Question

2x1-x2+3x3=-4, x1+3x2-x3=11, x1-2x2+2x3=-7 нужно решить с помощью обратной матрицы. По формулам: detA=/0 (не равен 0), А^-1×А*, Х=А^-1×В

Answer 1

$\displaystyle\begin{cases}2x_1-x_2+3x_3=-4\\x_1+3x_2-x_3=11\\x_1-2x_2+2x_3=-7\end{cases}\\\mathcal4=\left[\begin{array}{ccc}2&-1&3\\1&3&-1\\1&-2&2\end{array}\right]=12+1-6-9+2-4=-4\neq0 \\M_{11}=\left[\begin{array}{cc}3&-1\\-2&2\end{array}\right] =6-2=4\\M_{12}=\left[\begin{array}{cc}1&-1\\1&2\end{array}\right] =2+1=3\\M_{13}=\left[\begin{array}{cc}1&3\\1&-2\end{array}\right] =-2-3=-5\\M_{21}=\left[\begin{array}{cc}-1&3\\-2&2\end{array}\right] =-2+6=4$

$M_{22}=\left[\begin{array}{cc}2&3\\1&2\end{array}\right] =4-3=1\\M_{23}=\left[\begin{array}{cc}2&-1\\1&-2\end{array}\right] =-4+1=-3\\M_{31}=\left[\begin{array}{cc}-1&3\\3&-1\end{array}\right] =1-9=-8\\M_{32}=\left[\begin{array}{cc}2&3\\1&-1\end{array}\right] =-2-3=-5\\M_{33}=\left[\begin{array}{cc}2&-1\\1&3\end{array}\right] =6+1=7$

$\displaystyle A=\left[\begin{array}{ccc}4&-3&-5\\-4&1&3\\-8&5&7\end{array}\right] \\A^T=\left[\begin{array}{ccc}4&-4&-8\\-3&1&5\\-5&3&7\end{array}\right]\\A^{-1}=\frac{1}{\mathcal4}A^T=-\frac{1}{4}*\left[\begin{array}{ccc}4&-4&-8\\-3&1&5\\-5&3&7\end{array}\right]\\X=A^{-1}B=-\frac{1}{4}*\left[\begin{array}{ccc}4&-4&-8\\-3&1&5\\-5&3&7\end{array}\right]*\left[\begin{array}{c}-4\\11\\-7\end{array}\right]=-\frac{1}{4}\left[\begin{array}{c}-16-44+56\\12+11-35\\20+33-49\end{array}\right]=$

$=-\frac{1}{4}\left[\begin{array}{c}-4\\-12\\4\end{array}\right]=\left[\begin{array}{c}1\\3\\-1\end{array}\right]\\\\\left[\begin{array}{c}x\\y\\z\end{array}\right]=\left[\begin{array}{c}1\\3\\-1\end{array}\right]$