Đáp án:
$A^{-1} = \left(\begin{array}{c}-16&-11&3\\\dfrac72&\dfrac52&-\dfrac12\\-\dfrac52&\dfrac32&\dfrac12 \end{array}\right)$
Giải thích các bước giải:
Ta có:
$\quad (A|I)= \left(\begin{array}{ccc|ccc}1&2&-4&1&0&0\\-1&-1&5&0&1&0\\ 2&7&-3&0&0&1 \end{array} \right)$
$\xrightarrow{\begin{array}{l} r_2 + r_1 \to r_2\\r_3 - 2r_2 \to r_3\end{array}}\left(\begin{array}{ccc|ccc}1&2&-4&1&0&0\\0&1&1&1&1&0\\ 0&3&5&-2&0&1 \end{array}\right)$
$\xrightarrow{r_3 - 3r_2 \to r_3} \left(\begin{array}{ccc|ccc}1&2&-4&1&0&0\\0&1&1&1&1&0\\ 0&0&2&-5&3&1 \end{array}\right)$
$\xrightarrow{\ \tfrac12r_3 \to r_3\ } \left(\begin{array}{ccc|ccc}1&2&-4&1&0&0\\0&1&1&1&1&0\\ 0&0&1&-\dfrac52&\dfrac32&\dfrac12 \end{array}\right)$
$\xrightarrow{\begin{array}{l}r_2- r_3 \to r_2\\ r_1 +4 r_3\to r_1\end{array} } \left(\begin{array}{ccc|ccc}1&2&0&-9&-6&2\\0&1&0&\dfrac72&\dfrac52&-\dfrac12\\ 0&0&1&-\dfrac52&\dfrac32&\dfrac12 \end{array}\right)$
$\xrightarrow{r_1- 2r_2\to r_1} \left(\begin{array}{ccc|ccc}1&0&0&-16&-11&3\\0&1&0&\dfrac72&\dfrac52&-\dfrac12\\ 0&0&1&-\dfrac52&\dfrac32&\dfrac12 \end{array}\right) = (I|A^{-1})$
Vậy $A^{-1} = \left(\begin{array}{c}-16&-11&3\\\dfrac72&\dfrac52&-\dfrac12\\-\dfrac52&\dfrac32&\dfrac12 \end{array}\right)$
