• Matéria: Matemática
  • Autor: Ryansss
  • Perguntado 9 anos atrás

Atividade de matemática 2 questões

Anexos:

Respostas

respondido por: deividsilva784
0
 \\ X-3B =  l_{3} 
 \\ 
 \\ X = 3B+ l_{3}
-----------------------------

Substituindo os valores das matrizes:

 \\ X = 3  \left[\begin{array}{ccc}1&5&0\\-1&-2&3\\7&0&2\end{array}\right] +  \left[\begin{array}{ccc}1&0&0\\0&1&0\\0&0&1\end{array}\right] 
 \\ 
 \\ 
 \\ X =   \left[\begin{array}{ccc}3&15&0\\-3&-6&9\\21&0&6\end{array}\right] + \left[\begin{array}{ccc}1&0&0\\0&1&0\\0&0&1\end{array}\right] 
 \\ 
 \\ 
 \\ X =   \left[\begin{array}{ccc}4&15&0\\-3&-5&9\\21&0&7\end{array}\right]

2)

A₃ₓ₃ = j -i

a₁₁ = 1-1 =0              a₁₂ = 2- 1 = 1             a₁₃ = 3-1 =2
a₁₂ = 2-1 = 1             a₂₂ = 2-2 =0                a₂₃ = 3-2 =1
a₁₃ = 3-1 = 2             a₃₂ = 2-3 = -1              a₃₃ = 3-3 = 0

 \\ -X =  \frac{1}{2} (A- l_{3} )
 \\ 
 \\ X = -\frac{1}{2} (A- l_{3} )
 \\ 
 \\ 
 \\ X =  -\frac{1}{2}(   \left[\begin{array}{ccc}0&1&3\\1&0&-1\\2&1&0\end{array}\right] +  \left[\begin{array}{ccc}1&0&0\\0&1&0\\0&0&1\end{array}\right] )
 \\ 
 \\ 
 \\ X =  -\frac{1}{2}  \left[\begin{array}{ccc}1&1&3\\1&1&-1\\2&1&1\end{array}\right] 
 \\ 
 \\ 
 \\ X =   \left[\begin{array}{ccc} -1/2 &-1/2&-3/2\\-1/2&-1/2&1/2\\-1&-\frac{1}{2}&-1/2\end{array}\right]
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