Question

Preciso disso urgente

Answer 1

Resposta:

Explicação passo-a-passo:

8) Termos que

$M=\left[\begin{array}{cc}a&0\\b&-a\end{array}\right]=>M^{2}=\left[\begin{array}{cc}a&0\\b&-a\end{array}\right].\left[\begin{array}{cc}a&0\\b&-a\end{array}\right]=\left[\begin{array}{cc}a.a+0.b&a.0+0.(-a)\\b.a-a.b&b.0+(-a).(-a)\end{array}\right]=\left[\begin{array}{cc}a^{2} &0\\0&a^{2} \end{array}\right]$

Assim

$\left[\begin{array}{cc}a^{2} &0\\0&a^{2} \end{array}\right]=\left[\begin{array}{cc}8 &0\\0&8 \end{array}\right]$ => a² = 8 => a = √8 => a = 2√2, alternativa b)

9) Temos que

$A.B=\left[\begin{array}{cc}2&1\\x&2\end{array}\right].\left[\begin{array}{cc}1&y\\1&2\end{array}\right]=\left[\begin{array}{cc}2.1+1.1&2.y+1.2\\x.1+2.1&x.y+2.2\end{array}\right]=\left[\begin{array}{cc}3&2y+2\\x+2&x.y+4\end{array}\right]$

Então

$\left[\begin{array}{cc}3&2y+2\\x+2&x.y+4\end{array}\right]=\left[\begin{array}{cc}3&0\\5&z\end{array}\right]$

E assim

x + 2 = 5 => x = 5 - 2 => x = 3

2y + 2 = 0 => 2y = -2 => y = -2/2 => y = -1

x.y + 4 = z => z = 3.(-1) + 4 => z = -3 + 4 => z = 1

Logo,

x + y + z = 3 - 1 + 1 = 3, alternativa c).