Question

Soma subtração e multiplicação de matrizes. Deverá fazer
A-B
A+B
B-A
A.B
B.A

Answer 1

$\tt{A}=\begin{pmatrix}\sf{5}&\sf{7}&\sf{-1}\\\sf{6} &\sf{0} &\sf{-3}\\\sf{-4}&\sf{3}&\sf{0}\end{pmatrix}$

$\tt{B}=\begin{pmatrix}\sf{0}&\sf{3}&\sf{-5}\\\sf{2} &\sf{0} &\sf{0}\\\sf{-1}&\sf{-5}&\sf{3}\end{pmatrix}$

$\tt{A+B}\begin{pmatrix}\sf{5+0}&\sf{7+3}&\sf{-1-5}\\\sf{6+2}&\sf{0+0}&\sf{-3+0}\\\sf{-4-1}&\sf{3-5}&\sf{0+3}\end{pmatrix}$

$\tt{A+B}=\begin{pmatrix}\sf{5}&\sf{10}&\sf{-6}\\\sf{8}&\sf{0} &\sf{-3}\\\sf{-5}&\sf{-2}&\sf{3}\end{pmatrix}$

$\tt{A-B}=\begin{pmatrix}\sf{5-0}&\sf{7-3}&\sf{-1+5}\\\sf{6-2} &\sf{0-0} &\sf{-3-0}\\\sf{-4+1}&\sf{3+5}&\sf{0+3}\end{pmatrix}$

$\tt{A-B}=\begin{pmatrix}\sf{5}&\sf{4}&\sf{4}\\\sf{4} &\sf{0} &\sf{-3}\\\sf{-3}&\sf{8}&\sf{3}\end{pmatrix}$

$\tt{B-A=-(A-B)}\begin{pmatrix}\sf{-5}&\sf{-4}&\sf{-4}\\\sf{-4} &\sf{0} &\sf{3}\\\sf{3}&\sf{-8}&\sf{-3}\end{pmatrix}$

$\sf{A\cdot B=}\begin{pmatrix}\sf{5\cdot0+7\cdot2-1\cdot(-1)}&\sf{5\cdot3+7\cdot0-1\cdot(-5)}&\sf{5\cdot(-5)+7\cdot0-1\cdot3}\\\sf{6\cdot0+0\cdot2-3\cdot(-1)}&\sf{6\cdot3+0\cdot0-3\cdot(-5)}&\sf{6\cdot(-5)+0\cdot0-3\cdot3}\\\sf{-4\cdot0+3\cdot2+0\cdot(-1)}&\sf{-4\cdot3+3\cdot0+0\cdot(-5)}&\sf{-4\cdot(-5)+3\cdot0+0\cdot3}\end{pmatrix}$

$\tt{A\cdot B}\begin{pmatrix}\sf{15}&\sf{20}&\sf{-28}\\\sf{3}&\sf{33}&\sf{-39}\\\sf{6}&\sf{-12} &\sf{20}\end{pmatrix}$

$\tt{B\cdot A}=\begin{pmatrix}\sf{0\cdot5+3\cdot6-5\cdot(-4)}&\sf{0\cdot7+3\cdot0-5\cdot3}&\sf{0\cdot(-1)+3\cdot(-3)-5\cdot0}\\\sf{2\cdot5+0\cdot6+0\cdot(-4)}&\sf{2\cdot7+0\cdot0+0\cdot3}&\sf{2\cdot(-1)+0\cdot(-3)+0\cdot0}\\\sf{-1\cdot5-5\cdot6+3\cdot(-4)}&\sf{-1\cdot7-5\cdot0+3\cdot3}&\sf{-1\cdot(-1)-5\cdot(-3)+3\cdot0}\end{pmatrix}$

$\tt{B\cdot A}=\begin{pmatrix}\sf{38}&\sf{-15}&\sf{9}\\\sf{10}&\sf{14}&\sf{-2}\\\sf{-47}&\sf{2}&\sf{16}\end{pmatrix}$