Question

va rrogggg! dau coroana ​

Answer 1

Răspuns:

Explicație pas cu pas:

a) detB = 1*3 - 8*1 = 3 - 8 = -5

b) detA = 3*a - 8*1 = 3a-8

daca detA = 0, atunci 3a = 8, a = 8/3 ∉ Z

deci detA ≠ 0, ∀ a ∈ Z

c)

$A^{-1} =\left[\begin{array}{ccc}a&1\\8&3\end{array}\right]^{-1} =\frac{1}{detA}*C^{T} =\frac{1}{3a-8} *\left[\begin{array}{ccc}3&-1\\-8&a\end{array}\right]=\left[\begin{array}{ccc}\frac{3}{3a-8}&\frac{-1}{3a-8}\\\frac{-8}{3a-8}&\frac{a}{3a-8}\end{array}\right]$

$C^{T}=\left[\begin{array}{ccc}C11&C21\\C12&C22\end{array}\right]$

$C11=(+1)*|3|=3\\C12=(-1)*|8|=-8\\C21=(-1)*|1|=-1\\C22=(+1)*|a|=a$

Pentru ca matricea inversa sa aiba toate elementele numere intregi, trebuie pusa conditia ca: (3a - 8) sa divida 3, -1, -8 si a. Deoarece 3 si 8 sunt prime intre ele, poate fi 1 sau -1.

$3a-8=1\\3a=9\\a=3$

$3a-8=-1\\3a=7\\a=\frac{7}{3}$

=> a = 3