Question

a solução da equação abaixo é:

[2 3 -2]
[0 1 x] =2
[2 x -3]​

Answer 1

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$\begin{bmatrix}\sf{2}&\sf{3}&\sf{-2}\\\sf{0}&\sf{1}&\sf{x}\\\sf{2}&\sf{x}&\sf{-3}\end{bmatrix}=\sf{2}\\\sf{2}\cdot\begin{bmatrix}\sf{1}&\sf{x}\\\sf{x}&\sf{-3}\end{bmatrix}-\sf{3}\cdot\begin{bmatrix}\sf{0}&\sf{x}\\\sf{2}&\sf{-3}\end{bmatrix}-\sf{2}\cdot\begin{bmatrix}\sf{0}&\sf{1}\\\sf{2}&\sf{x}\end{vmatrix}=\sf{2}\\\sf{2\cdot(-3-x^2)-3\cdot(-2x)-2\cdot(-2)=2}\\\sf{-6-2x^2+6x+4=2}\\\sf{-2x^2+6x-2-2=0}\\\sf{-2x^2+6x}-4=0\div(-2)}\\\sf{x^2-3x+2=0}$

$\sf{\Delta=b^2-4ac}\\\sf{\Delta=(-3)^2-4\cdot1\cdot2}\\\sf{\Delta=9-8}\\\sf{\Delta=1}\\\sf{x=\dfrac{-b\pm\sqrt{\Delta}}{2a}}\\\sf{x=\dfrac{-(-3)\pm\sqrt{1}}{2\cdot1}}\\\sf{x=\dfrac{3\pm1}{2}}\begin{cases}\sf{x_1=\dfrac{3+1}{2}=\dfrac{4}{2}=2}\\\sf{x_2=\dfrac{3-1}{2}=\dfrac{2}{2}=1}\end{cases}\\\huge\boxed{\boxed{\boxed{\boxed{\sf{s=\{1,2\}}}}}}$