Question

Sabendo que:
a) 0
b) 12
c) 18
d) - 6
e) - 9

Answer 1

$\Large\boxed{\begin{array}{l}\sf Sabendo~que~\begin{pmatrix}\sf2&\sf1\\\sf3&\sf5\end{pmatrix}\cdot\begin{pmatrix}\sf x\\\sf y\end{pmatrix}=\begin{pmatrix}\sf-3\\\sf6\end{pmatrix},\\\sf o~valor~de~x^2+y^2~\acute e~igual~a:\\\tt a)~\sf0\\\tt b)~\sf12\\\tt c)~\sf18\\\tt d)~\sf-6\\\tt e)~\sf-9\end{array}}$

$\Large\boxed{\begin{array}{l}\sf\begin{pmatrix}\sf2&\sf1\\\sf3&\sf5\end{pmatrix}\cdot\begin{pmatrix}\sf x\\\sf y\end{pmatrix}=\begin{pmatrix}\sf-3\\\sf 6\end{pmatrix}\\\sf\begin{pmatrix}\sf 2x+y\\\sf 3x+5y\end{pmatrix}=\begin{pmatrix}\sf-3\\\sf6\end{pmatrix}\\\\\begin{cases} \sf2x+y=-3\\\sf 3x+5y=6\end{cases}\\\\+\underline{\begin{cases}\sf-10x-\diagup\!\!\!\!\!\!5y=15\\\sf 3x+\diagup\!\!\!\!\!\!5y=6\end{cases}}\\\sf -7x=21\cdot(-1)\\\sf 7x=-21\\\sf x=-\dfrac{21}{7}\\\\\sf x=-3\end{array}}$

$\Large\boxed{\begin{array}{l}\sf 2x+y=-3\\\sf 2\cdot(-3)+y=-3\\\sf -6+y=-3\\\sf y=6-3\\\sf y=3\\\sf x^2+y^2=(-3)^2+(3)^2\\\sf x^2+y^2=9+9\\\sf x^2+y^2=18\\\huge\boxed{\boxed{\boxed{\boxed{\sf\maltese~alternativa~c}}}}\end{array}}$