Question

Se considera maticea A= ( 2 3 )
4 5
Demonstrati ca A la a doua - 7A = 2I indice 2

Answer 1

$\displaystyle \mathtt{A= \left(\begin{array}{ccc}\mathtt2&\mathtt3\\\mathtt4&\mathtt5\end{array}\right)}\\ \\ \mathtt{A^2-7A=2I_2}\\ \\ \mathtt{A^2= A \cdot A= \left(\begin{array}{ccc}\mathtt2&\mathtt3\\\mathtt4&\mathtt5\end{array}\right) \cdot \left(\begin{array}{ccc}\mathtt2&\mathtt3\\\mathtt4&\mathtt5\end{array}\right) = \left(\begin{array}{ccc}\mathtt{2 \cdot 2+3 \cdot 4}&\mathtt{2 \cdot 3+3 \cdot 5}\\\mathtt{4 \cdot 2+5 \cdot 4}&\mathtt{4 \cdot 3+5 \cdot 5}\end{array}\right)= }$
$\displaystyle\mathtt{=\left(\begin{array}{ccc}\mathtt{4+12}&\mathtt{6+15}\\\mathtt{8+20}&\mathtt{12+25}\end{array}\right)=\left(\begin{array}{ccc}\mathtt{16}&\mathtt{21}\\\mathtt{33}&\mathtt{37}\end{array}\right)}\\\\ \mathtt{A^2=\left(\begin{array}{ccc}\mathtt{16}&\mathtt{21}\\\mathtt{28}&\mathtt{37}\end{array}\right)}$
$\displaystyle \mathtt{7A=7\cdot\left(\begin{array}{ccc}\mathtt2&\mathtt3\\\mathtt4&\mathtt5\end{array}\right) =\left(\begin{array}{ccc}\mathtt{7 \cdot 2}&\mathtt{7 \cdot 3}\\\mathtt{7 \cdot 4}&\mathtt{7 \cdot 5}\end{array}\right)=\left(\begin{array}{ccc}\mathtt{14}&\mathtt{21}\\\mathtt{28}&\mathtt{35}\end{array}\right) }\\ \\ \mathtt{7A=\left(\begin{array}{ccc}\mathtt{14}&\mathtt{21}\\\mathtt{28}&\mathtt{35}\end{array}\right)}$
$\displaystyle \mathtt{A^2-7A=\left(\begin{array}{ccc}\mathtt{16}&\mathtt{21}\\\mathtt{28}&\mathtt{37}\end{array}\right) -\left(\begin{array}{ccc}\mathtt{14}&\mathtt{21}\\\mathtt{28}&\mathtt{35}\end{array}\right)=\left(\begin{array}{ccc}\mathtt{16-14}&\mathtt{21-21}\\\mathtt{28-28}&\mathtt{37-35}\end{array}\right)=}\\ \\ \mathtt{=\left(\begin{array}{ccc}\mathtt2&\mathtt0\\\mathtt0&\mathtt2\end{array}\right)}\\ \\ \mathtt{A^2-7A=\left(\begin{array}{ccc}\mathtt2&\mathtt0\\\mathtt0&\mathtt2\end{array}\right)}$
$\displaystyle \mathtt{I_2=\left(\begin{array}{ccc}\mathtt1&\mathtt0\\\mathtt0&\mathtt1\end{array}\right)}\\ \\ \mathtt{2I_2=2 \cdot \left(\begin{array}{ccc}\mathtt1&\mathtt0\\\mathtt0&\mathtt1\end{array}\right)=\left(\begin{array}{ccc}\mathtt{2 \cdot 1}&\mathtt{2 \cdot 0}\\\mathtt{2 \cdot 0}&\mathtt{2 \cdot 1}\end{array}\right)=\left(\begin{array}{ccc}\mathtt2&\mathtt0\\\mathtt0&\mathtt2\end{array}\right)} \\ \\ \mathtt{2I_2=\left(\begin{array}{ccc}\mathtt2&\mathtt0\\\mathtt0&\mathtt2\end{array}\right)}$
$\displaystyle \mathtt{\Rightarrow A^2-7A=2I_2}$