Question

punctul b ............

Answer 1

$\displaystyle \mathtt{b)~AX=B\Rightarrow X=A^{-1}B~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~A^{-1}= \frac{1}{det(A)}\cdot A^* }\\ \\ \mathtt{det(A)=\left|\begin{array}{ccc}\mathtt2&\mathtt1\\\mathtt{-1}&\mathtt2\end{array}\right|=2\cdot2-1\cdot(-1)=4+1=5}$

$\displaystyle\mathtt{ A=\left(\begin{array}{ccc}\mathtt2&\mathtt1\\\mathtt{-1}&\mathtt2\end{array}\right)\Rightarrow A^{tr}=\left(\begin{array}{ccc}\mathtt2&\mathtt{-1}\\\mathtt1&\mathtt2\end{array}\right)}$

$\displaystyle \mathtt{D_{11}=(-1)^{1+1}\cdot 2=1\cdot2=2}\\ \\ \mathtt{D_{12}=(-1)^{1+2}\cdot1=(-1)\cdot1=-1}\\ \\ \mathtt{D_{21}=(-1)^{2+1}\cdot(-1)=(-1)\cdot(-1)=1}\\ \\ \mathtt{D_{22}=(-1)^{2+2}\cdot2=1\cdot2=2}$

$\displaystyle \mathtt{A^*=\left(\begin{array}{ccc}\mathtt2&\mathtt{-1}\\\mathtt1&\mathtt2\end{array}\right)}\\ \\ \mathtt{A^{-1}= \frac{1}{5}\cdot \left(\begin{array}{ccc}\mathtt2&\mathtt{-1}\\\mathtt1&\mathtt2\end{array}\right)=\left(\begin{array}{ccc}\mathtt{ \frac{2}{5} }&\mathtt{- \frac{1}{5}} \\\mathtt{ \frac{1}{5} }&\mathtt{ \frac{2}{5} }\end{array}\right)}$

$\displaystyle \mathtt{A^{-1}B=\left(\begin{array}{ccc}\mathtt{ \frac{2}{5} }&\mathtt{- \frac{1}{5}} \\\mathtt{ \frac{1}{5} }&\mathtt{ \frac{2}{5} }\end{array}\right)\cdot \left(\begin{array}{ccc}\mathtt{-1}&\mathtt0\\\mathtt3&\mathtt5\end{array}\right)=}$

$\displaystyle \mathtt{=\left(\begin{array}{ccc}\mathtt{ \frac{2}{5}\cdot(-1)+\left(- \frac{1}{5}\right)\cdot3 }& \mathtt{ \frac{2}{5}\cdot0+\left(- \frac{1}{5}\right)\cdot5 }\\\mathtt{ \frac{1}{5}\cdot(-1)+ \frac{2}{5}\cdot3 }&\mathtt{ \frac{1}{5}\cdot0+ \frac{2}{5}\cdot5 }\end{array}\right)=}$

$\displaystyle \mathtt{= \left(\begin{array}{ccc}\mathtt{- \frac{2}{5}- \frac{3}{5} }&\mathtt{0- \frac{5}{5} }\\\mathtt{- \frac{1}{5}+ \frac{6}{5} }&\mathtt{0+ \frac{10}{5} }\end{array}\right)= \left(\begin{array}{ccc}\mathtt{- \frac{5}{5} }&\mathtt{-1}\\\mathtt{ \frac{5}{5} }&\mathtt2\end{array}\right)= \left(\begin{array}{ccc}\mathtt{-1}&\mathtt{-1}\\\mathtt1&\mathtt2\end{array}\right)}\\ \\ \mathtt{X=\left(\begin{array}{ccc}\mathtt{-1}&\mathtt{-1}\\\mathtt1&\mathtt2\end{array}\right)}$