Question

Cum se rezolva aceste exercitii? Help :/​

Answer 1

Explicație pas cu pas:

a) det(A) = 0

det(A) = (x - 3)² - 1

=> (x - 3)² - 1 = 0 <=> (x - 3)² = 1

x - 3 = 1 => x = 4

x - 3 = -1 => x = 2

b)

$A^{2} = \left(\begin{array}{ccc}x-3&1\\1&x-3\end{array}\right)\cdot \left(\begin{array}{ccc}x-3&1\\1&x-3\end{array}\right) =$

$= \left(\begin{array}{ccc}(x-3)^{2}+1&x-3+x-3\\x-3+x-3&(x-3)^{2}+1\end{array}\right)$

$= \left(\begin{array}{ccc}x^{2}-6x+10&2(x-3)\\2(x-3)&x^{2}-6x+10\end{array}\right)$

$(2x-6)\cdot A - (x^{2}-6x+8)\cdot I_{2} =$

$= (2x-6)\cdot \left(\begin{array}{ccc}x-3&1\\1&x-3\end{array}\right) - (x^{2}-6x+8)\cdot \left(\begin{array}{ccc}1&0\\0&1\end{array}\right)$

$= \left(\begin{array}{ccc}(2x-6)(x-3)&2x-6\\2x-6&(2x-6)(x-3)\end{array}\right) - \left(\begin{array}{ccc}x^{2}-6x+8&0\\0&x^{2}-6x+8\end{array}\right)$

$= \left(\begin{array}{ccc}2x^{2} - 12x + 18&2x-6\\2x-6&2x^{2} - 12x + 18\end{array}\right) - \left(\begin{array}{ccc}x^{2}-6x+8&0\\0&x^{2}-6x+8\end{array}\right)$

$= \left(\begin{array}{ccc}2x^{2} - 12x + 18 - (x^{2}-6x+8)&2x-6\\2x-6&2x^{2} - 12x + 18 - (x^{2}-6x+8)\end{array}\right)$

$= \left(\begin{array}{ccc}x^{2} - 6x + 10&2(x-3)\\2(x-3)&x^{2} - 6x + 10\end{array}\right)$

=> relația se verifică

c)

A² = 2•A

$\left(\begin{array}{ccc}x^{2}-6x+10&2(x-3)\\2(x-3)&x^{2}-6x+10\end{array}\right) = 2\left(\begin{array}{ccc}x-3&1\\1&x-3\end{array}\right)$

$\left(\begin{array}{ccc}x^{2}-6x+10&2(x-3)\\2(x-3)&x^{2}-6x+10\end{array}\right) = \left(\begin{array}{ccc}2(x-3)&2\\2&2(x-3)\end{array}\right)$

=>

$\left \{ {{x^{2}-6x+10=2(x-3)} \atop {2(x-3)=2}} \right.$

$\left \{ {{x^{2}-6x+10-2x+6=0} \atop {x-3=1}} \right.$

$\left \{ {{x^{2}-8x+16=0} \atop {x=4}} \right.\iff \left \{ {{(x-4)^{2}=0} \atop {x=4}} \right.$

$\left \{ {x=4 \atop {x=4}} \right.\implies x = 4$