Question

a matriz de B=(bij)3x3 tal que bij= [i+2j, se i$\neq$j]
                                                  [i-3j, se i=j]

Answer 1

$\left[\begin{array}{ccc}b11&b12&b13\\b21&b22&b23\\b31&b32&b33\end{array}\right] \\ \\ i \neq j: b12 , b13 , b21 , b23 , b31 , b32 \\ i=j :b11 , b22 , b33 \\ \\ \\bij = i + 2j (i \neq j)\\ b12 = 1 + 2.2 = 1 + 4 = 5 \\ b13 = 1 + 2 . 3 = 1 + 6 = 7 \\ b21 = 2 + 2 . 1 = 2 + 2 = 4 \\ b23 = 2 + 2 .3 = 2 + 6 = 8 \\ b31 = 3 + 2 . 1 = 3 + 2 = 5 \\ b32 = 3 + 2 . 2 = 3 + 4 = 7$

$bij = i - 3j ( i = j ) \\ b11=1 - 3 . 1 = 1 - 3 = -2 \\ b22 = 2 - 3 . 2 = 2 - 6 = -4 \\ b33 = 3 - 3.3 = 3 - 9 = -6$


$matriz : \\ \\ \left[\begin{array}{ccc}-2&5&7\\4&-4&8\\5&7&-6\end{array}\right]$