Question

calcule o determinante

Answer 1

$\left[\begin{array}{ccc}cos\: 45&sen\: 30\\sen\: 60&cos\: 60\end{array}\right]_2 \\ \\ \\ Determinante \\ \\ \left|\begin{array}{ccc} \frac{ \sqrt{2} }{2} & \frac{1}{2} \\ \frac{ \sqrt{3} }{2} & \frac{1}{2} \end{array}\right| \\ \\ \\ det\: A = \frac{ \sqrt{2} }{4} - \frac{ \sqrt{3} }{4} \\ \\ \\ \left[\begin{array}{ccc}x&3&-1\\x&y&z\\-1+x&4+x&2+c\end{array}\right]_3 \\ \\ \\ Determinante \\ \\ \left|\begin{array}{ccc}x&3&-1\\x&y&z\\-1+x&4+x&2+c\end{array}\right|$

Aplicarei o Teorema de Laplace (Cofator):

$A_i_j = (- 1)^{i + j}\: . \: D_i_j \\ \\ A_1_1 = 1\: .\: \left|\begin{array}{ccc}y&z\\4+ y&2+c\end{array}\right| \\ \\ A_1_1 = 4z + zy - 2y + cy \\ \\ \\ A_1_2 = (-1)^{1 + 2}\: . \: D_1_ 2 \\ \\ A_1_2 = - 1\: . \: \left|\begin{array}{ccc}x&z\\-1+x&2+c\end{array}\right| \\ \\ A_1_2 = (-1).(2x + cx + z - x) \\ A_1_2 = - 2x - cx - z + x \\ \\ \\ A_1_3 = 1\: . \left|\begin{array}{ccc}x&y\\-1+x&4+y\end{array}\right| \\ \\ A_1_3 = 4x + \cancel{xy} + y - \cancel{xy} \\ A_1_3 = 4x + y$

det A = a11.A11 + a12.A12 + a13.A13

Faça o cálculo