Question

Dadas as matrizes $A\left[\begin{array}{ccc}x-1&2&x\\0&1&-1\\3x&x+1&2x\end{array}\right]$ e $B\left[\begin{array}{ccc}3x&2x\\4&-x\\\end{array}\right]$ , determine o valor de x para que se tenha detA= detB.

Answer 1

Os valores de x para que se tenha $det~A=det~B$ são $\dfrac{\sqrt{3}}{3}$ e $-\dfrac{\sqrt{3}}{3}$

• Resolvendo o problema

Para facilitar o cálculo do det A, vamos copiar as duas primeiras colunas da matriz A e colocá-las à direita das já existentes (ver imagem anexa) obtendo

$\left[\begin{array}{rrrrr}x-1&2&x&x-1&2\\0&1&-1&0&1\\3x&x+1&2x&3x&x+1\end{array}\right]$

Assim,

$\begin{array}{lclclclc}det~A&=&(x-1)~.~1~.~2x&+&2~.~-1~.~3x&+&x~.~0~.~(x+1)&-\\&\left[ \right.&x~.~1~.~3x&+&(x-1)~.~-1~.~(x+1)&+& 2~.~0~.~2x&\left.\right]\end{array}\\\\\\\begin{array}{lclclclc}det~A&=&2x^2-2x&-&6x&+&0&-\\&\left[ \right.&3x^2&-&(x-1)~.~(x+1)&+&0&\left.\right]\end{array}\\\\\\det~A=2x^2-2x-6x-[3x^2-(x-1).(x+1)]\\\\det~A=2x^2-8x-3x^2+(x-1).(x+1)\\\\det~A=2x^2-8x-3x^2+x^2+x-x-1\\\\det~A=2x^2-3x^2+x^2-8x+x-x-1\\\\det~A=-8x-1$

Para B temos

$det~B=3x~.-x-(2x~.~4)\\\\det~B=-3x^2-8x$

Igualando os dois valores

$det~A= det~B\\\\-8x-1=-3x^2-8x\\\\-8x-1+3x^2+8x=0\\\\-1+3x^2=0\\\\3x^2=1\\\\x^2=\dfrac{1}{3}\\\\x=\pm\sqrt{\dfrac{1}{3}}\\\\x=\pm\dfrac{1}{\sqrt{3}}\\\\x=\pm\dfrac{1}{\sqrt{3}}~.~\dfrac{\sqrt{3}}{\sqrt{3}}\\\\\\\boxed{\boxed{x=\pm\dfrac{\sqrt{3}}{3}}}$

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