Question

determine a matriz A=(aij) 4×2 sendo aij= 3i-j+1?​

Answer 1

Resposta:

$\sf A=(a_{ij})_{4x2}$

Na qual:

$\sf a_{ij}=3i-j+1$

Nossa matriz tem a seguinte forma:

$\sf A=\left[\begin{array}{cc}\sf a_{11}&\sf a_{12}\\\sf a_{21}&\sf a_{22}\\\sf a_{31}&\sf a_{32}\\\sf a_{41}&\sf a_{42}\end{array}\right]$

Pela lei de formação imposta:

$\sf a_{11}=3\cdot1-1+1=3-0=3$

$\sf a_{12}=3\cdot1-2+1=3-1=2$

$\sf a_{21}=3\cdot2-1+1=6-0=6$

$\sf a_{22}=3\cdot2-2+1=6-1=5$

$\sf a_{31}=3\cdot3-1+1=9-0=9$

$\sf a_{32}=3\cdot3-2+1=9-1=8$

$\sf a_{41}=3\cdot4-1+1=12-0=12$

$\sf a_{42}=3\cdot4-2+1=12-1=11$

Portanto:

$\red{\boldsymbol{\sf A=\left[\begin{array}{cc}\sf 3&\sf 2\\\sf 6&\sf 5\\\sf 9&\sf 8\\\sf 12&\sf 11\end{array}\right]}}$

Answer 2

Resposta no final:

$A=\left[\begin{array}{cccc}a__1_1&a_1_2&a_1_3&a_1_4\\a_2_1&a_2_2&a_2_3&a_2_4\\\end{array}\right] \\\\\\a_1_1=3.1-1+1=3\\\\a_1_2=3.1-2+1=3-1=2\\\\a_1_3=3.1-3+1=3-3+1=1\\\\a_1_4=3.1-4+1=3-4+1=0\\\\A_2_1=3.2-1+1=6\\\\a_2_2=3.2-2+1=6-1=5\\\\a_2_3=3.2-3+1=6-2=4\\\\a_2_4=3.2-4+1=6-3=3\\\\\\A=\left[\begin{array}{cccc}3&2&1&0\\6&5&4&3\\\end{array}\right]$