Question

Se consideră matricele $A=\left(\begin{array}{ll}2 & 0 \\ 3 & 1\end{array}\right), B=\left(\begin{array}{ll}2 & x \\ 3 & 0\end{array}\right)$ şi $C=\left(\begin{array}{ll}x & 0 \\ 0 & 1\end{array}\right)$, unde $x$ este număr real.

5p 1. Arătați că det $A=2$.

5 2 2. Determinați numărul real $x$ pentru care $B+C=A$.

5p 3. Determinați numărul real $x$ pentru care $\operatorname{det}(B-C)=0$.

5p 4. Demonstrați că $\operatorname{det}(B \cdot C-C \cdot B)=3 x(x-1)^{2}$, pentru orice număr real $x$.

5p 5. Pentru $x=1$, arătați că inversa matricei $B$ este matricea $\left(\begin{array}{cc}0 & \frac{1}{3} \\ 1 & -\frac{2}{3}\end{array}\right) .$

5p 6. Pentru $x=1$, rezolvați în $\mathcal{M}_{2}(\mathbb{R})$ ecuaţia $B \cdot X \cdot C=A$.

Answer 1

$A=\left(\begin{array}{ll}2 & 0 \\ 3 & 1\end{array}\right)$

$B=\left(\begin{array}{ll}2 & x \\ 3 & 0\end{array}\right)$

$C=\left(\begin{array}{ll}x & 0 \\ 0 & 1\end{array}\right)$

1)

Calculam detA, facand diferenta dintre produsul diagonalelor

$detA=\left|\begin{array}{ll}2 & 0 \\ 3 & 1\end{array}\right|=2-0=1$

2)

B+C=A

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

2+x=2

x=0

3)

det(B-C)=0

$\left(\begin{array}{ll}2 & x \\ 3 & 0\end{array}\right)-\left(\begin{array}{ll}x & 0 \\ 0 & 1\end{array}\right)=\left(\begin{array}{ll}2-x & x \\ 3 & -1\end{array}\right)\\\\\left|\begin{array}{ll}2-x & x \\ 3 & -1\end{array}\right|=0$

-(2-x)-3x=0

-2+x-3x=0

-2-2x=0

2x=-2

x=-1

4)

det(BC-CB)=3x(x-1)²

$B\cdot C=\left(\begin{array}{ll}2 & x \\ 3 & 0\end{array}\right)\cdot\left(\begin{array}{ll}x & 0 \\ 0 & 1\end{array}\right)=\left(\begin{array}{ll}2x & x \\ 3x& 0\end{array}\right)$

$C\cdot B=\left(\begin{array}{ll}x & 0 \\ 0 & 1\end{array}\right)\cdot \left(\begin{array}{ll}2 & x \\ 3 & 0\end{array}\right)=\left(\begin{array}{ll}2x & x^2 \\ 3& 0\end{array}\right)$

$BC-CB=\left(\begin{array}{ll}2x & x \\ 3x& 0\end{array}\right)-\left(\begin{array}{ll}2x & x^2 \\ 3& 0\end{array}\right)=\left(\begin{array}{ll}0 &x- x^2 \\ 3x-3& 0\end{array}\right)$

$\left|\begin{array}{ll}0 &x- x^2 \\ 3x-3& 0\end{array}\right|=0-(x-x^2)(3x-3)=-3x(1-x)(x-1)=3x(x-1)^2$

5)

x=1

Inversa unei matrice:

$B^{-1}=\frac{1}{detB}\cdot B^*$

detB=2×0-3×1=-3

Transpusa matricei B

$B^t=\left(\begin{array}{ccc}2&3\\1&0\\\end{array}\right)$

$B^*=\left(\begin{array}{ccc}0&-1\\-3&2\\\end{array}\right)$

$(-1)^{i+j}$× elementul care ramane dupa ce taiem linia si coloana pe care se afla termenul respectiv

$B^{-1}=\frac{1}{-3} \left(\begin{array}{ccc}0&-1\\-3&2\\\end{array}\right)= \left(\begin{array}{ccc}0&\frac{1}{3} \\1&-\frac{2}{3} \\\end{array}\right)$

6)

x=1

B·X·C=A

$C=\left(\begin{array}{ccc}1&0\\0&1\end{array}\right)=I_2\\\\B\cdot X\cdot C=A\\\\B\cdot X=A\\\\X=B^{-1}\cdot A$

$B^{-1}\cdot A=\left(\begin{array}{ccc}0&\frac{1}{3} \\1&-\frac{2}{3} \\\end{array}\right)\cdot \left(\begin{array}{ccc}2&0 \\3&1 \\\end{array}\right)=\left(\begin{array}{ccc}1&\frac{1}{3} \\0&-\frac{2}{3} \\\end{array}\right)$

Un alt exercitiu cu matrice gasesti aici: https://brainly.ro/tema/1857848

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