Question

Matrizes (Matriz transposta) Me ajudem por favor ! 55 pontos$Sendo = (A.B)^t Sendo = A^t . B^t  \left[\begin{array}{ccc}3&2&\\5&1&\\&&\end{array}\right] \left[\begin{array}{ccc}3&-1&\\2&0&\\&&\end{array}\right]$

Answer 1

matriz transposta: basta inverter i com j (trocar linhas pelas colunas):
$A= \left[\begin{array}{cc}a&b\\c&d\end{array}\right] \\\\A^t=\left[\begin{array}{cc}a&c\\b&d\end{array}\right]$
tendo:
$A=\left[\begin{array}{cc}3&2\\5&1\end{array}\right] \\\\B=\left[\begin{array}{cc}3&-1\\2&0\end{array}\right]$
então:
a)
$(A\cdot B)^t$
$\left(\left[\begin{array}{cc}3&2\\5&1\end{array}\right] \cdot\left[\begin{array}{cc}3&-1\\2&0\end{array}\right]\right)^t =\left[\begin{array}{cc}([3\cdot3]+[2\cdot2])&([3\cdot-1]+[2\cdot0])\\([5\cdot3]+[1\cdot2])&([5\cdot-1]+[1\cdot0])\end{array}\right]^t=\\\\\left[\begin{array}{cc}(9+4)&(-3+0)\\(15+2)&(-5+0)\end{array}\right]^t=\left[\begin{array}{cc}13&-3\\17&-5\end{array}\right]^t=\boxed{(A\cdot B)^t=\left[\begin{array}{cc}13&17\\-3&-5\end{array}\right]}$