Question

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Answer 1

Explicação passo-a-passo:

9)

$\left|\begin{array}{ccc}x+2&2x-1\\3&4\\\end{array}\right| =\left|\begin{array}{ccc}x&2\\8&3\\\end{array}\right|\\\\4.(x+2)-3.(2x-1)=3x-8.2\\\\4x+8-6x+3=3x-16\\\\-2x+11=3x-16\\\\3x+2x=11+16\\\\5x=27\\\\x=\dfrac{27}{5}\\\\\boxed{\boxed{x=5,4}}$

10)Temos as seguintes matrizes:

$A=\left[\begin{array}{ccc}-1&3\\7&4\\\end{array}\right]\ \ \ \text e\ \ \ B= \left[\begin{array}{ccc}2&1\\-3&0\\\end{array}\right]$

a)Calculando A+B:

$A+B=\left[\begin{array}{ccc}-1&3\\7&4\\\end{array}\right]+\left[\begin{array}{ccc}2&1\\-3&0\\\end{array}\right]\\\\A+B=\left[\begin{array}{ccc}-1+2&3+1\\7+(-3)&4+0\\\end{array}\right]\\\\\boxed{A+B=\left[\begin{array}{ccc}1&4\\4&4\\\end{array}\right]}$

Calculando o determinante:

$det(A+B)=\left|\begin{array}{ccc}1&4\\4&4\\\end{array}\right|\\\\det(A+B)=1.4-4.4\\\\det(A+B)=4-16\\\\\boxed{\boxed{det(A+B)=-12}}$

b)As matrizes transpostas de A e de B são:

$\boxed{A^{t} =\left[\begin{array}{ccc}-1&7\\3&4\\\end{array}\right]\ \ \ \text e\ \ \ B^{t} = \left[\begin{array}{ccc}2&-3\\1&0\\\end{array}\right]}$

Calculando o determinante:

$det(A^{t})+det(B^{t})=\left|\begin{array}{ccc}-1&7\\3&4\\\end{array}\right|+\left|\begin{array}{ccc}2&-3\\1&0\\\end{array}\right|\\\\det(A^{t})+det(B^{t})=-1.4-7.3+2.0-(-3).1\\\\det(A^{t})+det(B^{t})=-4-21+0+3\\\\\boxed{\boxed{det(A^{t})+det(B^{t})=-22}}$

c)Calculando a matriz 3.A:

$3.A=3.\left[\begin{array}{ccc}-1&3\\7&4\\\end{array}\right]\\\\3.A=\left[\begin{array}{ccc}3.(-1)&3.3\\3.7&3.4\\\end{array}\right]\\\\\boxed{3.A=\left[\begin{array}{ccc}-3&9\\21&12\\\end{array}\right]}$

Calculando o determinante:

$det(3.A)=\left|\begin{array}{ccc}-3&9\\21&12\\\end{array}\right|\\\\det(3.A)=-3.12-9.21\\\\det(3.A)=-36-189\\\\\boxed{\boxed{det(3.A)=-225}}$

d)Calculando o determinante:

$detA+detB=\left|\begin{array}{ccc}-1&3\\7&4\\\end{array}\right|+\left|\begin{array}{ccc}2&1\\-3&0\\\end{array}\right|\\\\detA+detB=-1.4-3.7+2.0-1.(-3)\\\\detA+detB=-4-21+0+3\\\\\boxed{\boxed{detA+detB=-22}}$

e)Calculando a transposta de A+B(item a):

$(A+B)^{t} =\left[\begin{array}{ccc}1&4\\4&4\\\end{array}\right]$

Calculando o determinante:

$det(A+B)^{t} =\left|\begin{array}{ccc}1&4\\4&4\\\end{array}\right|\\\\det(A+B)^{t}=1.4-4.4\\\\det(A+B)^{t}=4-16\\\\\boxed{\boxed{det(A+B)^{t}=-12}}$