Question

considere as seguintes matrizes : ​

Answer 1

Resposta:

Explicação passo a passo:

$\left[\begin{array}{ccc}1&0&0\\a&2&1\\-1&3&a\end{array}\right] \times\:\left[\begin{array}{ccc}1&2\\0&1\\2&0\end{array}\right] =\left[\begin{array}{ccc}1.1+0.0+0.2&1.2+0.1+0.0\\a.1+2.0+1.2&a.2+2.1+1.0\\(-1).1+3.0+a.2&(-1).2+3.1+a.0\end{array}\right] =$

$\left[\begin{array}{ccc}1&2\\a+2&2a+2\\2a-1&1\end{array}\right]=>A\times\:B$

$A\times\:B=C$

$Portanto...$

$\left[\begin{array}{ccc}1&2\\a+2&2a+2\\2a-1&1\end{array}\right]=\left[\begin{array}{ccc}1&2\\9&16\\13&1\end{array}\right]$

$Por\;comparacao\:temos...$

$a+2=9=>a=9-2=>a=7$

$2a+2=16=>2a=16-2=>2a=14=>a=14/2=>a=7$

$2a-1=13=>2a=13+1=>2a=14=>a=14/2=>a=7$

$Portanto\:\:esta\:provado\:que\:a=7\:ok!$

$7\:=>numero\:\:inteiro, impar\:e\;primo!$

$Resposta:(a)$