Question

Me ajudem??? Questão sobre matriz!
É URGENTE!!!!!!

Answer 1

Olá



Calcule o determinante de ambas os lados e em seguida iguale os determinantes e ache o valor de X.


A)

Calcularei primeiro o determinante da primeira matriz, em seguida o da segunda matriz, e depois igualarei os dois.


Por sarrus


$\mathsf{ \left[\begin{array}{ccc}1&0&x\\x&0&1\\1&1&x\end{array}\right] }\\\\\\\mathsf{\left|\begin{array}{cccc}1 ~ ~~~~ ~~ & 0~ ~~~~ ~~ & x~ ~~~~ ~~ & 1 ~ ~~~~ ~~ 0\\x~~~~~~~&0~~~~~ ~~ & 1~ ~~~~ ~~ & x~ ~~~~ ~~ 0 \\ 1 ~ ~~~~ ~~ &1~ ~~~~~~&x~ ~~~~ ~~ & 1 ~ ~~~~ ~~ 1\\ \end{array}\right|}\\\\\\\\\\\mathsf{\mathsf{\underbrace{(\mathsf{1\cdot0\cdot x~+~0\cdot1\cdot1~+~x\cdot x\cdot1})}_{diag.~principal}-\underbrace{(\mathsf{0\cdot x\cdot x~+~1\cdot1\cdot1~+~x\cdot0\cdot1})}_{diag.~secund\'aria}}}$


$\mathsf{(0+0+x^2)~-~(0+1+0)}\\\\\\\boxed{\mathsf{x^2-1}}$



Calculando o determinante da segunda matriz


$\mathsf{ \left[\begin{array}{ccc}x&7\\1&5\\\end{array}\right] }\\\\\\\mathsf{\mathsf{\underbrace{(\mathsf{x\cdot5})}_{diag.~principal}-\underbrace{(\mathsf{1\cdot 7})}_{diag.~secund\'aria}}}}\\\\\\\boxed{\mathsf{5x-7}}$



Igualando o determinante das duas matrizes.


$\displaystyle\mathsf{x^2-1~=~5x-7}\\\\\\\mathsf{x^2-5x+6=0}\\\\\\\text{Resolvendo por bhaskara}\\\\\\\mathsf{\Delta=(-5)^2-4\cdot1\cdot 6}\\\\\mathsf{\Delta=1}\\\\\\\\\mathsf{X= \frac{5\pm \sqrt{1}}{2\cdot1} }\\\\\\\boxed{\mathsf{x'=2}}\qquad\qquad\Longleftarrow\quad\text{Resposta 1}\\\\\boxed{\mathsf{x''=3}}\qquad\qquad\Longleftarrow\quad\text{Resposta 2}$







B)


Mesmo principio do item A), porém, vamos igualar o determinante da primeira matriz a 1.


$\mathsf{ \left[\begin{array}{ccc}1&2&-1\\0&1&x\\1&x&-1\end{array}\right] ~=~1}\\\\\\\mathsf{\left|\begin{array}{cccc}1~~~~~ ~~ & 2~ ~~~~ ~~ &-1~~~~~ ~~&1~~~~~~~2\\0~~~~~~~&1~~~~~ ~~ & x~~~~~ ~~ & 0~ ~~~~ ~~1\\1~~~~~~~&x~ ~~~~~~&-1~ ~~~~ ~~ & 1 ~ ~~~~ ~~ x\\ \end{array}\right|~=~1}\\\\\\\\\\\mathsf{\mathsf{\underbrace{(\mathsf{1\cdot1\cdot(-1)+2\cdot x\cdot1-1\cdot0\cdot x})}_{diag.~principal}-\underbrace{(\mathsf{-1\cdot 0\cdot 2~+~1\cdot x\cdot x-1\cdot1\cdot1})}_{diag.~secund\'aria}}~=~1}$


$\displaystyle \mathsf{(-1+2x+0)-(0+x^2-1)~=~1}\\\\\mathsf{-1+2x-x^2+1-1=0}\\\\\mathsf{-x^2+2x-1=0}\\\\\\\text{Por bhaskara}\\\\\\\mathsf{\Delta=2^2-4\cdot(-1)\cdot-1}\\\\\mathsf{\Delta=0}\\\\\\\mathsf{X= \frac{-2\pm \sqrt{0} }{2\cdot(-1)} }\\\\\\\boxed{\mathsf{x'=x''=1}}\qquad\qquad\Longleftarrow\quad\text{Resposta}$







C)



$\mathsf{ \left[\begin{array}{ccc}2&3&-2\\0&1&x\\2&x&-3\end{array}\right] }\\\\\\\mathsf{\left|\begin{array}{cccc}2~~~~~ ~~ & 3~ ~~~~ ~~ &-2~~~~~ ~~&2~~~~~~~3\\0~~~~~~~&1~~~~~ ~~ & x~~~~~ ~~ & 0~ ~~~~ ~~1\\2~~~~~~~&x~ ~~~~~~&-3~ ~~~~ ~~ & 2 ~~~~ ~~ x\\ \end{array}\right|}\\\\\\\\\\\mathsf{\mathsf{\underbrace{(\mathsf{2\cdot1\cdot(-3)+3\cdot x\cdot2-2\cdot0\cdot x})}_{diag.~principal}-\underbrace{(\mathsf{-3\cdot 0\cdot3~+~x\cdot x\cdot 2-2\cdot1\cdot2})}_{diag.~secund\'aria}}}$


$\mathsf{(-6+6x+0)~-~(0+2x^2-4)}\\\\\boxed{\mathsf{-2x^2+6x-2}}$