1. 크래머 공식을 이용하여 다음 방정식을 풀어라.
$$ \small (1).\ \left\{ \begin{array} {c{}c{}c{}c{}} 2x_{1}&{}+{}&x_{2}&{}+{}&x_{3}&=&{}{}&5\\x_{1}&{}{}&&{}+{}&x_{3}&=&{}{}&7\\x_{1}&{}-{}&2x_{2}&{}{}&&=&{}{}&15 \end{array} \right. $$
$$ \small (2).\ \left\{ \begin{array}{c{}c{}c{}c{}} x_{1}&{}+{}&x_{2}&{}+{}&4x_{3}&=&{}{}&3 \\x_{1}&{}+{}&2x_{2}&{}-{}&x_{3}&=&{}{}&6\\2x_{1}&{}+{}&2x_{2}&{}+{}&x_{3}&=&{}{}&9 \end{array} \right. $$
$$ \small (3).\ \left\{ \begin{array}{c{}c{}c{}c{}} x_{1}&{}+{}&x_{2}&{}+{}&2x_{3}&=&{}{}&2\\5x_{1}&{}+{}&3x_{2}&{}+{}&6x_{3}&=&{}-{}&2\\2x_{1}&{}+{}&2x_{2}&{}+{}&3x_{3}&=&{}{}&3 \end{array} \right. $$
(1).
$$ \small \begin{vmatrix}2&1&1\\1&0&1\\1&-2&0\end{vmatrix}\ =\ 3 $$
$$ \small x_{1}\ =\ \frac{1}{3}\begin{vmatrix}5&1&1\\7&0&1\\15&-2&0\end{vmatrix}\ =\ \frac{11}{3} $$
$$ \small x_{2}\ =\ \frac{1}{3}\begin{vmatrix}2&5&1\\1&7&1\\1&15&0\end{vmatrix}\ =\ -\frac{17}{3}$$
$$ \small x_{3}\ =\ \frac{1}{3}\begin{vmatrix}2&1&5\\1&0&7\\1&-2&15\end{vmatrix}\ =\ \frac{10}{3}$$
(2).
$$ \small \begin{vmatrix}1&1&4\\1&2&-1\\2&2&1\end{vmatrix}\ =\ -7 $$
$$ \small x_{1}\ =\ -\frac{1}{7}\begin{vmatrix}3&1&4\\6&2&-1\\9&2&1\end{vmatrix}\ =\ \frac{27}{7}$$
$$ \small x_{2}\ =\ -\frac{1}{7}\begin{vmatrix}1&3&4\\1&6&-1\\2&9&1\end{vmatrix}\ =\ \frac{6}{7}$$
$$ \small x_{3}\ =\ -\frac{1}{7}\begin{vmatrix}1&1&3\\1&2&6\\2&2&9\end{vmatrix}\ =\ -\frac{3}{7}$$
(3).
$$ \small \begin{vmatrix}1&1&2\\5&3&6\\2&2&3\end{vmatrix}\ =\ 2$$
$$ \small x_{1}\ =\ \frac{1}{2}\begin{vmatrix}2&1&2\\-2&3&6\\3&2&3\end{vmatrix}\ =\ -4$$
$$ \small x_{2}\ =\ \frac{1}{2}\begin{vmatrix}1&2&2\\5&-2&6\\2&3&3\end{vmatrix}\ =\ 4$$
$$ \small x_{3}\ =\ \frac{1}{2}\begin{vmatrix}1&1&2\\5&3&-2\\2&2&3\end{vmatrix}\ =\ 1$$
2. 다음 행렬에 대해 역행렬이 있으면 행렬식을 통해 역행렬을 구하라.
$$ \small (1).\ A=\begin{pmatrix}7&-2\\3&1\end{pmatrix} \qquad B=\begin{pmatrix}7&-6\\-6&5\end{pmatrix} $$
$$ \small (2).\ A=\begin{pmatrix}1&1&2\\3&2&6\\2&7&5\end{pmatrix} \qquad B=\begin{pmatrix}1&2&3\\1&1&2\\1&1&1\end{pmatrix} $$
$$ \small A^{-1}\ =\ \frac{adjA}{|A|} $$
(1).
$$ \small |A|\ =\ ad-bc\ =\ (7\cdot 1)-(-2 \cdot 3)\ =\ 7-(-6)\ =\ 13$$
$$ \small A^{-1}\ =\ \begin{pmatrix}\frac{1}{13}&\frac{2}{13}\\-\frac{3}{13}&\frac{7}{13}\end{pmatrix}$$
$$ \small |B|\ =\ ad-bc\ =\ (7 \cdot 5)-(-6 \cdot -6)\ =\ 35-36\ =\ -1$$
$$ \small B^{-1}\ =\ \begin{pmatrix}-5&-6\\-6&-7\end{pmatrix} $$
(2).
$$ \small |A|\ =\ \begin{vmatrix}1&1&2\\3&2&6\\2&7&6\end{vmatrix}\ =\ -1 $$
$$ \small adjA=\begin{pmatrix} +\begin{vmatrix}2&6\\7&5\end{vmatrix} & -\begin{vmatrix}3&6\\2&5\end{vmatrix} & +\begin{vmatrix}3&2\\2&7\end{vmatrix} \\ -\begin{vmatrix}1&2\\7&5\end{vmatrix} & +\begin{vmatrix}1&2\\2&5\end{vmatrix} & -\begin{vmatrix}1&1\\2&7\end{vmatrix} \\ +\begin{vmatrix}1&2\\2&6\end{vmatrix} & -\begin{vmatrix}1&2\\3&6\end{vmatrix} & +\begin{vmatrix}1&1\\3&2\end{vmatrix} \end{pmatrix}^{T}$$
$$ \small adjA=\begin{pmatrix}-32&-3&17\\9&1&-5\\2&0&-1\end{pmatrix}^{T}=\begin{pmatrix}-32&9&2\\-3&1&0\\17&-5&-1\end{pmatrix} $$
$$ \small A^{-1}=\frac{adjA}{|A|}=\begin{pmatrix}32&-9&-2\\3&-1&0\\-17&5&1\end{pmatrix}$$
$$ \small |B|\ =\ \begin{vmatrix}1&2&3\\1&1&2\\1&1&1\end{vmatrix}\ =\ 1$$
$$ \small adjB=\begin{pmatrix} +\begin{vmatrix}1&2\\1&1\end{vmatrix} & -\begin{vmatrix}1&2\\1&1\end{vmatrix} & +\begin{vmatrix}1&1\\1&1\end{vmatrix} \\ -\begin{vmatrix}2&3\\1&1\end{vmatrix} & +\begin{vmatrix}1&3\\1&1\end{vmatrix} & -\begin{vmatrix}1&2\\1&1\end{vmatrix} \\ +\begin{vmatrix}2&3\\1&2\end{vmatrix} & -\begin{vmatrix}1&3\\1&2\end{vmatrix} & +\begin{vmatrix}1&2\\1&1\end{vmatrix} \end{pmatrix}^{T}$$
$$ \small adjB=\begin{pmatrix}-1&1&0\\1&-2&1\\1&1&-1\end{pmatrix}^{T}=\begin{pmatrix}-1&1&1\\1&-2&1\\0&1&-1\end{pmatrix}$$
$$ \small B^{-1}=\frac{adjB}{|B|}=\begin{pmatrix}-1&1&1\\1&-2&1\\0&1&-1\end{pmatrix}$$
3. 다음 연립 방정식이 유일한 해를 가지는 실수 a의 조건과 해를 구하라.
$$ \small \left\{ \begin{array}{c{}c{}c{}c{}}x_{1}&{}+{}&x_{2}&{}-{}&x_{3}&=&{}{}&3\\x_{1}&{}-{}&2x_{2}&{}-{}&3x_{3}&=&{}-{}&7\\2x_{1}&{}+{}&2x_{2}&{}+{}&ax_{3}&=&{}{}&4 \end{array} \right. $$
방정식이 유일한 해를 가질 조건으로는 행렬식의 값이 0이 아니어야 한다.
$$ \small |A| = \begin{vmatrix}1&1&-1\\1&-2&-3\\2&2&a\end{vmatrix}\ =\ -3a-6$$
$$ \small |A| = a \ne -2$$
$$ \small x_{1}\ =\ \frac{1}{-3a-6}\begin{vmatrix}3&1&-1\\-7&-2&-3\\4&2&a\end{vmatrix}= -\frac{a+12}{3(a+2)} $$
$$ \small x_{2}\ =\ \frac{1}{-3a-6}\begin{vmatrix}1&3&-1\\1&-7&-3\\2&4&a\end{vmatrix}= \frac{-10a-24}{-3a-6} = \frac{2(5a+12)}{3(a+2)}$$
$$ \small x_{3}\ =\ \frac{1}{-3a-6}\begin{vmatrix}1&1&3\\1&-2&-7\\2&2&4\end{vmatrix}=-\frac{6}{3(a+2)} = -\frac{2}{a+2}$$
4. 다음 연립방정식이 유일한 해를 가지는지 판단하고 가지면 역행렬을 이용해 구하라.
$$ \small (1).\ \left\{ \begin{array}{c{}c{}c{}c{}}x_{1}&{}-{}&3x_{2}&{}-{}&2x_{3}&=&{}{}&0\\2x_{1}&{}-{}&2x_{2}&{}-{}&4x_{3}&=&{}{}&0\\3x_{1}&{}+{}&5x_{2}&{}-{}&x_{3}&=&{}{}&0 \end{array} \right. $$
$$ \small (2).\ \left\{ \begin{array}{c{}c{}c{}c{}}4x_{1}&{}+{}&5x_{2}&{}+{}&8x_{3}&=&{}{}&10\\2x_{1}&{}+{}&3x_{2}&{}+{}&4x_{3}&=&{}{}&6\\6x_{1}&{}+{}&2x_{2}&{}+{}&7x_{3}&=&{}{}&4 \end{array} \right. $$
(1).
$$ \small \begin{vmatrix}1&-3&-2\\2&-2&-4\\3&5&-1\end{vmatrix} = 20 $$
$$ \small AX=B이고,\ X=A^{-1}B이다.\ 이때,\ 상수항의\ 값이\ 0\ 이므로\ 자명한\ 해를 갖는다.$$
(2).
$$ \small \begin{vmatrix}4&5&8\\2&3&4\\6&2&7\end{vmatrix} = -10 $$
$$ \small adjA=\begin{pmatrix} +\begin{vmatrix}3&4\\2&7\end{vmatrix} & -\begin{vmatrix}2&4\\6&7\end{vmatrix} & +\begin{vmatrix}2&3\\6&2\end{vmatrix} \\ -\begin{vmatrix}5&8\\2&7\end{vmatrix} & +\begin{vmatrix}4&8\\6&7\end{vmatrix} & -\begin{vmatrix}4&5\\6&2\end{vmatrix} \\ +\begin{vmatrix}5&8\\3&4\end{vmatrix} & -\begin{vmatrix}4&8\\2&4\end{vmatrix} & +\begin{vmatrix}4&5\\2&3\end{vmatrix} \end{pmatrix}^{T}$$
$$ \small adjA=\begin{pmatrix}13&10&-14\\-19&-20&22\\-4&0&2\end{pmatrix}^{T}=\begin{pmatrix}13&-19&-4\\10&-20&0\\-14&22&2\end{pmatrix} $$
$$ \small A^{-1}=-\frac{1}{10}\begin{pmatrix}13&-19&-4\\10&-20&0\\-14&22&2\end{pmatrix} $$
$$ \small X=A^{-1}B =-\frac{1}{10}\begin{pmatrix}13&-19&-4\\10&-20&0\\-14&22&2\end{pmatrix} \cdot \begin{pmatrix}10\\6\\4\end{pmatrix} $$
$$ \small X=-\frac{1}{10}\begin{pmatrix}0\\-20\\0\end{pmatrix} = \begin{pmatrix}0\\2\\0\end{pmatrix}$$