Question

alguém ajuda com essas questões de matemática por favor!​

Answer 1

Resposta:

Basta somar ou subtrair cada elemento com o seu correspondente que ocupa a mesma posição na outra matriz:

1)

a) $\left[\begin{array}{cccc}3+6&(-5)+8&2+9&7+3\end{array}\right] =\left[\begin{array}{cccc}9&3&11&10\end{array}\right]$

b) $\left[\begin{array}{ccc}(-4)-8&7-1\\10-3&(-6)-(-2)\end{array}\right] =\left[\begin{array}{ccc}-12&6\\7&-4\end{array}\right]$

c) $\left[\begin{array}{ccc}12+10&(-1)+6&8+(-8)\\15+1&9+11&(-13)+5\end{array}\right] =\left[\begin{array}{ccc}22&5&0\\16&20&-8\end{array}\right]$

d) $\left[\begin{array}{ccc}19-20&26-16&14-3\\11-7&6-(-6)&32-5\\45-15&12-3&7-(-11)\end{array}\right] =\left[\begin{array}{ccc}-1&10&11\\4&12&27\\30&9&18\end{array}\right]$

2)

a)$\left[\begin{array}{ccc}3-1&x+2x\\4+y&z^{2}-1 \end{array}\right]= \left[\begin{array}{ccc}2&6\\-1&0\end{array}\right]$

x + 2x = 6 => x = 2

4 + y = -1 => y = -5

z² - 1 = 0 => z = 1

b) $\left[\begin{array}{ccc}-2+3x-0\\4+6-9\\y+y^{2}-6 \\z+0-1\end{array}\right] =\left[\begin{array}{ccc}x\\1\\2\\4\end{array}\right]$

-2 + 3x = x => x = 1

y² + y - 8 = 0 => $y_{1}=\frac{-1+\sqrt[]{33} }{2}, y_{2}=\frac{-1-\sqrt[]{33} }{2}$

z - 1 = 4 => z = 5