Question

Se consideră matricele $I_{2}=\left(\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right)$ şi $A(a)=\left(\begin{array}{cc}5-a & 10 \\ -2 & -4-a\end{array}\right)$, unde $a$ este număr real.

5p a) Arătați că $\operatorname{det}(A(0))=0$.

$5 \mathbf{p}$ b) Determinați numărul real $[ a$, știind că $A(a) \cdot A(a)=A(0)$.

$5 p$ c) Determinaţi matricea $X \in \mathcal{M}_{2}(\mathbb{R})$ cu proprietatea $A(-1) \cdot X=A(0)$.

Answer 1

$A(a)=\left(\begin{array}{cc}5-a & 10 \\ -2 & -4-a\end{array}\right)$

a)

Calculam det(A(0)), inlocuind pe a cu 0, facem diferenta dintre produsul diagonalelor

detA=5×(-4)-(-20)=-20+20=0

b)

$A(a)\cdot A(a)=\left(\begin{array}{cc}5-a & 10 \\ -2 & -4-a\end{array}\right)\cdot \left(\begin{array}{cc}5-a & 10 \\ -2 & -4-a\end{array}\right)=\\\\=\left(\begin{array}{cc}(5-a)^2-20 & 50-10a-40-10a \\ -10+2a+8+2a &-20+(-a-4)^2\end{array}\right)=\\\\=\left(\begin{array}{cc}25-10a+a^2-20 & 50-10a-40-10a \\ -10+2a+8+2a &-20+a^2+8a+16\end{array}\right)=\\\\=\left(\begin{array}{cc}a^2-10a+5 &-20a+10 \\4a-2&a^2+8a-4\end{array}\right)=\left(\begin{array}{cc}5 & 10 \\ -2 & -4\end{array}\right)$

-20a+10=10

a=0

c)

$X=A^{-1}(-1)\cdot A(0)$

$det(A(-1)=6\cdot (-3)-(-20)=-18+20=2$

$A^t(-1)=\left(\begin{array}{ccc}6&-2\\10&-3\\\end{array}\right)\\\\A^*=\left(\begin{array}{ccc}-3&-10\\2&6\\\end{array}\right)\\\\A^{-1}= \left(\begin{array}{ccc}\frac{-3}{2} &-5\\1&3\\\end{array}\right)$

$X=A^{-1}(-1)\cdot A(0)$

$X=\left(\begin{array}{ccc}\frac{-3}{2} &-5\\1&3\\\end{array}\right)\cdot \left(\begin{array}{ccc}5 &10\\-2&-4\\\end{array}\right)=\left(\begin{array}{ccc}\frac{5}{2} &5\\-1&-2\\\end{array}\right)$

Un alt exercitiu cu matrice gasesti aici: https://brainly.ro/tema/9918882

#BAC2022

#SPJ4