Question

Dacă $A=\left([tex]\begin{array}{ll}1 & 2 \\ 3 & 4\end{array}\right)$, atunci $A^{2}-3 A$ este matricea : a) $\left.\left(\begin{array}{cc}4 & 4 \\ 6 & 10\end{array}\right) ; b\right)\left(\begin{array}{cc}6 & 10 \\ 4 & 4\end{array}\right) ;$ c $)\left(\begin{array}{cc}4 & -4 \\ 6 & -10\end{array}\right) ;$ d) $\left(\begin{array}{cc}2 & 4 \\ 6 & 8\end{array}\right)$;

e) $\left(\begin{array}{rr}1 & 2 \\ -1 & 3\end{array}\right) \cdot(0,5$ puncte $)$[/tex]

Answer 1

Răspuns:

$A^2=\begin{pmatrix}7 & 10\\15 & 14\end{pmatrix}$

$A^2-3A=\begin{pmatrix}7 & 10\\15 & 22\end{pmatrix}-\begin{pmatrix}3 & 6\\9 & 12\end{pmatrix}=\begin{pmatrix}4 & 4\\6 & 10\end{pmatrix}$

Explicație pas cu pas: