Question

Se consideră matricele $A=\left(\begin{array}{rr}4 & -6 \\ 2 & -3\end{array}\right), I_{2}=\left(\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right)$ şi $M(x)=I_{2}+x A$, unde $x$ este număr real.

$5 \mathbf{a}$ a) Arătați că det $A=0$.

$5 \mathbf{p}$ b) Demonstrați că $M(x) M(y)=M(x+y+x y)$, pentru orice numere reale $x$ și $y$.

$5 \mathbf{p}$ c) Determinați perechile de numere naturale $(m, n)$ pentru care $M(m) M(n)=M(6)$.

Answer 1

$A=\left(\begin{array}{rr}4 & -6 \\ 2 & -3\end{array}\right)$

a)

Calculam detA, inmultind prima diagonala si o scadem pe a doua

$detA=\left|\begin{array}{rr}4 & -6 \\ 2 & -3\end{array}\right|=-12-(-12)=-12+12=0$

b)

M(x)=I₂+xA

$M(x)=\left(\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right)+x\left(\begin{array}{rr}4 & -6 \\ 2 & -3\end{array}\right)=\left(\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right)+\left(\begin{array}{rr}4x & -6x \\ 2x & -3x\end{array}\right)\\\\M(x)=\left(\begin{array}{rr}1+4x & -6x \\ 2x & 1-3x\end{array}\right)$

$M(y)=\left(\begin{array}{rr}1+4y & -6y \\ 2y & 1-3y\end{array}\right)$

$M(x)M(y)=\left(\begin{array}{rr}1+4x & -6x \\ 2x & 1-3x\end{array}\right)\times \left(\begin{array}{rr}1+4y & -6y \\ 2y & 1-3y\end{array}\right)=\\\\=\left(\begin{array}{rr}(1+4x)(1+4y)-12xy & -6y-24xy-6x+18xy \\ 2x+8xy+2y-6xy & -12xy+(1-3x)(1-3y)\end{array}\right)$

$M(x)M(y)=\left(\begin{array}{rr}1+4x+4y+4xy & -6x-6y-6xy \\ 2x+2y+2xy & 1-3x-3y-3xy\end{array}\right)\\\\M(x)M(y)=\left(\begin{array}{rr}1+4(x+y+xy) & -6(x+y+xy) \\ 2(x+y+xy) & 1-3(x+y+xy)\end{array}\right)=M(x+y+xy)$

c)

Ne folosim de punctul b, stim ca M(x)M(y)=M(x+y+xy)

M(m)M(n)=M(m+n+mn)

M(m+n+mn)=M(6)

m+n+mn=6

m+n+mn+1-1=6

m+1+n(m+1)=7

(m+1)(n+1)=7

I.

m+1=7

m=6

n+1=1

n=0

II.

m+1=1

m=0

n+1=7

n=6

Un exercitiu similar de bac gasesti aici: https://brainly.ro/tema/2494494

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