Question

matriz inversa 3x3
l a b c
l d e f
l g h i
_______________________________
1 2 3 l
3 4 5 l
3 2 1 l

Answer 1

Hallemos la matriz de cofactores

$C_{11}=(-1)^{1+1}(1)\left|\begin{matrix}4&5\\2&1\end{matrix}\right|=-6\\ \\ C_{12}=(-1)^{1+2}(2)\left|\begin{matrix}3&5\\3&1\end{matrix}\right|=24\\ \\ C_{13}=(-1)^{1+3}(3)\left|\begin{matrix}3&4\\3&2\end{matrix}\right|=-18\\ \\ C_{21}=(-1)^{2+1}(3)\left|\begin{matrix}2&3\\2&1\end{matrix}\right|=12\\ \\ C_{22}=(-1)^{2+2}(4)\left|\begin{matrix}1&3\\3&1\end{matrix}\right|=-32\\ \\ C_{23}=(-1)^{2+3}(5)\left|\begin{matrix}1&2\\3&2\end{matrix}\right|=20$

$C_{31}=(-1)^{3+1}(3)\left|\begin{matrix}2&3\\4&5\end{matrix}\right|=-6\\ \\ C_{32}=(-1)^{3+2}(2)\left|\begin{matrix}1&3\\3&5\end{matrix}\right|=8\\ \\ C_{33}=(-1)^{3+3}(1)\left|\begin{matrix}1&2\\3&4\end{matrix}\right|=-2$

Entonces la matriz de cofactores es
$C=\left[\begin{matrix} -6&24&-18\\ 12&-32&20\\ -6&8&-2 \end{matrix}\right]$

Hallemos el determinante de la matriz dada

$\left|\begin{matrix} 1&2&3\\ 3&4&5\\ 3&2&1 \end{matrix}\right| = -6-2(-12)+3(-6)=0$

NO TIENE INVERSA