Question

L am rezolvat îs prost ma candesm ca tre înmulțit cu inversa

Answer 1

Explicație pas cu pas:

rezolvarea in fotografie

Answer 2

$\it AX-BX=I_2-X \Rightarrow AX-BX+X=I_2 \Rightarrow (A-B+I_2)\cdot X=I_2\ \ \ \ \ (1)\\ \\ A-B+I_2=\begin{pmatrix} \it3&-\it2\\ \\ \it2&\it1\end{pmatrix}-\begin{pmatrix} \it4&-\it3\\ \\ \it3&\it1\end{pmatrix}+\begin{pmatrix} \it1&\it0\\ \\ \it0&\it1\end{pmatrix}=\begin{pmatrix} \it0&\it1\\ \\ \it-1&\it1\end{pmatrix}\ \ \ \ \ (2)\\ \\ \\ (1),\ (2) \Rightarrow \begin{pmatrix} \it0&\it1\\ \\ \it-1&\it1\end{pmatrix}\cdot X=I_2\ \ \ \ \ (3)$

$\it Inversa\ matricei\ \begin{pmatrix} \it0&\it1\\ \\ \it-1&\it1\end{pmatrix}\ este\ matricea\ \begin{pmatrix} \it1&\it-1\\ \\ \it1&\it0\end{pmatrix}\\ \\ \\ \hat Inmul\c{\it t}im\ (3),\ la \ st\hat anga, \ cu \ matricea\ \begin{pmatrix} \it1&\it-1\\ \\ \it1&\it0\end{pmatrix}\ \d si\ rezult\breve a:\\ \\ \\ X=\begin{pmatrix} \it1&\it-1\\ \\ \it1&\it0\end{pmatrix}$