Question

Se consideră matricele $A=\left(\begin{array}{cc}1 & -1 \\ 3 & 0\end{array}\right), I_{2}=\left(\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right)$ şi $M(x, y)=\left(\begin{array}{ll}x & y \\ 3 & 4\end{array}\right)$, unde $x$[ şi $y$ sunt numere reale.

$5 p$ 1. Arătați că det $A=3$.

5p 2. Determinaţi numerele reale $x$ şi $y$ astfel încât $M(x, y)=A+4 I_{2}$.

5p 3. Determinați numărul real y pentru care $\operatorname{det}(M(0, y))=9$.

5p 4. Arătaţi că $A \cdot A \cdot A-A \cdot A=-3 A$.

5p 5. Determinaţi numerele reale $x$ şi $y$, ştiind că $A \cdot M(x, y)=M(x, y) \cdot A$.

5p 6. Demonstrați că, dacă $m$ şi $n$ sunt numere întregi pentru care $M(m,-n) \cdot M(-m, n)=\left(\begin{array}{cc}-1 & 0 \\ 0 & 1\end{array}\right)$, atunci numărul $N=m-n$ este pătratul unui număr natural.

Answer 1

$A=\left(\begin{array}{cc}1 & -1 \\ 3 & 0\end{array}\right)$

$M(x, y)=\left(\begin{array}{ll}x & y \\ 3 & 4\end{array}\right)$

1)

Calculam detA, facem diferenta dintre produsul diagonalelor

detA=0-(-3)=3

2)

$M(x, y)=\left(\begin{array}{ll}x & y \\ 3 & 4\end{array}\right)=\left(\begin{array}{ll}5 & -1 \\ 3 &4\end{array}\right)$

x=5

y=-1

3)

det(M(0,y))=9

$det(M(0, y))=\left|\begin{array}{ll}0 & y \\ 3 & 4\end{array}\right|=0-3y=-3y$

-3y=9

y=-3

4)

$A\cdot A =\left(\begin{array}{cc}1 & -1 \\ 3 & 0\end{array}\right)\cdot \left(\begin{array}{cc}1 & -1 \\ 3 & 0\end{array}\right)=\left(\begin{array}{cc}-2 & -1 \\ 3 & -3\end{array}\right)\\\\A\cdot A\cdot A=\left(\begin{array}{cc}-2 & -1 \\ 3 & -3\end{array}\right)\cdot \left(\begin{array}{cc}1 & -1 \\ 3 & 0\end{array}\right)=\left(\begin{array}{cc}-5 & 2 \\ -6 & -3\end{array}\right)$

$A\cdot A\cdot A-A\cdot A=\left(\begin{array}{cc}-5 & 2 \\ -6 & -3\end{array}\right)-\left(\begin{array}{cc}-2 & -1 \\ 3 & -3\end{array}\right)=\left(\begin{array}{cc}-3 & 3\\ -9 & 0\end{array}\right)=-3A$

5)

$A\cdot M(x,y)=\left(\begin{array}{cc}1 & -1 \\ 3 & 0\end{array}\right)\cdot \left(\begin{array}{ll}x & y \\ 3 & 4\end{array}\right)=\left(\begin{array}{ll}x-3 & y-4 \\ 3x & 3y\end{array}\right)$

$M(x,y)\cdot A= \left(\begin{array}{ll}x & y \\ 3 & 4\end{array}\right)\cdot \left(\begin{array}{cc}1 & -1 \\ 3 & 0\end{array}\right)=\left(\begin{array}{ll}x+3y & -x \\ 15 & -3\end{array}\right)$

Egalam termenii si obtinem:

3x=15

x=5

3y=-3

y=-1

6)

$M(m, -n)\cdot M(-m,n)=\left(\begin{array}{ll}m & -n \\ 3 & 4\end{array}\right)\cdot \left(\begin{array}{ll}-m & n \\ 3 & 4\end{array}\right)=\left(\begin{array}{ll}-m^2-3n &mn-4n \\ -3m+12 & 3n+16\end{array}\right)\\\\\left(\begin{array}{ll}-m^2-3n &mn-4n \\ -3m+12 & 3n+16\end{array}\right)=\left(\begin{array}{ll}-1&0\\0 & 1\end{array}\right)$

Egalam termenii si obtinem:

-m²-3n=-1

m²+3n=1

mn-4n=0

mn=4n

-3m+12=0

3m=12

m=4

3n+16=1

3n=-15

n=-5

N=m-n=4+5=9=3², patrat perfect

Un alt exercitiu cu matrice gasesti aici: https://brainly.ro/tema/9928505

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