Exercício de Multiplicação de Matrizes

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Exercício de Multiplicação de Matrizes 
Seja 𝐴 = (
𝑐𝑜𝑠15° −𝑠𝑒𝑛15°
𝑠𝑒𝑛15° 𝑐𝑜𝑠15°
), calcule o valor de 2(𝐴 ⋅ 𝐴). 
Solução: 
Tomando o produto 𝐴 ⋅ 𝐴 = (
𝑐𝑜𝑠15° −𝑠𝑒𝑛15°
𝑠𝑒𝑛15° 𝑐𝑜𝑠15°
) ⋅ (
𝑐𝑜𝑠15° −𝑠𝑒𝑛15°
𝑠𝑒𝑛15° 𝑐𝑜𝑠15°
), tal produto (𝐴 ⋅ 𝐴) 
será a matriz (
𝑎11 𝑎12
𝑎21 𝑎22
). 
 
Fazemos: 
 
𝐴 ⋅ 𝐴 = (
𝑐𝑜𝑠15° −𝑠𝑒𝑛15°
𝑠𝑒𝑛15° 𝑐𝑜𝑠15°
) ⋅ (
𝑐𝑜𝑠15° −𝑠𝑒𝑛15°
𝑠𝑒𝑛15° 𝑐𝑜𝑠15°
) 
 
 
Seguindo o esquema acima para calcular o produto 𝐴 ⋅ 𝐴, temos: 
 
𝑎11 = [(𝑐𝑜𝑠15°) ⋅ (𝑐𝑜𝑠15°)] + [(−𝑠𝑒𝑛15°) ⋅ (𝑠𝑒𝑛15°)] 
𝑎11 = 𝑐𝑜𝑠
215° + (−𝑠𝑒𝑛215) 
𝑎11 = 𝑐𝑜𝑠
215° − 𝑠𝑒𝑛215° 
Sabendo que 𝑐𝑜𝑠2𝜃 − 𝑠𝑒𝑛2𝜃 = 𝑐𝑜𝑠2𝜃, temos: 
𝑎11 = 𝑐𝑜𝑠2 ⋅ 15° 
𝑎11 = 𝑐𝑜𝑠30° 
 
𝑎12 = [(𝑐𝑜𝑠15°) ⋅ (−𝑠𝑒𝑛15°)] + [(−𝑠𝑒𝑛15°) ⋅ (𝑐𝑜𝑠15°)] 
𝑎12 = −𝑠𝑒𝑛15°𝑐𝑜𝑠15° + (−𝑠𝑒𝑛15°𝑐𝑜𝑠15°) 
𝑎12 = −𝑠𝑒𝑛15°𝑐𝑜𝑠15° − 𝑠𝑒𝑛15°𝑐𝑜𝑠15° 
𝑎12 = −2𝑠𝑒𝑛15°𝑐𝑜𝑠15° 
Sabendo que 2 ⋅ 𝑠𝑒𝑛𝜃 ⋅ 𝑐𝑜𝑠𝜃 = 𝑠𝑒𝑛2𝜃, temos: 
𝑎12 = −𝑠𝑒𝑛2 ⋅ 15° 
𝑎12 = −𝑠𝑒𝑛30° 
 
 
𝑎21 = [(𝑠𝑒𝑛15°) ⋅ (𝑐𝑜𝑠15°)] + [(𝑐𝑜𝑠15°) ⋅ (𝑠𝑒𝑛15°)] 
𝑎21 = 𝑠𝑒𝑛15°𝑐𝑜𝑠15° + 𝑠𝑒𝑛15°𝑐𝑜𝑠15° 
𝑎21 = 2𝑠𝑒𝑛15°𝑐𝑜𝑠15° 
Sabendo que 2 ⋅ 𝑠𝑒𝑛𝜃 ⋅ 𝑐𝑜𝑠𝜃 = 𝑠𝑒𝑛2𝜃, temos: 
𝑎21 = 𝑠𝑒𝑛2 ⋅ 15° 
𝑎21 = 𝑠𝑒𝑛30° 
 
𝑎22 = [(𝑠𝑒𝑛15°)(−𝑠𝑒𝑛15°)] ⋅ [(𝑐𝑜𝑠15°) ⋅ (𝑐𝑜𝑠15°)] 
𝑎22 = 𝑐𝑜𝑠
215° − 𝑠𝑒𝑛215° 
Sabendo que 𝑐𝑜𝑠2𝜃 − 𝑠𝑒𝑛2𝜃 = 𝑐𝑜𝑠2𝜃, temos: 
𝑎22 = 𝑐𝑜𝑠2 ⋅ 15° 
𝑎22 = 𝑐𝑜𝑠30° 
 
Neste caso, temos: 
𝐴 ⋅ 𝐴 = (
𝑎11 𝑎12
𝑎21 𝑎22
) ⟹ 𝐴 ⋅ 𝐴 = (
𝑐𝑜𝑠30° −𝑠𝑒𝑛30°
𝑠𝑒𝑛30° 𝑐𝑜𝑠30°
) 
 
Sabendo que e , temos: 
 
𝐴 ⋅ 𝐴 = (
𝑐𝑜𝑠30° −𝑠𝑒𝑛30°
𝑠𝑒𝑛30° 𝑐𝑜𝑠30°
) ⟹ 𝐴 ⋅ 𝐴 =
(
 
 
√3
2
−
1
2
1
2
√3
2 )
 
 
 
 
Calculando, agora, o valor de 2(𝐴 ⋅ 𝐴), temos: 
 
2(𝐴 ⋅ 𝐴) = 2 ⋅
(
 
 
√3
2
−
1
2
1
2
√3
2 )
 
 
=
(
 
 2 ⋅
√3
2
2 ⋅ (−
1
2
)
2 ⋅
1
2
2 ⋅
√3
2 )
 
 
=
(
 
 2 ⋅
√3
2
2 ⋅ (−
1
2
)
2 ⋅
1
2
2 ⋅
√3
2 )
 
 
= (√
3 −1
1 √3
) 
𝑐𝑜𝑠30° =
√3
2
 𝑠𝑒𝑛30° =
1
2
 
Portanto, 2(𝐴 ⋅ 𝐴) = (√
3 −1
1 √3
).