Question

Dadas as matrizes A=[ 0 4 -2], B=[-3 6 9] e C=[0 -1 0], Calcule 2A - B +3C
[ 6 2 8] [ 12-60] [1 -1 2]

Answer 1

$\mathbf{A}=\left[ \begin{array}{ccc} 0&4&-2\\ 6&2&8 \end{array} \right],\,\mathbf{B}=\left[ \begin{array}{ccc} -3&6&9\\ 12&-6&0 \end{array} \right],\,\mathbf{C}=\left[ \begin{array}{ccc} 0&-1&0\\ 1&-1&2 \end{array} \right]$

_________

Calculando

$\bullet\;\;2\mathbf{A}=2\cdot \left[ \begin{array}{ccc} 0&4&-2\\ 6&2&8 \end{array} \right]\\\\\\ 2\mathbf{A}=\left[\begin{array}{ccc} 2\cdot 0~&~2\cdot 4~&~2\cdot (-2)\\ 2\cdot 6~&~2\cdot 2~&~2\cdot 8 \end{array}\right]\\\\\\ 2\mathbf{A}=\left[\begin{array}{ccc} 0&8&-4\\ 12&4&16 \end{array}\right]$


$\bullet\;\;-\mathbf{B}=(-1)\cdot \left[ \begin{array}{ccc} -3&6&9\\ 12&-6&0 \end{array} \right]\\\\\\ -\mathbf{B}=\left[ \begin{array}{ccc} (-1)\cdot (-3)~&~(-1)\cdot 6~&~(-1)\cdot 9\\ (-1)\cdot 12~&~(-1)\cdot (-6)~&~(-1)\cdot 0 \end{array} \right]\\\\\\ -\mathbf{B}=\left[ \begin{array}{ccc} 3&6&-9\\ -12&6&0 \end{array} \right]$


$\bullet\;\;3\mathbf{C}=3\cdot \left[ \begin{array}{ccc} 0&-1&0\\ 1&-1&2 \end{array} \right]\\\\\\ 3\mathbf{C}=\left[ \begin{array}{ccc} 3\cdot 0~&~3\cdot (-1)~&~3\cdot 0\\ 3\cdot 1~&~3\cdot (-1)~&~3\cdot 2 \end{array} \right]\\\\\\ 3\mathbf{C}=\left[ \begin{array}{ccc} 0&-3&0\\3&-3&6 \end{array}\right]$


_______

Sendo assim,

$2\mathbf{A}-\mathbf{B}+3\mathbf{C}\\\\\\ =\left[\begin{array}{ccc} 0&8&-4\\ 12&4&16 \end{array}\right]+\left[ \begin{array}{ccc} 3&6&-9\\ -12&6&0 \end{array} \right]+\left[ \begin{array}{ccc} 0&-3&0\\3&-3&6 \end{array}\right]\\\\\\ =\left[\begin{array}{ccc} 0+3+0~&~8+6-3~&~-4-9+0\\ 12-12+3~&~4+6-3~&~16+0+6 \end{array}\right]\\\\\\ =\left[\begin{array}{ccc} 3&11&-13\\ 3&7&22 \end{array}\right]~~~~~~\checkmark$


Bons estudos! :-)