Respuesta :
[tex]\bf A=
\begin{bmatrix}
5&2\\3&0
\end{bmatrix}\qquad B=
\begin{bmatrix}
4&3\\-1&6
\end{bmatrix}\qquad C=
\begin{bmatrix}
2&8\\0&5
\end{bmatrix}\\\\
-------------------------------\\\\
2B\implies 2\begin{bmatrix}
4&3\\-1&6
\end{bmatrix}\implies \begin{bmatrix}
2\cdot 4&2\cdot 3\\2\cdot -1&2\cdot 6
\end{bmatrix}\implies
\begin{bmatrix}
8&6\\-2&12
\end{bmatrix}[/tex]
[tex]\bf \stackrel{A-2B}{\begin{bmatrix} 5&2\\3&0 \end{bmatrix}-\begin{bmatrix} 8&6\\-2&12 \end{bmatrix}}\implies \begin{bmatrix} -3&-4\\5&-12 \end{bmatrix} \\\\\\ \stackrel{A-2B+C}{\begin{bmatrix} -3&-4\\5&-12 \end{bmatrix}+\begin{bmatrix} 2&8\\0&5 \end{bmatrix}}\implies \begin{bmatrix} -1&4\\5&-7 \end{bmatrix}[/tex]
[tex]\bf \stackrel{A-2B}{\begin{bmatrix} 5&2\\3&0 \end{bmatrix}-\begin{bmatrix} 8&6\\-2&12 \end{bmatrix}}\implies \begin{bmatrix} -3&-4\\5&-12 \end{bmatrix} \\\\\\ \stackrel{A-2B+C}{\begin{bmatrix} -3&-4\\5&-12 \end{bmatrix}+\begin{bmatrix} 2&8\\0&5 \end{bmatrix}}\implies \begin{bmatrix} -1&4\\5&-7 \end{bmatrix}[/tex]
It is convenient to use the matrix capability of your calculator to do this. The result is ...
[tex]\left[ \begin{array}{cc}-1&4\\5&-7\end{array} \right][/tex]
[tex]\left[ \begin{array}{cc}-1&4\\5&-7\end{array} \right][/tex]
