Question

me ajudem com mais essa por favor​

Answer 1

Explicação passo-a-passo:

$det$ $(\frac{A^{2}}{7}-2A)$

$sendo$  $A=\left[\begin{array}{ccc}2&-1\\4&5\\\end{array}\right]$

Vamos calcular por partes

cálculo do A²

    $A^{2}=A.A=\left[\begin{array}{ccc}2&-1\\4&5\\\end{array}\right].\left[\begin{array}{ccc}2&-1\\4&5\\\end{array}\right]$

    $A^{2}=\left[\begin{array}{ccc}2.2+(-1).4&2.(-1)+(-1).5\\4.2+5.4&4.(-1)+5.5\\\end{array}\right]$

    $A^{2}=\left[\begin{array}{ccc}4-4&-2-5\\8+20&-4+25\\\end{array}\right]$

    $A^{2}=\left[\begin{array}{ccc}0&-7\\28&21\\\end{array}\right]$

cálculo do 2A

    $2A=2.\left[\begin{array}{ccc}2&-1\\4&5\\\end{array}\right]$

    $2A=\left[\begin{array}{ccc}2.2&2.(-1)\\2.4&2.5\\\end{array}\right]$

    $2A=\left[\begin{array}{ccc}4&-2\\8&10\\\end{array}\right]$

agora o cálculo do $(\frac{A^{2}}{7}-2A)$

    $(\frac{\left[\begin{array}{ccc}0&-7\\28&21\\\end{array}\right]}{7}-\left[\begin{array}{ccc}4&-2\\8&10\\\end{array}\right])$

    $(\left[\begin{array}{ccc}0&-7\\28&21\\\end{array}\right]:7-\left[\begin{array}{ccc}4&-2\\8&10\\\end{array}\right])$

    $\left[\begin{array}{ccc}0:7&-7:7\\28:7&21:7\\\end{array}\right]-\left[\begin{array}{ccc}4&-2\\8&10\\\end{array}\right]$

    $\left[\begin{array}{ccc}0&-1\\4&3\\\end{array}\right]-\left[\begin{array}{ccc}4&-2\\8&10\\\end{array}\right]$

    $\left[\begin{array}{ccc}0-4&-1-(-2)\\4-8&3-10\\\end{array}\right]$

    $\left[\begin{array}{ccc}-4&1\\-4&-7\\\end{array}\right]$

agora o cálculo do determinante

   $det \left[\begin{array}{ccc}-4&1\\-4&-7\\\end{array}\right]$

   $det=-4.(-7)-1.(-1)$

   $det=28+4$

   $det=32$