad E_{1}=\begin{pmatrix}1&0\\-3&1\end{pmatrix}$$
$$ \small \underrightarrow{R_{2}(-\frac{1}{2})} \begin{pmatrix}1&2\\0&1\end{pmatrix},\ \qquad E_{2}=\begin{pmatrix}1&0\\0&-\frac{1}{2}\end{pmatrix} $$
$$ \small \underrightarrow{R_{2,\ 1}(-2)}\begin{pmatrix}1&0\\0&1\end{pmatrix},\ \qquad E_{3}=\begin{pmatrix}1&-2\\0&1\end{pmatrix} $$
$$ \small B=E_{1}^{-1}E_{2}^{-1}E_{3}^{-1} $$
$$ \small E_{1}^{-1}=\begin{pmatrix}1&0\\3&1\end{pmatrix} $$
$$ \small E_{2}^{-1}=\begin{pmatrix}1&0\\0&-2\end{pmatrix} $$
$$ \small E_{3}^{-1}=\begin{pmatrix}1&2\\0&1\end{pmatrix} $$
$$ \small B=\begin{pmatrix}1&0\\3&1\end{pmatrix} \begin{pmatrix}1&0\\0&-2\end{pmatrix} \begin{pmatrix}1&2\\0&1\end{pmatrix} $$
(3)
$$ \small C=\begin{pmatrix}1&0&0\\2&3&0\\6&1&2\end{pmatrix}$$
$$ \small \underrightarrow{R_{1,\ 2}(-2)}\begin{pmatrix}1&0&0\\0&3&0\\6&1&2\end{pmatrix},\ \qquad E_{1}=\begin{pmatrix}1&0&0\\-2&1&0\\0&0&1\end{pmatrix} $$
$$ \small \underrightarrow{R_{1,\ 3}(-6)}\begin{pmatrix}1&0&0\\0&3&0\\0&1&2\end{pmatrix},\ \qquad E_{2}=\begin{pmatrix}1&0&0\\0&1&0\\-6&0&1\end{pmatrix} $$
$$ \small \underrightarrow{R_{2}(\frac{1}{3})}\begin{pmatrix}1&0&0\\0&1&0\\0&1&2\end{pmatrix},\ \qquad E_{3}=\begin{pmatrix}1&0&0\\0&\frac{1}{3}&0\\0&0&1\end{pmatrix} $$
$$ \small \underrightarrow{R_{2,\ 3}(-1)} \begin{pmatrix}1&0&0\\0&1&0\\0&0&2\end{pmatrix},\ \qquad E_{4}=\begin{pmatrix}1&0&0\\0&1&0\\0&-1&1\end{pmatrix} $$
$$ \small \underrightarrow{R_{2}(\frac{1}{2})} \begin{pmatrix}1&0&0\\0&1&0\\0&0&1\end{pmatrix},\ \qquad E_{5}=\begin{pmatrix}1&0&0\\0&1&0\\0&0&\frac{1}{2}\end{pmatrix}$$
$$ \small C=E_{1}^{-1}E_{2}^{-1}E_{3}^{-1}E_{4}^{-1}E_{5}^{-1} $$
$$ \small E_{1}^{-1}=\begin{pmatrix}1&0&0\\2&1&0\\0&0&1\end{pmatrix}$$
$$ \small E_{2}^{-1}=\begin{pmatrix}1&0&0\\0&1&0\\6&0&1\end{pmatrix}$$
$$ \small E_{3}^{-1}=\begin{pmatrix}1&0&0\\0&3&0\\0&0&1\end{pmatrix}$$
$$ \small E_{4}^{-1}=\begin{pmatrix}1&0&0\\0&1&0\\0&1&1\end{pmatrix}$$
$$ \small E_{5}^{-1}=\begin{pmatrix}1&0&0\\0&1&0\\0&0&2\end{pmatrix}$$
$$ \small C=\begin{pmatrix}1&0&0\\2&1&0\\0&0&1\end{pmatrix}\begin{pmatrix}1&0&0\\0&1&0\\6&0&1\end{pmatrix}\begin{pmatrix}1&0&0\\0&3&0\\0&0&1\end{pmatrix}\begin{pmatrix}1&0&0\\0&1&0\\0&1&1\end{pmatrix}\begin{pmatrix}1&0&0\\0&1&0\\0&0&2\end{pmatrix}$$
(4)
$$ \small D=\begin{pmatrix}1&3&2\\1&2&1\\2&4&1\end{pmatrix} $$
$$ \small \underrightarrow{R_{1,\ 2}(-1)}\begin{pmatrix}1&3&2\\0&2&1\\2&4&1\end{pmatrix},\ \qquad E_{1}=\begin{pmatrix}1&0&0\\-1&1&0\\0&0&1\end{pmatrix}$$
$$ \small \underrightarrow{R_{1,\ 2}(-2)} \begin{pmatrix}1&3&2\\0&2&1\\0&4&1\end{pmatrix},\ \qquad E_{2}=\begin{pmatrix}1&0&0\\0&1&0\\-2&0&1\end{pmatrix}$$
$$ \small \underrightarrow{R_{2}(\frac{1}{2})}\begin{pmatrix}1&3&2\\0&1&1\\0&4&1\end{pmatrix},\ \qquad E_{3}=\begin{pmatrix}1&0&0\\0&\frac{1}{2}&0\\0&0&1\end{pmatrix}$$
$$ \small \underrightarrow{R_{2,\ 1}(-3)}\begin{pmatrix}1&0&2\\0&1&1\\0&4&1\end{pmatrix},\ \qquad E_{4}=\begin{pmatrix}1&-3&0\\0&1&0\\0&0&1\end{pmatrix}$$
$$ \small \underrightarrow{R_{2,\ 3}(-4)}\begin{pmatrix}1&0&2\\0&1&1\\0&0&1\end{pmatrix},\ \qquad E_{5}=\begin{pmatrix}1&0&0\\0&1&0\\0&-4&1\end{pmatrix}$$
$$ \small \underrightarrow{R_{3,\ 1}(-2)}\begin{pmatrix}1&0&0\\0&1&1\\0&0&1\end{pmatrix},\ \qquad E_{6}=\begin{pmatrix}1&0&-2\\0&1&0\\0&0&1\end{pmatrix}$$
$$ \small \underrightarrow{R_{3,\ 2}(-1)}\begin{pmatrix}1&0&0\\0&1&0\\0&0&1\end{pmatrix},\ \qquad E_{7}=\begin{pmatrix}1&0&0\\0&1&-1\\0&0&1\end{pmatrix}$$
$$ \small D=E_{1}^{-1}E_{2}^{-1}E_{3}^{-1}E_{4}^{-1}E_{5}^{-1}E_{6}^{-1}E_{7}^{-1} $$
$$ \small E_{1}^{-1}=\begin{pmatrix}1&0&0\\1&1&0\\0&0&1\end{pmatrix},\ \qquad E_{2}^{-1}=\begin{pmatrix}1&0&0\\0&1&0\\2&0&1\end{pmatrix}$$
$$ \small E_{3}^{-1}=\begin{pmatrix}1&0&0\\0&2&0\\0&0&1\end{pmatrix},\ \qquad E_{4}^{-1}=\begin{pmatrix}1&3&0\\0&1&0\\0&0&1\end{pmatrix}$$
$$ \small E_{5}^{-1}=\begin{pmatrix}1&0&0\\0&1&0\\0&4&1\end{pmatrix},\ \qquad E_{6}^{-1}=\begin{pmatrix}1&0&2\\0&1&0\\0&0&1\end{pmatrix}$$
$$ \small E_{7}^{-1}=\begin{pmatrix}1&0&0\\0&1&1\\0&0&1\end{pmatrix}$$
$$ \small D=\begin{pmatrix}1&0&0\\1&1&0\\0&0&1\end{pmatrix}\begin{pmatrix}1&0&0\\0&1&0\\2&0&1\end{pmatrix}\begin{pmatrix}1&0&0\\0&2&0\\0&0&1\end{pmatrix}\begin{pmatrix}1&3&0\\0&1&0\\0&0&1\end{pmatrix}\begin{pmatrix}1&0&0\\0&1&0\\0&4&1\end{pmatrix}\begin{pmatrix}1&0&2\\0&1&0\\0&0&1\end{pmatrix}\begin{pmatrix}1&0&0\\0&1&1\\0&0&1\end{pmatrix}$$
$$ \small 5.\ A^{-1}과\ B가\ 다음과\ 같을\ 때\ 행렬방정식\ AX=B를\ 풀어라.$$
$$ \small (1)\ A^{-1}=\begin{pmatrix}3&-2\\1&0\end{pmatrix},\ B=\begin{pmatrix}3\\-4\end{pmatrix}$$
$$ \small (2)\ A^{-1}=\begin{pmatrix}1&3\\5&1\end{pmatrix},\ B=\begin{pmatrix}1\\0\end{pmatrix}$$
$$ \small (3)\ A^{-1}=\begin{pmatrix}1&0&3\\-1&2&4\\0&5&1\end{pmatrix},\ B=\begin{pmatrix}5\\-3\\2\end{pmatrix}$$
$$ \small X=A^{-1}B $$
(1)
$$ \small X=\begin{pmatrix}3&-2\\1&0\end{pmatrix}\begin{pmatrix}3\\-4\end{pmatrix} $$
$$ \small X=\begin{pmatrix}17\\3\end{pmatrix}$$
(2)
$$ \small X=\begin{pmatrix}1&3\\5&1\end{pmatrix}\begin{pmatrix}1\\0\end{pmatrix} $$
$$ \small X=\begin{pmatrix}1\\5\end{pmatrix} $$
(3)
$$ \small X=\begin{pmatrix}1&0&3\\-1&2&4\\0&5&1\end{pmatrix}\begin{pmatrix}5\\-3\\2\end{pmatrix}$$
$$ \small X=\begin{pmatrix}11\\-3\\-13\end{pmatrix} $$
$$ \small 6.\ A와\ B가\ 다음과\ 같을\ 때\ AX=B를\ 풀어라$$
$$ \small A=\begin{pmatrix}7&3\\5&2\end{pmatrix},\ B=\begin{pmatrix}3\\4\end{pmatrix}$$
$$ \small A=\begin{pmatrix}3&4\\5&7\end{pmatrix},\ B=\begin{pmatrix}4\\2\end{pmatrix}$$
$$ \small A=\begin{pmatrix}1&2&1\\3&1&4\\1&-2&2\end{pmatrix},\ B=\begin{pmatrix}1\\2\\1\end{pmatrix}$$
(1)
먼저 A의 역행렬을 구하자.
$$ \small (A|I_{2})=\left( \begin{array}{cc|cc}7&3&1&0\\5&2&0&1\end{array} \right) $$
$$ \small \underrightarrow{R_{1}(\frac{1}{7})} \left( \begin{array}{cc|cc}1&\frac{3}{7}&\frac{1}{7}&0\\5&2&0&1\end{array} \right) $$
$$ \small \underrightarrow{R_{1,\ 2}(-5)} \left( \begin{array}{cc|cc}1&\frac{3}{7}&\frac{1}{7}&0\\0&-\frac{1}{7}&-\frac{5}{7}&1\end{array} \right) $$
$$ \small \underrightarrow{R_{2,\ 1}(3)} \left( \begin{array}{cc|cc}1&0&-2&3\\0&-\frac{1}{7}&-\frac{5}{7}&1\end{array} \right) $$
$$ \small \underrightarrow{R_{2}(-7)} \left( \begin{array}{cc|cc}1&0&-2&3\\0&1&5&-7\end{array} \right) $$
$$ \small X=A^{-1}B이므로, $$
$$ \small X=\begin{pmatrix}-2&3\\5&-7\end{pmatrix}\begin{pmatrix}3\\4\end{pmatrix} $$
$$ \small X=\begin{pmatrix}6\\-13\end{pmatrix} $$
(2)
$$ \small (A|I_{2})=\left( \begin{array}{cc|cc}3&4&1&0\\5&7&0&1\end{array} \right) $$
$$ \small \underrightarrow{R_{1}(\frac{1}{3})} \left( \begin{array}{cc|cc}1&\frac{4}{3}&\frac{1}{3}&0\\5&7&0&1\end{array} \right) $$
$$ \small \underrightarrow{R_{1,\ 2}(-5)} \left( \begin{array}{cc|cc}1&\frac{4}{3}&\frac{1}{3}&0\\0&\frac{1}{3}&-\frac{5}{3}&1\end{array} \right) $$
$$ \small \underrightarrow{R_{2,\ 1}(-4)} \left( \begin{array}{cc|cc}1&0&7&-4\\0&\frac{1}{3}&-\frac{5}{3}&1\end{array} \right) $$
$$ \small \underrightarrow{R_{2}(3)} \left( \begin{array}{cc|cc}1&0&7&-4\\0&1&-5&3\end{array} \right) $$
$$ \small X=A^{-1}B이므로,$$
$$ \small X=\begin{pmatrix}7&-4\\-5&3\end{pmatrix}\begin{pmatrix}4\\2\end{pmatrix}$$
$$ \small X=\begin{pmatrix}20\\-16\end{pmatrix}$$
(3)
$$ \small (A|I_{3})=\left( \begin{array}{ccc|ccc}1&2&1&1&0&0\\3&1&4&0&1&0\\1&-2&2&0&0&1\end{array} \right) $$
$$ \small \underrightarrow{R_{1, 2}(-3)} \left( \begin{array}{ccc|ccc}1&2&1&1&0&0\\0&-5&1&-3&1&0\\1&-2&2&0&0&1\end{array} \right) $$
$$ \small \underrightarrow{R_{1,\ 3}(-1)}\left( \begin{array}{ccc|ccc}1&2&1&1&0&0\\0&-5&1&-3&1&0\\0&-4&1&-1&0&1\end{array} \right) $$
$$ \small \underrightarrow{R_{2}(-\frac{1}{5})} \left( \begin{array}{ccc|ccc}1&2&1&1&0&0\\0&1&-\frac{1}{5}&\frac{3}{5}&-\frac{1}{5}&0\\0&-4&1&-1&0&1\end{array} \right) $$
$$ \small \underrightarrow{R_{2,\ 3}(4)} \left( \begin{array}{ccc|ccc}1&2&1&1&0&0\\0&1&-\frac{1}{5}&\frac{3}{5}&-\frac{1}{5}&0\\0&0&\frac{1}{5}&\frac{7}{5}&-\frac{4}{5}&1\end{array} \right) $$
$$ \small \underrightarrow{R_{2,\ 1}(-2)} \left( \begin{array}{ccc|ccc}1&0&\frac{7}{5}&-\frac{1}{5}&\frac{2}{5}&0\\0&1&-\frac{1}{5}&\frac{3}{5}&-\frac{1}{5}&0\\0&0&\frac{1}{5}&\frac{7}{5}&-\frac{4}{5}&1\end{array} \right) $$
$$ \small \underrightarrow{R_{3,\ 1}(-7)} \left( \begin{array}{ccc|ccc}1&0&0&-10&6&-7\\0&1&-\frac{1}{5}&\frac{3}{5}&-\frac{1}{5}&0\\0&0&\frac{1}{5}&\frac{7}{5}&-\frac{4}{5}&1\end{array} \right) $$
$$ \small \underrightarrow{R_{3,\ 2}(1)} \left( \begin{array}{ccc|ccc}1&0&0&-10&6&-7\\0&1&0&2&-1&1\\0&0&\frac{1}{5}&\frac{7}{5}&-\frac{4}{5}&1\end{array} \right) $$
$$ \small \underrightarrow{R_{3}(5)} \left( \begin{array}{ccc|ccc}1&0&0&-10&6&-7\\0&1&0&2&-1&1\\0&0&1&7&-4&5\end{array} \right) $$
$$ \small X=A^{-1}B이므로,$$
$$ \small X=\begin{pmatrix}-10&6&-7\\2&-1&1\\7&-4&5\\\end{pmatrix}\begin{pmatrix}5\\-3\\2\end{pmatrix}$$
$$ \small X=\begin{pmatrix}-82\\15\\57\end{pmatrix}$$
$$ \small 7.\ [예제\ 4.8]과\ 관련된\ 문제이다.\ 즉,\ 행렬방정식\ AX=B에\ 대해$$
$$ \small 확대행렬\ (A|B)를\ 소거행제형\ 행렬\ (C|D)로\ 변환한\ 경우 $$
$$ \small 미지수행렬\ X와\ (C|D)가\ 각각\ 다음과\ 같을\ 때$$
$$ \small 방정식의\ 해는\ 어떤\ 형태로\ 표시할\ 수\ 있는가?$$
$$ \small (1)\ X=\begin{pmatrix}x\\y\\z\\w\end{pmatrix},\ (C|D)=\Biggl(\begin{array}{cccc|c}1&2&0&-2&4\\0&0&1&3&2\\0&0&0&0&1\end{array}\Biggr)$$
$$ \small(2)\ X=\begin{pmatrix}a\\b\\c\\d\end{pmatrix},\ (C|D)=\Biggl( \begin{array}{cccc|c}1&0&0&1&2\\0&1&0&2&1\\0&0&1&1&3\end{array} \Biggr)$$
$$ \small (3)\ X=\begin{pmatrix}p\\g\\r\end{pmatrix},\ (C|D)=\Biggl( \begin{array}{ccc|c}1&0&-2&4\\0&1&-1&2\\0&0&0&0\end{array} \Biggr)$$
(1)
(C|D)의 행 중에서 세번째 행에 0 형태의 행이 있으므로
해를 갖지 못한다.
(2)
방정식의 수가 미지수의 수보다 적으므로, 무수히 많은 해를 가진다.
(3)
미지수 r이 자유변수가 되어 무수히 많은 해를 가지며,
(2t+4, t+2, t)
8. 다음 동차연립방정식이 자명하지 않은 해를 갖는지 확인하라.
$$ \small (1)\ \begin{cases}\begin{array}{r@{}r@{}r@{}r@{}}x&{}+{}&3y&{}+{}&2z&=0\\x&{}+{}&2y&{}-{}&2z&=0\\4x&{}+{}&y&{}+{}&7z&=0\end{array}\end{cases}$$
$$ \small (2)\ \begin{cases}\begin{array}{r@{}r@{}r@{}r@{}}2x&{}-{}&y&{}+{}&z&=0\\x&{}+{}&2y&{}-{}&3z&=0\\6x&{}-{}&3y&{}+{}&3z&=0\end{array}\end{cases}$$
$$ \small (3)\ \begin{cases}\begin{array}{r@{}r@{}r@{}r@{}}x&{}+{}&3y&{}+{}&3z&=0\\3x&{}+{}&y&{}+{}&z&=0\\-2x&{}-{}&y&{}-{}&z&=0\end{array}\end{cases}$$
$$ \small (4)\ \begin{cases}\begin{array}{r@{}r@{}r@{}r@{}}4x&{}+{}&7y&{}-{}&2z&=0\\-3x&{}+{}&5y&{}-{}&2z&=0\\5x&{}-{}&4y&{}+{}&6z&=0\end{array}\end{cases}$$
$$ \small 자명하지\ 않은\ 해를\ 가지기\ 위해서는\ I_{3}과\ 행상등하지\ 않아야\ 한다.$$
(1)
$$ \small A=\begin{pmatrix}1&3&2\\1&2&-2\\4&1&7\end{pmatrix}$$
$$ \small \underrightarrow{R_{1,\ 2}(-1)}\begin{pmatrix}1&3&2\\0&-1&-4\\4&1&7\end{pmatrix}$$
$$ \small \underrightarrow{R_{1,\ 3}(-4)}\begin{pmatrix}1&3&2\\0&-1&-4\\0&-11&-1\end{pmatrix}$$
$$ \small \underrightarrow{R_{2}(-1)}\begin{pmatrix}1&3&2\\0&1&4\\0&-11&-1\end{pmatrix}$$
$$ \small \underrightarrow{R_{2,\ 1}(-3)}\begin{pmatrix}1&0&-10\\0&1&4\\0&-11&-1\end{pmatrix}$$
$$ \small \underrightarrow{R_{2,\ 3}(11)}\begin{pmatrix}1&0&-10\\0&1&4\\0&0&43\end{pmatrix}$$
$$ \small \underrightarrow{R_{3}(\frac{1}{43})}\begin{pmatrix}1&0&-10\\0&1&4\\0&0&1\end{pmatrix}$$
$$ \small \underrightarrow{R_{3,\ 1}(10)}\begin{pmatrix}1&0&0\\0&1&4\\0&0&1\end{pmatrix}$$
$$ \small \underrightarrow{R_{3,\ 2}(-4)}\begin{pmatrix}1&0&0\\0&1&0\\0&0&1\end{pmatrix}$$
$$ \small I_{3}와\ 행상등\ 하므로, 자명하는\ 해를\ 갖는다.$$
(2)
$$ \small A=\begin{pmatrix}2&-1&1\\1&2&-3\\6&-3&3\end{pmatrix}$$
$$ \small \underrightarrow{R_{2, 1}}\begin{pmatrix}1&2&-3\\2&-1&1\\6&-3&3\end{pmatrix}$$
$$ \small \underrightarrow{R_{1,\ 2}(-2)}\begin{pmatrix}1&2&3\\0&-5&7\\6&-3&3\end{pmatrix}$$
$$ \small \underrightarrow{R_{1,\ 3}(-6)}\begin{pmatrix}1&2&3\\0&-5&7\\0&-15&-15\end{pmatrix}$$
$$ \small \underrightarrow{R_{2}(-\frac{1}{5})}\begin{pmatrix}1&2&3\\0&1&-\frac{7}{5}\\0&-15&-15\end{pmatrix}$$
$$ \small \underrightarrow{R_{2,\ 1}(-2)}\begin{pmatrix}1&0&\frac{29}{5}\\0&1&-\frac{7}{5}\\0&-15&-15\end{pmatrix}$$
$$ \small \underrightarrow{R_{2,\ 3}(15)}\begin{pmatrix}1&0&\frac{29}{5}\\0&1&-\frac{7}{5}\\0&0&-36\end{pmatrix}$$
$$ \small \underrightarrow{R_{3}(-\frac{1}{36})}\begin{pmatrix}1&0&\frac{29}{5}\\0&1&-\frac{7}{5}\\0&0&1\end{pmatrix}$$
$$ \small \underrightarrow{R_{3,\ 1}(-\frac{29}{5})}\begin{pmatrix}1&0&0\\0&1&-\frac{7}{5}\\0&0&1\end{pmatrix}$$
$$ \small \underrightarrow{R_{3,\ 2}(\frac{7}{5})}\begin{pmatrix}1&0&0\\0&1&0\\0&0&1\end{pmatrix}$$
$$ \small I_{3}와\ 행상등\ 하므로, 자명하는\ 해를\ 갖는다.$$
(3)
$$ \small A=\begin{pmatrix}1&3&3\\3&1&1\\-2&-1&-1\end{pmatrix}$$
$$ \small \underrightarrow{R_{1,\ 2}(-3)}\begin{pmatrix}1&3&3\\0&-8&-8\\-2&-1&-1\end{pmatrix}$$
$$ \small \underrightarrow{R_{1,\ 3}(2)}\begin{pmatrix}1&3&3\\0&-8&-8\\0&5&5\end{pmatrix}$$
$$ \small \underrightarrow{R_{2}(-\frac{1}{8})}\begin{pmatrix}1&3&3\\0&1&1\\0&5&5\end{pmatrix}$$
$$ \small \underrightarrow{R_{2,\ 1}(-3)}\begin{pmatrix}1&0&0\\0&1&1\\0&5&5\end{pmatrix}$$
$$ \small \underrightarrow{R_{2,\ 3}(-5)}\begin{pmatrix}1&0&0\\0&1&1\\0&0&0\end{pmatrix}$$
$$ \small I_{3}와\ 행상등\ 하지\ 않으므로,\ 자명하지\ 않는\ 해를\ 갖는다.$$
(4)
$$ \small A=\begin{pmatrix}4&7&-2\\-3&5&-2\\5&-4&6\end{pmatrix}$$
$$ \small \underrightarrow{R_{1}(\frac{1}{4})}\begin{pmatrix}1&\frac{7}{4}&-\frac{1}{2}\\-3&5&-2\\5&-4&6\end{pmatrix}$$
$$ \small \underrightarrow{R_{1,\ 2}(3)}\begin{pmatrix}1&\frac{7}{4}&-\frac{1}{2}\\0&\frac{41}{4}&-\frac{7}{2}\\5&-4&6\end{pmatrix}$$
$$ \small \underrightarrow{R_{1,\ 3}(-5)}\begin{pmatrix}1&\frac{7}{4}&-\frac{1}{2}\\0&\frac{41}{4}&-\frac{7}{2}\\0&-\frac{51}{4}&\frac{17}{2}\end{pmatrix}$$
$$ \small \underrightarrow{R_{2}(\frac{4}{41})}\begin{pmatrix}1&\frac{7}{4}&-\frac{1}{2}\\0&1&\frac{14}{41}\\0&-\frac{51}{4}&\frac{17}{2}\end{pmatrix}$$
$$ \small \underrightarrow{R_{2,\ 1}(-\frac{7}{4})}\begin{pmatrix}1&0&-\frac{57}{82}\\0&1&\frac{14}{41}\\0&-\frac{51}{4}&\frac{17}{2}\end{pmatrix}$$
$$ \small \underrightarrow{R_{2,\ 3}(\frac{51}{4})}\begin{pmatrix}1&0&-\frac{57}{82}\\0&1&\frac{14}{41}\\0&0&\frac{1411}{82}\end{pmatrix}$$
$$ \small \underrightarrow{R_{3}(\frac{82}{1411})}\begin{pmatrix}1&0&-\frac{57}{82}\\0&1&\frac{14}{41}\\0&0&1\end{pmatrix}$$
$$ \small \underrightarrow{R_{3,\ 1}(\frac{57}{82})}\begin{pmatrix}1&0&0\\0&1&\frac{14}{41}\\0&0&1\end{pmatrix}$$
$$ \small \underrightarrow{R_{3,\ 2}(-\frac{14}{41})}\begin{pmatrix}1&0&0\\0&1&0\\0&0&1\end{pmatrix}$$
$$ \small I_{3}와\ 행상등\ 하므로, 자명하는\ 해를\ 갖는다.$$
9. 다음 동차연립방정식이 자명하지 않은 해를 갖도록 t의 값을 정하라.
$$ \small \begin{cases}\begin{array}{r@{}r@{}r@{}r@{}}x&{}+{}&2y&{}-{}&2z&=0\\x&{}+{}&ty&{}+{}&3z&=0\\2x&{}+{}&3y&{}+{}&3x&=0\end{array}\end{cases}$$
$$ \small A=\begin{pmatrix}1&2&-2\\1&t&3\\2&3&3\end{pmatrix}$$
$$ \small \underrightarrow{R_{1,\ 2}(-1)}\begin{pmatrix}1&2&-2\\0&t-2&5\\2&3&3\end{pmatrix}$$
$$ \small \underrightarrow{R_{1,\ 3}(-2)}\begin{pmatrix}1&2&-2\\0&t-2&5\\0&-1&7\end{pmatrix}$$
$$ \small \underrightarrow{R_{2,\ 3}}\begin{pmatrix}1&2&-2\\0&-1&7\\0&t-2&5\end{pmatrix}$$
$$ \small \underrightarrow{R_{2}(-1)}\begin{pmatrix}1&2&-2\\0&1&-7\\0&t-2&5\end{pmatrix}$$
$$ \small \underrightarrow{R_{2,\ 1}(-2)}\begin{pmatrix}1&0&12\\0&1&-7\\0&t-2&5\end{pmatrix}$$
$$ \small \underrightarrow{R_{2,\ 3}(-t+2)}\begin{pmatrix}1&0&12\\0&1&-7\\0&0&7t-9\end{pmatrix}$$
위의 결과값에서 행상등하지 않기 위해서는
$$ \small 7t-9=0,\ 즉\ t=\frac{9}{7}이면\ 된다.$$