Dadas as matrizes abaixo, calcule o que se pede:
A=\left[\begin{array}{cccc} 1 & -2 & 3 & 0 \\\\ 4 & 5 & -1 & 2 \\\\ -3 & 0 & 6 & 1 \\\\ 2 & 4 & -5 & 3 \\\\ 0 & 1 & 2 & 4 \end{array}\right], \quad B=\left[\begin{array}{cccc} 6 & 1 & -2 & 3 \\\\ 1 & 0 & 4 & 2 \\\\ 3 & -4 & 1 & 0 \\\\ 5 & -2 & 0 & 1 \\\\ 2 & 3 & -1 & 5 \end{array}\right]
(b) A B^{T}

Passo 1

O enunciado pede o produto da matriz A (de ordem 5\times4) pela matriz transposta de B, isto é, calcular
C \;=\; A\,B^{T}.
O resultado será uma matriz C de ordem 5\times5. Vamos primeiro obter B^{T} e depois usar a definição de multiplicação de matrizes.

Passo 2

A matriz B é de ordem 5\times4:
B = \begin{pmatrix} 6 & 1 & -2 & 3 \\\\ 1 & 0 & 4 & 2 \\\\ 3 & -4 & 1 & 0 \\\\ 5 & -2 & 0 & 1 \\\\ 2 & 3 & -1 & 5 \end{pmatrix}.
Sua transposta B^{T} é 4\times5 e tem como linhas as colunas de B:
B^{T} = \begin{pmatrix} 6 & 1 & 3 & 5 & 2\\\\ 1 & 0 & -4 & -2 & 3\\\\ -2 & 4 & 1 & 0 & -1\\\\ 3 & 2 & 0 & 1 & 5 \end{pmatrix}.

Passo 3

Pela definição, para obter C = A\,B^{T} (onde C é 5\times5) usamos
c_{ij} \;=\;\sum_{k=1}^{4} a_{ik}\,b^{T}_{kj}.
  1. Primeira linha de A: \,(1,-2,3,0). Então
\begin{aligned} c_{11}&=1\cdot6+(-2)\cdot1+3\cdot(-2)+0\cdot3=-2,\\\ c_{12}&=1\cdot1+(-2)\cdot0+3\cdot4+0\cdot2=13,\\\ c_{13}&=1\cdot3+(-2)\cdot(-4)+3\cdot1+0\cdot0=14,\\\ c_{14}&=1\cdot5+(-2)\cdot(-2)+3\cdot0+0\cdot1=9,\\\ c_{15}&=1\cdot2+(-2)\cdot3+3\cdot(-1)+0\cdot5=-7. \end{aligned}
Logo, a linha 1 de C é \,(-2,\,13,\,14,\,9,\,-7).
  1. Segunda linha de A: \,(4,5,-1,2). Calculando:
\begin{aligned} c_{21}&=4\cdot6+5\cdot1+(-1)\cdot(-2)+2\cdot3=37,\\\ c_{22}&=4\cdot1+5\cdot0+(-1)\cdot4+2\cdot2=4,\\\ c_{23}&=4\cdot3+5\cdot(-4)+(-1)\cdot1+2\cdot0=-9,\\\ c_{24}&=4\cdot5+5\cdot(-2)+(-1)\cdot0+2\cdot1=12,\\\ c_{25}&=4\cdot2+5\cdot3+(-1)\cdot(-1)+2\cdot5=34. \end{aligned}
Assim, a linha 2 de C é \,(37,\,4,\,-9,\,12,\,34).

Passo 4

  1. Terceira linha de A: \,(-3,0,6,1).
\begin{aligned} c_{31}&=-3\cdot6+0\cdot1+6\cdot(-2)+1\cdot3=-27,\\\ c_{32}&=-3\cdot1+0\cdot0+6\cdot4+1\cdot2=23,\\\ c_{33}&=-3\cdot3+0\cdot(-4)+6\cdot1+1\cdot0=-3,\\\ c_{34}&=-3\cdot5+0\cdot(-2)+6\cdot0+1\cdot1=-14,\\\ c_{35}&=-3\cdot2+0\cdot3+6\cdot(-1)+1\cdot5=-7. \end{aligned}
Linha 3 de C: \,(-27,\,23,\,-3,\,-14,\,-7).
  1. Quarta linha de A: \,(2,4,-5,3).
\begin{aligned} c_{41}&=2\cdot6+4\cdot1+(-5)\cdot(-2)+3\cdot3=35,\\\ c_{42}&=2\cdot1+4\cdot0+(-5)\cdot4+3\cdot2=-12,\\\ c_{43}&=2\cdot3+4\cdot(-4)+(-5)\cdot1+3\cdot0=-15,\\\ c_{44}&=2\cdot5+4\cdot(-2)+(-5)\cdot0+3\cdot1=5,\\\ c_{45}&=2\cdot2+4\cdot3+(-5)\cdot(-1)+3\cdot5=36. \end{aligned}
Linha 4 de C: \,(35,\,-12,\,-15,\,5,\,36).

Passo 5

  1. Quinta linha de A: \,(0,1,2,4).
\begin{aligned} c_{51}&=0\cdot6+1\cdot1+2\cdot(-2)+4\cdot3=9,\\\ c_{52}&=0\cdot1+1\cdot0+2\cdot4+4\cdot2=16,\\\ c_{53}&=0\cdot3+1\cdot(-4)+2\cdot1+4\cdot0=-2,\\\ c_{54}&=0\cdot5+1\cdot(-2)+2\cdot0+4\cdot1=2,\\\ c_{55}&=0\cdot2+1\cdot3+2\cdot(-1)+4\cdot5=21. \end{aligned}
Linha 5 de C: \,(9,\,16,\,-2,\,2,\,21).
Portanto, o produto A\,B^{T} é
C \;=\; \begin{pmatrix} -2 & 13 & 14 & 9 & -7\\\\ 37 & 4 & -9 & 12 & 34\\\\ -27 & 23 & -3 & -14 & -7\\\\ 35 & -12 & -15 & 5 & 36\\\\ 9 & 16 & -2 & 2 & 21 \end{pmatrix}.

Resposta

A\,B^{T} = \begin{pmatrix} -2 & 13 & 14 & 9 & -7\\\\ 37 & 4 & -9 & 12 & 34\\\\ -27 & 23 & -3 & -14 & -7\\\\ 35 & -12 & -15 & 5 & 36\\\\ 9 & 16 & -2 & 2 & 21 \end{pmatrix}.

Exercícios relacionados