Question

me ajudem a resolver esse calculo.
Alternativas:
a. p = 3; q = 2
b. p = 3; q = 3
c. p = 2; q = 3
d. p = 6; q = 2
e. p = 9; q = 3

Answer 1

$\left[ \begin{array}{rrccc} p+q && -2 \\ 0 && 2p-q \end{array} \right]=\left[ \begin{array}{rrccc} 6 && -2 \\ 0 && 3 \end{array} \right]\\\\\\ p+q=6 \ e \ 2p-q=3\\\\ \begin{cases} p+q=6 \\ 2p-q=3 \end{cases}\\\\\\ p+q+2p-q=6+3\\\\ 3p=9$

$p= \frac{9}{3}\\\\ p=3\\\\ p+q=6\\\\ 3+q=6\\\\ q=6-3\\\\ q=3$