Question

ALGUÉM ME AJUDA PELO AMOR DE DEUS

Answer 1

1)
a) Matriz quadrada de ordem 2
b) Matriz quadrada de ordem 2, matriz nula 
c) Matriz quadrada de ordem 3, matriz identidade, matriz diagonal, matriz simétrica. 

2)
a)

$A = \left[\begin{array}{cccc} a_{11} &a_{12}&a_{13}&a_{14}\\a_{21}&a_{22}&a_{23}&a_{24}\\a_{31}&a_{32}&a_{33}&a_{34}\end{array}\right]$

$a_{11}= 1 + 2.1 = 1 + 2 = 3 \\ a_{12} = 2 + 2 = 4 \\ a_{13} = 2 + 3 = 5 \\ a_{14} = 2 + 4 = 6 \\ a_{21} = 2 + 1 = 3 \\ a_{22} = 1 + 2 . 2 = 1 + 4 = 5 \\ a_{23} = 2 + 3 = 5 \\ a_{24} = 2 + 4 = 6 \\ a_{31} = 2 + 1 = 3 \\ a_{32} = 2 + 2 = 4 \\ a_{33} = 1 + 2 . 3 = 1 + 6 = 7 \\ a_{34} = 2 + 4 = 6$

Logo: $A = \left[\begin{array}{cccc} 3 &4&5&6\\3&5&5&6\\3&4&7&6\end{array}\right]$

b) 

$B = \left[\begin{array}{cc} b_{11}&b_{12}\\b_{21}&b_{22}\\b_{31}&b_{32}\end{array}\right]$

$b_{11} = 1 - 1 = 0 \\ b_{12} = 2 . 2 - 1 = 4 - 1 = 3\\ b_{21} = 2 . 1 - 2 = 2 - 2 = 0 \\ b_{22}= 2 - 2 = 0 \\ b_{31} = 2 . 1 - 3 = 2 - 3 = -1 \\ b_{32} = 2 . 2 - 3 = 4 - 3 = 1$

Logo,  $B = \left[\begin{array}{cc} 0&3\\0&0\\-1&1\end{array}\right]$

3) A matriz será do tipo:

$A = \left[\begin{array}{cc} a_{11}&a_{12}\\a_{21}&a_{22}\\a_{31}&a_{32}\end{array}\right]$

onde:

$a_{12} = 5 \\ a_{31} = 7 \\ a_{11} = 1 \\ a_{22} = 1$

Os demais são iguais a zero. Assim, temos:

$A = \left[\begin{array}{cc} 1&5\\0&1\\7&0\end{array}\right]$

4)

$C = \left[\begin{array}{cc}-2&3\\7&-5\\ \frac{1}{2} &-1\\1&2,5\end{array}\right] \\ \\ 3a_{31} - 5a_{42} \\ 3 . \frac{1}{2} - 5.2,5 \\ \frac{3}{2} - 12,5 \\ 1,5 - 12,5 \\ -11$

5)

$C = \left[\begin{array}{cc} a_{11} &a_{12}\\a_{21}&a_{22}\\ \end{array}\right] \\ \\ a_{11} = 1 \\ a_{12} = 0 \\ a_{21} =0\\ a_{22} = 1 \\ \\ C = \left[\begin{array}{cc} 1 &0\\0&1\\ \end{array}\right]$

a) V
b) F
c) V
d) V

6) 

$M = \left[\begin{array}{ccc} a_{11} &a_{12}&a_{13}\\a_{21}&a_{22}&a_{23}\\a_{31}&a_{32}&a_{33}\end{array}\right] \\ \\ a_{11} = 1 + 1 = 2 \\ a_{12} = 0 \\ a_{13} = 0\\ a_{21} = 0\\ a_{22} = 2 + 2 = 4 \\ a_{23} = 0 \\ a_{31} = 0 \\ a_{32} = 0 \\ a_{33} = 3 + 3 = 6 \\ \\ M = \left[\begin{array}{ccc} 2 &0&0\\0&4&0\\0&0&6\end{array}\right]$

Matriz transposta é aquela onde as linhas se transformam em colunas:

$M^{T} =\left[\begin{array}{ccc} 2 &0&0\\0&4&0\\0&0&6\end{array}\right]$

7) 

Como bij = 3i-j² então

$b_{22} = 3 . 2 - 2^{2} = 6 - 4 = 2$

8)

$A = \left[\begin{array}{ccccc} a_{11} &a_{12}&a_{13}&a_{14}&a_{15}\\a_{21}&a_{22}&a_{23}&a_{24}&a_{25}\\a_{31}&a_{32}&a_{33}&a_{34}&a_{35}\\a_{41}&a_{42}&a_{43}&a_{44}&a_{45}\\a_{51}&a_{52}&a_{53}&a_{54}&a_{55}\end{array}\right] \\ \\ a_{11} = 1 - 1 = 0 \\ a_{22} = 2 - 2 = 0 \\ a_{33} = 3 - 3 = 0\\ a_{44} =4 - 4 = 0 \\ a_{55} = 5 - 5 = 0$

Como a matriz A é diagonal, es elementos fora da diagonal principal é igual a zero, assim:

$A = \left[\begin{array}{ccccc} 0 &0&0&0&0\\0&0&0&0&0\\0&0&0&0&0\\0&0&0&0&0\\0&0&0&0&0\end{array}\right]$

9)

$\left[\begin{array}{cccc}1&-2&3&4\end{array}\right] + \left[\begin{array}{cccc}5&3&-5&4\end{array}\right] = \left[\begin{array}{cccc}X&Y&Z&T\end{array}\right] \\ \\ \left[\begin{array}{cccc}1+5&-2+3&3-5&4+4\end{array}\right] = \left[\begin{array}{cccc}X&Y&Z&T\end{array}\right] \\ \\ \left[\begin{array}{cccc}6&1&-2&8\end{array}\right] = \left[\begin{array}{cccc}X&Y&Z&T\end{array}\right] \\ \\ X = 6 \\ Y = 1 \\ Z = -2 \\ T = 8$

Espero ter ajudado...