Question

Решите матричное уравнение y*( 1 строка 1 -2 1 ,2 строка (3 -1 1) 3 строка( 1 -3 2) = 1 строка( 2 1 4) 2 строка( -1 2 3)

Answer 1

$Y* \left(\begin{array}{ccc}1&-2&1\\3&-1&1\\1&-3&2\end{array}\right) = \left(\begin{array}{ccc}2 &1 &4\\-1 &2& 3\end{array}\right) \\ \\ Y*A=B \\ Y=B*A^{-1} \\ \\ 1)|A|= \left|\begin{array}{ccc}1&-2&1\\3&-1&1\\1&-3&2\end{array}\right|=-2-2-9-(-1-12-3)=-13-(-16) \\ \\ =-13+16=3 \\ \\ 2)\ A^*= \left(\begin{array}{ccc}A_{11}&A_{12}&A_{13}\\A_{21}&A_{22}&A_{23}\\A_{31}&A_{32}&A_{33}\end{array}\right) \\ \\ A_{11}=(-1)^{1+1}* \left|\begin{array}{ccc}-1&1\\-3&2\end{array}\right|=-2-(-3)=-2+3=1 \\ \\$

$A_{12}=(-1)^{1+2}* \left|\begin{array}{ccc}3&1\\1&2\end{array}\right|=-(6-1)=-5 \\ \\ A_{13}=(-1)^{1+3}* \left|\begin{array}{ccc}3&-1\\1&-3\end{array}\right|=-9-(-1)=-8 \\ \\ A_{21}=(-1)^{2+1}* \left|\begin{array}{ccc}-2&1\\-3&2\end{array}\right|=-(-4-(-3))=-(-1)=1 \\ \\ A_{22}=(-1)^{2+2}* \left|\begin{array}{ccc}1&1\\1&2\end{array}\right|=2-1=1 \\ \\ A_{23}=(-1)^{2+3}* \left|\begin{array}{ccc}1&-2\\1&-3\end{array}\right|=-(-3-(-2))=1 \\ \\$

$A_{31}=(-1)^{3+1}* \left|\begin{array}{ccc}-2&1\\-1&1\end{array}\right|=-2-(-1)=-1 \\ \\ A_{32}=(-1)^{3+2}* \left|\begin{array}{ccc}1&1\\3&1\end{array}\right|=-(1-3)=2 \\ \\ A_{33}=(-1)^{3+3}* \left|\begin{array}{ccc}1&-2\\3&-1\end{array}\right|=-1-(-6)=5 \\ \\ \\ (A^*)^T = \left(\begin{array}{ccc}1&-5&-8\\1&1&1\\-1&2&5\end{array}\right) ^T= \left(\begin{array}{ccc}1&1&-1\\-5&1&2\\-8&1&5\end{array}\right) \\ \\ \\ \\$

$A^{-1}= \frac{1}{|A|} *(A^*)^T= \frac{1}{3} * \left(\begin{array}{ccc}1&1&-1\\-5&1&2\\-8&1&5\end{array}\right) \\ \\ \\ Y= \frac{1}{3} *\left(\begin{array}{ccc}2 &1 &4\\-1 &2& 3\end{array}\right) *\left(\begin{array}{ccc}1&1&-1\\-5&1&2\\-8&1&5\end{array}\right)= \\ \\ \\ \frac{1}{3} *\left(\begin{array}{ccc}(2-5-32) &(2+1+4) &(-2+2+20)\\(-1-10-24) &(-1+2+3)& (1+4+15)\end{array}\right) = \\ \\ \\$

$= \frac{1}{3}* \left(\begin{array}{ccc}-35&7 &20\\-35 &4& 20\end{array}\right)=\left(\begin{array}{ccc} -\frac{35}{3} & \frac{7}{3} & \frac{20}{3} \\ \\ -\frac{35}{3}& \frac{4}{3} & \frac{20}{3}\end{array}\right) \\ \\ \\ OTBET: Y=\left(\begin{array}{ccc} -\frac{35}{3} & \frac{7}{3} & \frac{20}{3} \\ \\ -\frac{35}{3}& \frac{4}{3} & \frac{20}{3}\end{array}\right)$