Question

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

1. Arătați că det $A=0$.

2. Determinaţi numerele reale $x$ pentru care $\operatorname{det}(M(x))=16$.

3. Arătați că $M(-1)+M(0)+M(1)=3 A$.

4. Demonstrați că $M(x) \cdot M(y)=x y I_{2}+(x+y) A$, pentru orice numere reale $x$ şi [tex]$y$[tex].

5. Determinaţi mulţimea valorilor reale ale lui $x$ pentru care $\operatorname{det}(M(x)-x A) \leq 3 x-2$.

6. Determinaţi numărul natural $n$ pentru care $M(1)+M(2)+\ldots+M(n)=9 M(5)$.

Answer 1

$A=\left(\begin{array}{cc}2 & 1 \\ -4 & -2\end{array}\right)$

1)

Calculam detA, facem diferenta dintre produsul diagonalelor

detA=-4-(-4)=-4+4=0

2)

det(M)x))=16

$M(x)=\left(\begin{array}{cc}2+x & 1 \\ -4 & -2+x\end{array}\right)\\\\\left|\begin{array}{cc}2+x & 1 \\ -4 & -2+x\end{array}\right|=(2+x)(-2+x)+4=x^2-4+4=x^2\\\\x^2=16\\\\x=4\ si \ x=-4$

3)

$M(-1)+M(0)+M(1)=\left(\begin{array}{cc}1 & 1 \\ -4 & -3\end{array}\right)+\left(\begin{array}{cc}2& 1 \\ -4 & -2\end{array}\right)+\left(\begin{array}{cc}3 & 1 \\ -4 & -1\end{array}\right)=\left(\begin{array}{cc}6 & 3 \\ -12 &-6\end{array}\right)=3A$

4)

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

$xyI_2+(x+y)A=\left(\begin{array}{cc}xy &0 \\0 &xy\end{array}\right)+\left(\begin{array}{cc}2(x+y) & x+y \\ -4(x+y) & -2(x+y)\end{array}\right)=\\\left(\begin{array}{cc}xy+2(x+y) & x+y \\ -4 (x+y)&xy-2(x+y)\end{array}\right)$

Observam ca sunt egale

5)

det(M(x)-xA)≤3x-2

$M(x)-xA=\left(\begin{array}{cc}2+x & 1 \\ -4 & -2+x\end{array}\right)-\left(\begin{array}{cc}2x & x \\ -4x & -2x\end{array}\right)=\left(\begin{array}{cc}2-x & 1-x \\ 4x-4 & 3x-2\end{array}\right)$

$\left|\begin{array}{cc}2-x & 1-x \\ 4x-4 & 3x-2\end{array}\right|=(2-x)(3x-2)-(1-x)(4x-4)=6x-4-3x^2+2x-4x+4+4x^2-4x=x^2\\\\x^2\leq 3x-2$

$x^2-3x+2\leq 0\\\\\Delta=9-8=1\\\\x_1=\frac{3+1}{2} =2\\\\x_2=\frac{3-1}{2} =1$

Tabel semn

x                -∞          1           2            +∞

x²-3x+2    + + + + + 0 - - - - 0 + + + + +

x∈[1,2]

6)

Ne folosim de punctul 4

$M(1)+M(2)+...+M(n)=(1+2+..+n)I_2+nA=\frac{n(n+1)}{2} I_2+nA=n(\frac{n+1}{2}I_2+A)=nM(\frac{n+1}{2} ) \\\\nM(\frac{n+1}{2} ) =9M(5)$

n=9

Un alt exercitiu cu matrice gasesti aici: https://brainly.ro/tema/9928497

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