Tính ma trận nghịch đảo

Cho

$A = \left[ {\begin{array}{*{20}{c}} 1 & 2 & 3 \\ 2 & 5 & 3 \\ 1 & 0 & 8 \\ \end{array}} \right]$

Ta có:

$\det (A) = - 1 \ne 0$

${C_{11}} = 40,{C_{12}} = - 13,{C_{13}} = - 5$

${C_{21}} = - 16,{C_{22}} = 5,{C_{23}} = 2$

${C_{31}} = - 9,{C_{32}} = 3,{C_{33}} = 1$

Do đó:

$C = \left[ {\begin{array}{*{20}{c}} {40} & { - 13} & { - 5} \\ { - 16} & 5 & 2 \\ { - 9} & 3 & 1 \\ \end{array}} \right],{C^t} = \left[ {\begin{array}{*{20}{c}} {40} & { - 16} & { - 9} \\ { - 13} & 5 & 3 \\ { - 5} & 2 & 1 \\ \end{array}} \right]$

Vậy :

${A^{ - 1}} = \frac{1}{{\det A}}{C^t} = \frac{1}{{ - 1}}\left[ {\begin{array}{*{20}{c}} {40} & { - 16} & { - 9} \\ { - 13} & 5 & 3 \\ { - 5} & 2 & 1 \\ \end{array}} \right]$