Question

URGENTE GALERA, QUESTÃO DE MATRIZ!!!

Answer 1

$Matriz\: inversa : \\ \\ A\: .\: A^{- 1} = I_2 \\ \\ \left[\begin{array}{ccc}1&- 2\\3&5\end{array}\right] . \left[\begin{array}{ccc}a&b\\c&d\end{array}\right] = \left[\begin{array}{ccc}1&0\\0&1\end{array}\right] \\ \\ \left[\begin{array}{ccc}a - 2c&b - 2d\\3a + 5c&3b + 5d\end{array}\right] = \left[\begin{array}{ccc}1&0\\0&1\end{array}\right]$

{a - 2c = 1
{3a + 5c = 0

a = 1 + 2c

3(1 + 2c) + 5c = 0
3 + 6c + 5c = 0
11c = - 3
c = - 11/3

a = 1 - 2.11/3
a = 1 - 22/3 = - 19/3

{b - 2d = 0
{3b + 5d = 1

b = 2d

3.2d + 5d = 1
6d + 5d = 1
11d = 1
d = 1/11

b = 2/11

Temos:

$A^{- 1} = \left[\begin{array}{ccc}- \frac{19}{3} & \frac{2}{11} \\- \frac{11}{3} & \frac{1}{11} \end{array}\right]_2$

$A + B = C \\ \\ \left[\begin{array}{ccc}1&- 2\\3&5\end{array}\right] + \left[\begin{array}{ccc}2&3\\4&- 1\end{array}\right] = \left[\begin{array}{ccc}3&1\\7&4\end{array}\right] \\ \\ \\ det\: (A + B) = \left|\begin{array}{ccc}3&1\\7&4\end{array}\right| = 12 - 7 = 5$

$A^{- 1} + A^t \\ \\ \left[\begin{array}{ccc}- \frac{19}{3} & \frac{2}{11} \\- \frac{11}{3} & \frac{1}{11} \end{array}\right] + \left[\begin{array}{ccc}1&3\\- 2&5\end{array}\right] = \left[\begin{array}{ccc}- \frac{16}{3} & \frac{35}{11} \\- \frac{17}{3} & \frac{56}{11} \end{array}\right] \\ \\ \\ det\: (A^{- 1} + A^t) = \left|\begin{array}{ccc}- \frac{16}{3} & \frac{35}{11} \\- \frac{17}{3} & \frac{56}{11} \end{array}\right| = - \frac{880}{33} + \frac{595}{33} = - \frac{285}{33}$