Question

( Ufam) Considere a matriz A= [-4 0
7 2]. Os valores de k que tornam nulo o determinante da matriz A=kI,sendo I a matriz igualdade,são:
A) 0e 5
B) -2 e 4
C) 0 e 4
D) -4 e 2
E) -4 e 0

Obs: o gabarito diz que a resposta é a letra "d" mas preciso fazer as contas.

Answer 1

Olá


A matriz 'I', é a matriz identidade de mesma ordem que a matriz 'A'.


$\displaystyle\mathsf{ A= \left[\begin{array}{ccc}-4&0\\7&2\\\end{array}\right]} }\\\\\\\\\mathsf{I= \left[\begin{array}{ccc}1&0\\0&1\\\end{array}\right] }\\\\\\\mathsf{k = ?}$


Igualando as matrizes

A = kI

$\displaystyle\mathsf{\left[\begin{array}{ccc}-4&0\\7&2\\\end{array}\right]}~=~k\cdot \left[\begin{array}{ccc}1&0\\0&1\\\end{array}\right]}$


Aplica a distributiva do 'k' na matriz 'I'


$\displaystyle\mathsf{ \left[\begin{array}{ccc}-4&0\\7&2\\\end{array}\right]}~=~ \left[\begin{array}{ccc}k&0\\0&k\\\end{array}\right]}}}$


$\displaystyle\mathsf{ \left[\begin{array}{ccc}-4&0\\7&2\\\end{array}\right]} ~-~ \left[\begin{array}{ccc}k&0\\0&k\\\end{array}\right]}}=0}\\\\\\\\\mathsf{\left[\begin{array}{ccc}(-4-k)&\quad(0-0)\\(7-0)&\quad(2-k)\\\end{array}\right]~=~0}$


Calcula o determinante, e iguala a zero


$\displaystyle\mathsf{\underbrace{(\mathsf{(-4-k)\cdot(2-k)})}_{diag.~principal}~-~\underbrace{(\mathsf{(7)\cdot(0)})}_{diag.~secund\'aria}=0}\\\\\\\\\mathsf{(-8+4k-2k+k^2)~-~0=0}\\\\\\\\\mathsf{k^2+2k-8=0}$



Aplicando bhaskara para determinar os valores de k


k² + 2k - 8 = 0

Δ = 2² - 4.1.(-8)
Δ = 4 + 32
Δ = 36



$\displaystyle\mathsf{K= \frac{-2\pm \sqrt{36} }{2\cdot 1} }\\\\\\\\\mathsf{K_1= \frac{-2+ \sqrt{36} }{2\cdot 1} \qquad\Longrightarrow K_1= \frac{-2+ 6 }{2} \qquad\Longrightarrow \boxed{K_1=2}}\\\\\\\\\mathsf{K_2= \frac{-2- \sqrt{36} }{2\cdot 1} \qquad\Longrightarrow K_2= \frac{-2- 6 }{2} \qquad\Longrightarrow \boxed{K_2=-4}}$






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