Question

Salut de la Exercițiu 2 am și eu nevoie de un exemplu cum se calculează matricele alea din poza, dacă se poate doar subpunctul a) ca restu fac eu! Mulțumesc!

Answer 1

$\displaystyle \mathtt{a)A= \left(\begin{array}{ccc}\mathtt1&\mathtt1&\mathtt0\\\mathtt0&\mathtt1&\mathtt{-1}\\\mathtt0&\mathtt2&\mathtt1\end{array}\right);~B=\left(\begin{array}{ccc}\mathtt1&\mathtt0&\mathtt0\\\mathtt2&\mathtt1&\mathtt{-1}\\\mathtt3&\mathtt2&\mathtt2\end{array}\right)} \\ \\ \mathtt{det(AB)=det(A) \cdot det(B)}$

$\displaystyle \mathtt{ AB=\left(\begin{array}{ccc}\mathtt1&\mathtt1&\mathtt0\\\mathtt0&\mathtt1&\mathtt{-1}\\\mathtt0&\mathtt2&\mathtt1\end{array}\right) \cdot \left(\begin{array}{ccc}\mathtt1&\mathtt0&\mathtt0\\\mathtt2&\mathtt1&\mathtt{-1}\\\mathtt3&\mathtt2&\mathtt2\end{array}\right)=}$

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

$\displaystyle \mathtt{=\left(\begin{array}{ccc}\mathtt{1+2+0}&\mathtt{0+1+0}&{0-1+0}\\ \mathtt{0+2-3}&\mathtt{0+1-2}&\mathtt{0-1-2}\\\mathtt{0+4+3}&\mathtt{0+2+2}&\mathtt{0-2+2}\end{array}\right)= \left(\begin{array}{ccc}\mathtt3&\mathtt1&\mathtt{-1}\\\mathtt{-1}&\mathtt{-1}&\mathtt{-3}\\\mathtt7&\mathtt4&\mathtt0\end{array}\right)}\\ \\ \mathtt{AB=\left(\begin{array}{ccc}\mathtt3&\mathtt1&\mathtt{-1}\\\mathtt{-1}&\mathtt{-1}&\mathtt{-3}\\\mathtt7&\mathtt4&\mathtt0\end{array}\right) }$

$\displaystyle \mathtt{det(AB)=\left|\begin{array}{ccc}\mathtt3&\mathtt1&\mathtt{-1}\\\mathtt{-1}&\mathtt{-1}&\mathtt{-3}\\\mathtt7&\mathtt4&\mathtt0\end{array}\right|=3 \cdot (-1) \cdot 0+(-1) \cdot (-1) \cdot 4+}\\ \\ \mathtt{+1 \cdot (-3) \cdot 7-(-1) \cdot (-1) \cdot 7-1 \cdot (-1) \cdot 0-3 \cdot (-3) \cdot 4= 12}\\ \\ \mathtt{det(AB)=12}$

$\displaystyle \mathtt{det(A)= \left|\begin{array}{ccc}\mathtt1&\mathtt1&\mathtt0\\\mathtt0&\mathtt1&\mathtt{-1}\\\mathtt0&\mathtt2&\mathtt1\end{array}\right|=1 \cdot1 \cdot 1+0 \cdot 0 \cdot 2+1 \cdot (-1) \cdot 0-0 \cdot 1 \cdot 0-}\\ \\ \mathtt{-1 \cdot 0 \cdot 1-1 \cdot (-1) \cdot 2=3}\\ \\ \mathtt{det(A)=3}$

$\displaystyle \mathtt{det(B)=\left|\begin{array}{ccc}\mathtt1&\mathtt0&\mathtt0\\\mathtt2&\mathtt1&\mathtt{-1}\\\mathtt3&\mathtt2&\mathtt2\end{array}\right|=1 \cdot 1 \cdot 2+0 \cdot 2 \cdot 2+0 \cdot (-1) \cdot 3-0\cdot1\cdot 3-}\\\\ \mathtt{-0\cdot2\cdot 2-1\cdot (-1) \cdot 2=4}\\\\ \mathtt{det(B)=4}\\\\ \mathtt{det(A)\cdot det(B)=3 \cdot4=12}\\ \\ \mathtt{det(A) \cdot det(B)=12}\\ \\ \mathtt{\left\{{{det(AB)=12} \atop {det(A) \cdot det(B)=12}} \right. \Rightarrow \underline{det(AB)=det(A)\cdot det(B)}}$

$\displaystyle \mathtt{det(t_A)=det(A)}\\ \\ \mathtt{t_A=\left(\begin{array}{ccc}\mathtt1&\mathtt0&\mathtt0\\\mathtt1&\mathtt1&\mathtt{2}\\\mathtt0&\mathtt{-1}&\mathtt1\end{array}\right) } \\ \\ \mathtt{det(t_A)=\left|\begin{array}{ccc}\mathtt1&\mathtt0&\mathtt0\\\mathtt1&\mathtt1&\mathtt{2}\\\mathtt0&\mathtt{-1}&\mathtt1\end{array}\right|=1 \cdot 1 \cdot 1+0 \cdot 1 \cdot (-1)+0 \cdot 2 \cdot 0-0 \cdot 1 \cdot 0-}\\ \\ \mathtt{-0 \cdot 1 \cdot 1-1 \cdot 2 \cdot (-1)=3}\\ \\ \mathtt{det(t_A)=3}$

$\displaystyle \mathtt{ \left \{ {{det(t_A)=3} \atop {det(A)=3}} \right.\Rightarrow \underline{det(t_A)=det(A)} }$

$\displaystyle \mathtt{det(2A)=8det(A)}\\ \\ \mathtt{2A=2\cdot \left(\begin{array}{ccc}\mathtt1&\mathtt1&\mathtt0\\\mathtt0&\mathtt1&\mathtt{-1}\\\mathtt0&\mathtt2&\mathtt1\end{array}\right)= \left(\begin{array}{ccc}\mathtt{2 \cdot 1}&\mathtt{2 \cdot 1}&\mathtt{2 \cdot 0}\\\mathtt{2 \cdot 0}&\mathtt{2 \cdot 1}&\mathtt{2 \cdot (-1)}\\\mathtt{2 \cdot 0}&\mathtt{2 \cdot 2}&\mathtt{2 \cdot1}\end{array}\right)=\left(\begin{array}{ccc}\mathtt2&\mathtt2&\mathtt0\\\mathtt0&\mathtt2&\mathtt{-2}\\\mathtt0&\mathtt4&\mathtt2\end{array}\right)}$

$\displaystyle \mathtt{2A= \left(\begin{array}{ccc}\mathtt2&\mathtt2&\mathtt0\\\mathtt0&\mathtt2&\mathtt{-2}\\\mathtt0&\mathtt4&\mathtt2\end{array}\right)} \\ \\ \mathtt{det(2A)=\left|\begin{array}{ccc}\mathtt2&\mathtt2&\mathtt0\\\mathtt0&\mathtt2&\mathtt{-2}\\\mathtt0&\mathtt4&\mathtt2\end{array}\right|=2 \cdot 2 \cdot 2+0 \cdot 0 \cdot 4+2 \cdot (-2) \cdot 0-0 \cdot 2 \cdot 0-}\\ \\ \mathtt{-2 \cdot 0 \cdot 2-2 \cdot (-2) \cdot 4=24}\\ \\ \mathtt{det(2A)=24}$

$\displaystyle \mathtt{det(A)=3 \Rightarrow 8det(A)=8 \cdot 3=24}\\ \\ \mathtt{8det(A)=24}\\ \\ \mathtt{ \left \{ {{det(2A)=24} \atop {8det(A)=24}} \right.\Rightarrow \underline{det(2A)=8det(A)} }$