Question

da uma ajuda aí :-\ ajuda mt​

Answer 1

Resposta:

50

Explicação passo a passo:

$A_{i\times n} = A_{3\times2} = \left[\begin{array}{cc}2&-1\\-2&2\\0&1\end{array}\right]\\B_{n\times j} = A_{2\times3} = \left[\begin{array}{ccc}-1&2&3\\2&1&1\end{array}\right] \\N = 50 + det(AB)$

$AB = P_{ij}} = \left[\begin{array}{ccc}p_{11}&p_{12}&p_{13}\\p_{21}&p_{22}&p_{23}\\p_{31}&p_{32}&p_{33}\end{array}\right] \\p_{ij} = \sum\limits_{k=1}^{n}a_{ik}b_{kj}\\p_{11} =\sum\limits_{k=1}^{n}a_{1k}b_{k1} = a_{11}b_{11} + a_{12}b_{21}\\p_{11} = (2)(-1) + (-1)(2) = -2 -2 = -4\\p_{12} =\sum\limits_{k=1}^{n}a_{1k}b_{k2} = a_{11}b_{12} + a_{12}b_{22}\\p_{12} = (2)(2) + (-1)(1) = 4 - 1 = 3$

$p_{13} =\sum\limits_{k=1}^{n}a_{1k}b_{k3} = a_{11}b_{13} + a_{12}b_{23}\\p_{13} = (2)(3) + (-1)(1) = 6 - 1 = 5\\p_{21} =\sum\limits_{k=1}^{n}a_{2k}b_{k1} = a_{21}b_{11} + a_{22}b_{21}\\p_{21} = (-2)(-1) + (2)(2) = 2 + 4 = 6\\p_{22} =\sum\limits_{k=1}^{n}a_{2k}b_{k2} = a_{21}b_{12} + a_{22}b_{22}\\p_{22} = (-2)(2) + (2)(1) = -4 + 2 = -2\\p_{23} =\sum\limits_{k=1}^{n}a_{2k}b_{k3} = a_{21}b_{13} + a_{22}b_{23}\\p_{23} = (-2)(3) + (2)(1) = -6 + 2 = -4$

$p_{31} =\sum\limits_{k=1}^{n}a_{3k}b_{k1} = a_{31}b_{11} + a_{32}b_{21}\\p_{31} = (0)(-1) + (1)(2) = 0 + 2 = 2\\p_{32} =\sum\limits_{k=1}^{n}a_{3k}b_{k2} = a_{31}b_{12} + a_{32}b_{22}\\p_{32} = (0)(2) + (1)(1) = 0 + 1 = 1\\p_{33} =\sum\limits_{k=1}^{n}a_{3k}b_{k3} = a_{31}b_{13} + a_{32}b_{23}\\p_{33} = (0)(3) + (1)(1) = 0 + 1 = 1$

$P = \left[\begin{array}{ccc}-4&3&5\\6&-2&-4\\2&1&1\end{array}\right]\begin{array}{ccc}-4&3\\6&-2\\2&1\end{array}\\det(P) = -[(5)(-2)(2)] - [(-4)(-4)(1)]-[(3)(6)(1)]+[(-4)(-2)(1)]+[(3)(-4)(2)]+[(5)(6)(1)]\\det(P) = -[-20] - [16] - [18] + [8] + [-24] + [30]\\det(P) = 20 - 16 -18 + 8 - 24 + 30\\det(P) = 0$

$N = 50 + det(P)\\N = 50 + 0\\N = 50$