Question

Resuelve los siguientes sistemas de 3×3 con el método de determinantes.

Answer 1

1-) x/3 + y = 2z + 3

   x - y = 1

   x + z = y/4 + 11

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x + 3y - 6z = 9

x - y = 1

4x - y + 4z = 44

$D=\left[\begin{array}{ccc}1&3&-6\\1&-1&0\\4&-1&4\end{array}\right]$

D₁=$\left[\begin{array}{ccc}9&3&-6\\1&-1&0\\44&-1&4\end{array}\right]$

D₂=$\left[\begin{array}{ccc}1&9&-6\\1&1&0\\4&44&4\end{array}\right]$

D₃=$\left[\begin{array}{ccc}1&3&9\\1&-1&1\\4&-1&44\end{array}\right]$

D = -34

D₁ = -306 /34

D₂ = -272 /34

D₃ = -136 /34

x = 9

y = 8

z = 4

2-) 6x/5 + y + z - 4= 19/5

    z - 4 + 6x - 19/5 = -y

    4z + 3y = 3x - y

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6x + 5y + 5z = 39

6x + 5y + 5z = 39

-3x + 4y + 4z = 0

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$\left \{ {{6x+5y+5z=39} \atop {-3x+4y+4z=0}} \right.$

$\left[\begin{array}{ccc}6&5&5/39\\-3&4&4/0\\\end{array}\right]$

$\left[\begin{array}{ccc}0&13&13/39\\-3&4&4/0\\\end{array}\right]$

$\left[\begin{array}{ccc}0&1&1/3\\-3&0&0/-12\\\end{array}\right]$

$\left[\begin{array}{ccc}0&1&1/3\\1&0&0/4\\\end{array}\right]$

$\left \{ {{y+z=3} \atop {x=4}} \right.$

$\left \{ {{y=3-z} \atop {x=4}} \right.$

x = 4

x = 3-z

z= z

z ∈ R

Saludos :D