Question

Sa se calculeze 3A
A=1 -1
-3 3

Answer 1

$\[ A= \left( {\begin{array}{cc} 1 & -1 \\ -3 & 3 \\ \end{array} } \right) \] \\ \\ 3A=? \\ \\ 3A= 3 \cdot \left( {\begin{array}{cc} 1 & -1 \\ -3 & 3 \\ \end{array} } \right) = \left( {\begin{array}{cc} 3\cdot (1) & 3\cdot (-1) \\ 3\cdot (-3) & 3 \cdot (3) \\ \end{array} } \right) = \left( {\begin{array}{cc} 3 & -3 \\ -9 & 9 \\ \end{array} } \right) \\ \\ \text{Asadar, matricea ceruta arata asa}$

Answer 2

$A=\left(\begin{array}{cc}1&-1\\-3&3\end{array}\right)\\3A=3\cdot\left(\begin{array}{cc}1&-1\\-3&3\end{array}\right) = \left(\begin{array}{cc}1\cdot3&-1\cdot3\\-3\cdot3&3\cdot3\end{array}\right)=\\3A=\left(\begin{array}{cc}3&-3\\-9&9\end{array}\right)$