Question

Va rog, ajutați-mă. E urgent! Va rog!!!!!!
Exercițiul e atașat. ​

Answer 1

a)

[tex]\displaystyle\\ A(a)\cdot A(b)=\left(\begin{array}{cc}1-10a&8a\\-5a&1+4a\end{array}\right) \cdot\left(\begin{array}{cc}1-10b&8b\\-5b&1+4b\end{array}\right)\\\\ =\left(\begin{array}{cc}(1-10a)(1-10b)+8a\cdot(-5b)&(1-10a)\cdot8b+8a\cdot(1+4b)\\-5a(1-10b)+(1+4a)\cdot(-5b)&-5a\cdot8b+(1+4a)(1+4b)\end{array}\right)\\ =\left(\begin{array}{cc}1-10a-10b+60ab&8a+8b-48ab\\-5a-5b+30ab&1+4a+4b-24ab\end{array}\right)\\[/tex]

$=\left(\begin{array}{cc}1-10(a+b-6ab)&8(a+b-6ab)\\-5(a+b-6ab)&1+4(a+b-6ab)\end{array}\right)=A(a+b-6ab)$

b)

[tex]A(m)\cdot A(n)=A(6-5mn)\\ A(m)\cdot A(n)\overset{\text{din a)}}{=}A(m+n-6mn)\\ A(m+n-mn)=A(6-5mn)\Rightarrow m+n-6mn=6-5mn\\ m+n-mn -6=0[/tex]

[tex]\text{Am \^incercat s\u a fac o descompunere \^in factori.}\\ \text{\c Stiu ca m \c si n sunt numere naturale, o descompunere m-ar ajuta.}\\ m+n-mn-6=0\\ m-mn+n-6=0\\ m(1-n)+n-6=0\\ m(1-n)-(-n)-1-5=0\\ m(1-n)-(-n+1)-5=0\\ m(1-n)-(1-n)=5\\ (1-n)(m-1)=5\\ \text{Cum m \c si n sunt numere naturale, am doar dou\u a variante.}\\ \text{Varianta 1:} \displaystyle\left \{ {{1-n=1} \atop {m-1=5}} \right. \Leftrightarrow\left \{ {{n=0} \atop {m=6}} \right. [/tex]

[tex]\displaystyle \text{Varianta 2:}\left \{ {{1-n=5} \atop {m-1=1}} \right. \Leftrightarrow\left \{ {{n=-4\notin\mathbb N} \atop {m=0}} \right.\\ \text{Deci, }m=6, n=0[/tex]