Răspuns:
Explicație pas cu pas:
a. [tex]\left|\begin{array}{ccc}1&1\\2&1\\\end{array}\right|=1-2=-1[/tex]
b. [tex]\left(\begin{array}{ccc}1&1\\2&2\\\end{array}\right) \cdot \left(\begin{array}{ccc}1&a\\b&1\\\end{array}\right) =\left(\begin{array}{ccc}1+b&1+a\\2+2b&2a+2\\\end{array}\right)[/tex]
[tex]\left|\begin{array}{ccc}1+b&1+a\\2+2b&2a+2\\\end{array}\right|=[/tex]
(1+b)2(a+1)-(1+a)2(1+b)=0
c. [tex]\left(\begin{array}{ccc}1+b&1+a\\2+2b&2a+2\\\end{array}\right)=\left(\begin{array}{ccc}1+2a&1+2a\\2+b&b+2\\\end{array}\right)[/tex]
1+b=1+2a
b=2a
1+a=1+2a
a=2a
a=0
b=0
2. x°y=|x-y|+1
a. 3°5=|3-5|+1=2+1=3
b. a-b=?
a=(2°3)°4=2°4=3
b=2°(3°4)=2°2=1
a-b=3-1=2
c. m°n=m
|m-n|+1=m
caz 1 m>n
m-n+1=m
n=1
caz 2 m<n
n-m+1=m
n+1=2m
exista o infinitate de solutii naturale, n impar
