Question

Решить систему матричным методом. решение распишите подробнее. хочется разобраться 3x+4y+z=-1 3x+5y+3z=-1 6x+8y+z=-3

Answer 1

$AX=B\\X=A^{-1}B\\A^{-1}=\frac{1}{\mathcal4}A^T\\\mathcal4= \left[\begin{array}{ccc}3&4&1\\3&5&3\\6&8&1\end{array}\right] =15+72+24-30-12-72=-3\\A_{11}= (-1)^2 \left[\begin{array}{cc}5&3\\8&1\end{array}\right]=5-24=-19\\A_{12}= (-1)^3 \left[\begin{array}{cc}3&3\\6&1\end{array}\right]=-(3-18)=15\\A_{13}= (-1)^4 \left[\begin{array}{cc}3&5\\6&8\end{array}\right]=24-30=-6\\A_{21}= (-1)^3 \left[\begin{array}{cc}4&1\\8&1\end{array}\right]=-(4-8)=4\\$
$A_{22}= (-1)^4 \left[\begin{array}{cc}3&1\\6&1\end{array}\right]=3-6=-3\\A_{23}= (-1)^5 \left[\begin{array}{cc}3&4\\6&8\end{array}\right]=-(24-24)=0\\A_{31}= (-1)^4 \left[\begin{array}{cc}4&1\\5&3\end{array}\right]=12-5=7\\A_{32}= (-1)^5 \left[\begin{array}{cc}3&1\\3&3\end{array}\right]=-(9-3)=-6\\A_{33}= (-1)^6 \left[\begin{array}{cc}3&4\\3&5\end{array}\right]=15-12=3\\A^T= \left[\begin{array}{ccc}-19&4&7\\15&-3&-6\\-6&0&3\end{array}\right]$
$A^{-1}=-\frac{1}{3} \left[\begin{array}{ccc}-19&4&7\\15&-3&-6\\-6&0&3\end{array}\right] \\AA^{-1}=E\\-\frac{1}{3}\left[\begin{array}{ccc}3&4&1\\3&5&3\\6&8&1\end{array}\right]\left[\begin{array}{ccc}-19&4&7\\15&-3&-6\\-6&0&3\end{array}\right]= -\frac{1}{3} \left[\begin{array}{ccc}-3&0&0\\0&-3&0\\0&0&-3\end{array}\right] =\\=\left[\begin{array}{ccc}1&0&0\\0&1&0\\0&0&1\end{array}\right]$
$X= -\frac{1}{3} \left[\begin{array}{ccc}-19&4&7\\15&-3&-6\\-6&0&3\end{array}\right]\left[\begin{array}{c}1\\-1\\-3\end{array}\right]=-\frac{1}{3}\left[\begin{array}{c}-44\\36\\-15\end{array}\right]=\left[\begin{array}{c}\frac{44}{3}\\-12\\5\end{array}\right]$