Question

Salut, ma puteți ajuta va rog?!

Answer 1

$\displaystyle \mathtt{a)~B= \left(\begin{array}{ccc}\mathtt0&\mathtt1&\mathtt{-1}\\\mathtt2&\mathtt3&\mathtt0\\\mathtt1&\mathtt{-1}&\mathtt1\end{array}\right)}\\ \\ \mathtt{det(B)=\left|\begin{array}{ccc}\mathtt0&\mathtt1&\mathtt{-1}\\\mathtt2&\mathtt3&\mathtt0\\\mathtt1&\mathtt{-1}&\mathtt1\end{array}\right|=0\cdot3\cdot1+(-1)\cdot2\cdot(-1)+1\cdot0\cdot1-}\\ \\ \mathtt{-(-1)\cdot3\cdot1-1\cdot2\cdot1-0\cdot0\cdot(-1)=3}\\ \\ \mathtt{det(B)=3\ne 0 \Rightarrow Matricea~B~este~inversabil\u{a}}$

$\displaystyle \mathtt{b)~A=\left(\begin{array}{ccc}\mathtt1&\mathtt1&\mathtt2\\\mathtt0&\mathtt1&\mathtt3\\\mathtt0&\mathtt0&\mathtt1\end{array}\right)~~~~~~~~~~~~~~~~~~~~~~~~~~~~A^{-1}= \frac{1}{det(A)}\cdot A^* }$

$\displaystyle \mathtt{det(A)=\left|\begin{array}{ccc}\mathtt1&\mathtt1&\mathtt2\\\mathtt0&\mathtt1&\mathtt3\\\mathtt0&\mathtt0&\mathtt1\end{array}\right|=1\cdot1\cdot1+2\cdot0 \cdot0+1\cdot3\cdot0-2\cdot1\cdot0-}\\ \\ \mathtt{-1\cdot0\cdot1-1\cdot3\cdot0=1}\\ \\ \mathtt{~~~~~~~~~~~~~~~~~~~~~~~~~~det(A)=1}$

$\displaystyle \mathtt{A=\left(\begin{array}{ccc}\mathtt1&\mathtt1&\mathtt2\\\mathtt0&\mathtt1&\mathtt3\\\mathtt0&\mathtt0&\mathtt1\end{array}\right)\Rightarrow A^{tr}=\left(\begin{array}{ccc}\mathtt1&\mathtt0&\mathtt0\\\mathtt1&\mathtt1&\mathtt0\\\mathtt2&\mathtt3&\mathtt1\end{array}\right)}$

$\displaystyle \mathtt{D_{11}=(-1)^{1+1}\cdot \left|\begin{array}{ccc}\mathtt1&\mathtt0\\\mathtt3&\mathtt1\end{array}\right|=1 \cdot 1=1}\\ \\ \mathtt{D_{12}=(-1)^{1+2}\cdot \left|\begin{array}{ccc}\mathtt1&\mathtt0\\\mathtt2&\mathtt1\end{array}\right|=(-1) \cdot 1=-1}\\ \\ \mathtt{D_{13}=(-1)^{1+3}\cdot \left|\begin{array}{ccc}\mathtt1&\mathtt1\\\mathtt2&\mathtt3\end{array}\right|=1 \cdot 1=1}$
$\mathtt{D_{21}=(-1)^{2+1}\cdot \left|\begin{array}{ccc}\mathtt0&\mathtt0\\\mathtt3&\mathtt1\end{array}\right|=(-1)\cdot 0=0}\\ \\ \mathtt{D_{22}=(-1)^{2+2}\cdot \left|\begin{array}{ccc}\mathtt1&\mathtt0\\\mathtt2&\mathtt1\end{array}\right|=1\cdot1=1}\\ \\ \mathtt{D_{23}=(-1)^{2+3}\cdot \left|\begin{array}{ccc}\mathtt1&\mathtt0\\\mathtt2&\mathtt3\end{array}\right|=(-1)\cdot3=-3}$
$\displaystyle \mathtt{D_{31}=(-1)^{3+1}\cdot \left|\begin{array}{ccc}\mathtt0&\mathtt0\\\mathtt1&\mathtt0\end{array}\right|=1 \cdot 0=0}\\ \\ \mathtt{D_{32}=(-1)^{3+2}\cdot \left|\begin{array}{ccc}\mathtt1&\mathtt0\\\mathtt1&\mathtt0\end{array}\right|=(-1)\cdot 0=0}\\ \\ \mathtt{D_{33}=(-1)^{3+3}\cdot \left|\begin{array}{ccc}\mathtt1&\mathtt0\\\mathtt1&\mathtt1\end{array}\right|=1 \cdot 1=1}$

$\displaystyle \mathtt{A^*= \left(\begin{array}{ccc}\mathtt1&\mathtt{-1}&\mathtt1\\\mathtt0&\mathtt1&\mathtt{-3}\\\mathtt0&\mathtt0&\mathtt1\end{array}\right)}\\ \\ \mathtt{A^{-1}= \frac{1}{1}\cdot \left(\begin{array}{ccc}\mathtt1&\mathtt{-1}&\mathtt1\\\mathtt0&\mathtt1&\mathtt{-3}\\\mathtt0&\mathtt0&\mathtt1\end{array}\right)= \left(\begin{array}{ccc}\mathtt1&\mathtt{-1}&\mathtt1\\\mathtt0&\mathtt1&\mathtt{-3}\\\mathtt0&\mathtt0&\mathtt1\end{array}\right) }$

$\displaystyle \mathtt{c)~X\cdot A=B\Rightarrow X=BA^{-1}}\\ \\ \mathtt{X=\left(\begin{array}{ccc}\mathtt0&\mathtt1&\mathtt{-1}\\\mathtt2&\mathtt3&\mathtt0\\\mathtt1&\mathtt{-1}&\mathtt1\end{array}\right)\cdot \left(\begin{array}{ccc}\mathtt1&\mathtt{-1}&\mathtt1\\\mathtt0&\mathtt1&\mathtt{-3}\\\mathtt0&\mathtt0&\mathtt1\end{array}\right)}$

$\displaystyle \mathtt{X= \left(\begin{array}{ccc}\mathtt{0\cdot1+1\cdot0+(-1)\cdot0}&\mathtt{0\cdot(-1)+1\cdot1+(-1)\cdot0}&\mathtt{0\cdot1+1\cdot(-3)+(-1)\cdot1}\\\mathtt{2\cdot1+3\cdot0+0\cdot0}&\mathtt{2\cdot(-1)+3\cdot1+0\cdot0}&\mathtt{2\cdot1+3\cdot(-3)+0\cdot1}\\\mathtt{1\cdot1+(-1)\cdot0+1\cdot0}&\mathtt{1\cdot(-1)+(-1)\cdot1+1\cdot0}&\mathtt{1\cdot1+(-1)\cdot(-3)+1\cdot1}\end{array}\right)}$

$\displaystyle \mathtt{X= \left(\begin{array}{ccc}\mathtt{0+0+0}&\mathtt{0+1+0}&\mathtt{0-3-1}\\\mathtt{2+0+0}&\mathtt{-2+3+0}&\mathtt{2-9+0}\\\mathtt{1+0+0}&\mathtt{-1-1+0}&\mathtt{1+3+1}\end{array}\right)}\\ \\ \mathtt{X=\left(\begin{array}{ccc}\mathtt0&\mathtt1&\mathtt{-4}\\\mathtt2&\mathtt1&\mathtt{-7}\\\mathtt1&\mathtt{-2}&\mathtt5\end{array}\right)}$