Subiectul III

Se consideră matricele $A=\begin{pmatrix} 1 &1 \\ 2 & 0 \end{pmatrix}$ , $B=\begin{pmatrix} 4 &1 \\ 2 & 3 \end{pmatrix}$ și $I_2=\begin{pmatrix} 1 &0 \\ 0 & 1 \end{pmatrix}$ .

Calculați $\det A$ .

Calculăm $\det A$ .

$\begin{align*} \det A&=\begin{vmatrix} 1 &1 \\ 2 & 0 \end{vmatrix}\\ &=1\cdot 0 -2\cdot 1\\ &=0-2\\ &=-2 \end{align*}$

Rezultă că:

$\begin{align*} \det A=-2 \end{align*}$ .

Arătați că $A\cdot A+I_2=B$ .

Calculăm $A\cdot A+I_2$ .

$\begin{align*} A\cdot A+I_2&= \begin{pmatrix} 1 &1 \\ 2 & 0 \end{pmatrix} \cdot \begin{pmatrix} 1 &1 \\ 2 & 0 \end{pmatrix} +\begin{pmatrix} 1 &0 \\ 0 & 1 \end{pmatrix} \\\\ &=\begin{pmatrix} 1\cdot 1+1\cdot 2 &1\cdot 1+1\cdot 0 \\2\cdot 1+ 0\cdot 2 & 2\cdot 1+0\cdot 0 \end{pmatrix} +\begin{pmatrix} 1 &0 \\ 0 & 1 \end{pmatrix}\\\\ &=\begin{pmatrix} 1+2 &1+0 \\2+0 & 2+0 \end{pmatrix} +\begin{pmatrix} 1 &0 \\ 0 & 1 \end{pmatrix}\\\\ &=\begin{pmatrix} 3 &1 \\ 2 & 2 \end{pmatrix} +\begin{pmatrix} 1 &0 \\ 0 & 1 \end{pmatrix}\\ \\&=\begin{pmatrix}3+1 &1+0 \\2+0 & 2+1 \end{pmatrix}\\\\ &=\begin{pmatrix} 4&1 \\ 2 & 3 \end{pmatrix}\\ \end{align*}$

Am obținut că:

$\begin{align*} A\cdot A+I_2=B \end{align*}$ .

Verificați dacă $A\cdot B=B\cdot A$ .

Calculăm $A\cdot B$ .

$\begin{align*} A\cdot B&=\begin{pmatrix} 1 &1 \\ 2 & 0 \end{pmatrix} \cdot \begin{pmatrix} 4 &1 \\ 2 & 3 \end{pmatrix} \\\\ &=\begin{pmatrix} 1\cdot 4 +1\cdot 2&1\cdot 1+1\cdot 3 \\ 2\cdot 4+0\cdot 2 & 2\cdot 1+0\cdot 3 \end{pmatrix}\\ \\&=\begin{pmatrix} 4+2 &1+3 \\ 8+0 & 2+0 \end{pmatrix}\\\\ &=\begin{pmatrix} 6 &4 \\ 8 & 2 \end{pmatrix} \end{align*}$

Calculăm $B\cdot A$ .

$\begin{align*} B\cdot A&= \begin{pmatrix} 4 &1 \\ 2 & 3 \end{pmatrix}\cdot \begin{pmatrix} 1 &1 \\ 2 & 0 \end{pmatrix} \\\\ &=\begin{pmatrix} 4\cdot 1 +1\cdot 2&4\cdot 1+1\cdot 0 \\ 2\cdot 1+3\cdot 2 & 2\cdot 1+3\cdot 0 \end{pmatrix}\\\\ &=\begin{pmatrix} 4+2 &4+0 \\ 2+6 & 2+0 \end{pmatrix}\\\\ &=\begin{pmatrix} 6 &4 \\ 8 & 2 \end{pmatrix} \end{align*}$

Așadar, în acest caz, înmulțirea a două matrice este comutativă, adică:

$\begin{align*} A\cdot B=B\cdot A \end{align*}$ .

Arătați că matricea $C=\begin{pmatrix} 0 &\displaystyle\frac{1}{2} \\ 1 & -\displaystyle\frac{1}{2} \end{pmatrix}$ este inversa matricei .

Calculăm $A\cdot C$ .

$\begin{align*} A\cdot C&=\begin{pmatrix} 1 &1 \\ 2 & 0 \end{pmatrix}\cdot \begin{pmatrix} 0 &\displaystyle\frac{1}{2} \\ 1 & -\displaystyle\frac{1}{2} \end{pmatrix}\\\\ &=\begin{pmatrix} 1\cdot 0+1\cdot 1 &1\cdot\displaystyle\frac{1}{2}+1\cdot \Big(-\displaystyle\frac{1}{2}\Big) \\ 2\cdot 0+0\cdot 1 & 2\cdot \displaystyle\frac{1}{2}+0\cdot \Big(-\displaystyle\frac{1}{2}\Big) \end{pmatrix}\\\\ &=\begin{pmatrix} 0+1& \displaystyle\frac{1}{2}-\displaystyle\frac{1}{2}\\ 0+0&1-0 \end{pmatrix}\\\\ &=\begin{pmatrix} 1&0\\0&1 \end{pmatrix}\\\\ &=I_2 \end{align*}$

Calculăm $C\cdot A$ .

$\begin{align*} C\cdot A&=\begin{pmatrix} 0 &\displaystyle\frac{1}{2} \\ 1 & -\displaystyle\frac{1}{2} \end{pmatrix}\cdot \begin{pmatrix} 1 &1 \\ 2 & 0 \end{pmatrix}\\\\ &=\begin{pmatrix} 0\cdot 1+\displaystyle\frac{1}{2}\cdot 2&0\cdot 1+\displaystyle\frac{1}{2}\cdot 0 \\ 1\cdot 1+\Big(-\displaystyle\frac{1}{2}\Big)\cdot 2 & 1\cdot 1+\Big(-\displaystyle\frac{1}{2}\Big)\cdot 0\ \end{pmatrix}\\ \\&=\begin{pmatrix} 0+1& 0+0\\ 1-1&1+0 \end{pmatrix}\\\\ &=\begin{pmatrix} 1&0\\0&1 \end{pmatrix}\\\\ &=I_2 \end{align*}$

Am obținut, efectuând cele două înmulțiri de matrice, că:

$\begin{align*} A\cdot C=C\cdot A=I_2 \end{align*}$

Rezultă că matricea $\begin{align*} C \end{align*}$ este inversa matricei $\begin{align*} A \end{align*}$ .

Determinați numerele reale știind că $\det (A+aI_2)=10$ .

Calculăm A+aI_2 .

$\begin{align*} A+aI_2&=\begin{pmatrix} 1 &1 \\ 2 & 0 \end{pmatrix}+a\cdot \begin{pmatrix} 1 &0 \\ 0 & 1 \end{pmatrix}\\\\ &=\begin{pmatrix} 1 &1 \\ 2 & 0 \end{pmatrix}+\begin{pmatrix} a &0 \\ 0 & a \end{pmatrix}\\\\ &=\begin{pmatrix} 1+a &1+0 \\2+0 & 0+a \end{pmatrix}\\\\ &=\begin{pmatrix} 1+a &1 \\ 2 & a \end{pmatrix} \end{align*}$

Calculăm $\det (A+aI_2)$ .

$\begin{align*} \det (A+aI_2)&=\begin{vmatrix} 1+a &1 \\ 2 & a \end{vmatrix}\\ &=(1+a)\cdot a-2\cdot 1\\ &=a+a^2-2 \end{align*}$

Obținem că:

$\begin{align*} &\det (A+aI_2)=8\Leftrightarrow \\ &\Leftrightarrow a^2+a-2=8\\ &\Leftrightarrow a^2+a-2-8=0\\ &\Leftrightarrow a^2+a-12=0 \end{align*}$

Calculăm discriminantul ecuației de gradul al doilea obținute mai sus.

$\begin{align*} &\Delta =1^2-4\cdot 1\cdot (-12)\\ &\Delta =1+48\\ &\Delta =49 \end{align*}$

Atunci, soluțiile ecuației sunt:

$\begin{align*} a_1&=\frac{-1-\sqrt{49}}{2\cdot 1}\\\\ &=\frac{-1-7}{2}\\\\ &=\frac{-8}{2}\\\\ &=-4 \end{align*}$

$\begin{align*} a_2&=\frac{-1+\sqrt{49}}{2\cdot 1}\\\\ &=\frac{-1+7}{2}\\\\ &=\frac{6}{2}\\\\ &=3 \end{align*}$

Rezultă că $\begin{align*} a_1=-4 \end{align*}$ și $\begin{align*} a_2=3 \end{align*}$ .

Rezolvați în $\mathcal{M}_2(\mathbb{R})$ ecuația $A\cdot X=B$ .

Avem că:

$\begin{align*} &A\cdot X=B\ | \cdot A^{-1}\\ &\Leftrightarrow A^{-1}\cdot A\cdot X=A^{-1}\cdot B\\ &\Leftrightarrow I_2\cdot X=A^{-1}\cdot B\\ &\Leftrightarrow X=A^{-1}\cdot B \end{align*}$

Dar, de la punctul 4. avem că $\begin{align*} A^{-1}=C \end{align*}$ .

Rezultă că:

$\begin{align*} X&=C\cdot B\\\\ &=\begin{pmatrix} 0 &\displaystyle\frac{1}{2} \\\\ 1 & -\displaystyle\frac{1}{2} \end{pmatrix}\cdot \begin{pmatrix} 4 &1 \\ 2 & 3 \end{pmatrix}\\\\ &=\begin{pmatrix} 0\cdot 4+\displaystyle\frac{1}{2}\cdot 2 &0\cdot 1+\displaystyle\frac{1}{2}\cdot 3 \\\\ 1\cdot 4+\Big(-\displaystyle\frac{1}{2}\Big)\cdot 2 & 1\cdot 1+\Big(-\displaystyle\frac{1}{2}\Big)\cdot 3 \end{pmatrix}\\\\ &=\begin{pmatrix} 0+1& 0+\displaystyle\frac{3}{2}\\\\ 4-1 &1-\displaystyle\frac{3}{2} \end{pmatrix}\\\\ &=\begin{pmatrix} 1& \displaystyle\frac{3}{2}\\\\ 3 &-\displaystyle\frac{1}{2} \end{pmatrix}\\ \end{align*}$

Deci:

$\begin{align*} X=\begin{pmatrix} 1& \displaystyle\frac{3}{2}\\\\ 3 &-\displaystyle\frac{1}{2} \end{pmatrix}. \end{align*}$