Question

Subiectul al II-lea va rog

Answer 1

$\displaystyle \mathtt{Subiectul~al~II-lea}\\ \\ \mathtt{1.~~~A(a)= \left(\begin{array}{ccc}\mathtt{a-4}&\mathtt3\\\mathtt8&\mathtt{a-6}\end{array}\right)} \\ \\ \mathtt{a)~det~A(5)=?}\\ \\ \mathtt{A(5)=\left(\begin{array}{ccc}\mathtt{5-4}&\mathtt3\\\mathtt8&\mathtt{5-6}\end{array}\right)=\left(\begin{array}{ccc}\mathtt1&\mathtt3\\\mathtt8&\mathtt{-1}\end{array}\right)}\\ \\ \mathtt{det~A(5)=\left|\begin{array}{ccc}\mathtt1&\mathtt3\\\mathtt8&\mathtt{-1}\end{array}\right|=1\cdot(-1)-8\cdot3=-1-24=-25}$

$\displaystyle \mathtt{b)~det~A(a)=\left|\begin{array}{ccc}\mathtt{a-4}&\mathtt3\\\mathtt8&\mathtt{a-6}\end{array}\right|=(a-4)(a-6)-8\cdot3=}\\ \\ \mathtt{=a^2-6a-4a+24-24=a^2-10a=a(a-10)}\\ \\ \mathtt{det~A(a)\ne0\Rightarrow a(a-10)\ne0\Rightarrow a\ne0}\\ \\ \mathtt{~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\Rightarrow a\ne10}\\ \\ \mathtt{a \in \mathbb{R}-\{0,10\}}$

$\displaystyle \mathtt{c)~a=2~~~~~~~~~~~~~~~~~~~~A^2-3A=?}\\ \\ \mathtt{a=2\Rightarrow A=\left(\begin{array}{ccc}\mathtt{2-4}&\mathtt3\\\mathtt8&\mathtt{2-6}\end{array}\right)=\left(\begin{array}{ccc}\mathtt{-2}&\mathtt3\\\mathtt8&\mathtt{-4}\end{array}\right)}$

$\displaystyle \mathtt{A^2=A\cdot A=\left(\begin{array}{ccc}\mathtt{-2}&\mathtt3\\\mathtt8&\mathtt{-4}\end{array}\right)\left(\begin{array}{ccc}\mathtt{-2}&\mathtt3\\\mathtt8&\mathtt{-4}\end{array}\right)=}\\ \\ \mathtt{=\left(\begin{array}{ccc}\mathtt{(-2)\cdot(-2)+3\cdot8}&\mathtt{(-2)\cdot3+3\cdot(-4)}\\\mathtt{8\cdot(-2)+(-4)\cdot8}&\mathtt{8\cdot3+(-4)\cdot(-4)}\end{array}\right)=}$

$\displaystyle \mathtt{=\left(\begin{array}{ccc}\mathtt{4+24}&\mathtt{-6-12}\\\mathtt{-16-32}&\mathtt{24+16}\end{array}\right)=\left(\begin{array}{ccc}\mathtt{28}&\mathtt{-18}\\\mathtt{-48}&\mathtt{40}\end{array}\right)}$

$\displaystyle \mathtt{3A=3\cdot\left(\begin{array}{ccc}\mathtt{-2}&\mathtt3\\\mathtt8&\mathtt{-4}\end{array}\right) =\left(\begin{array}{ccc}\mathtt{3\cdot(-2)}&\mathtt{3\cdot3}\\\mathtt{3\cdot8}&\mathtt{3\cdot(-4)}\end{array}\right)=\left(\begin{array}{ccc}\mathtt{-6}&\mathtt9\\\mathtt{24}&\mathtt{-12}\end{array}\right)}$

$\displaystyle \mathtt{A^2-3A=\left(\begin{array}{ccc}\mathtt{28}&\mathtt{-18}\\\mathtt{-48}&\mathtt{40}\end{array}\right)-\left(\begin{array}{ccc}\mathtt{-6}&\mathtt9\\\mathtt{24}&\mathtt{-12}\end{array}\right)=}\\ \\ \mathtt{=\left(\begin{array}{ccc}\mathtt{28-(-6)}&\mathtt{-18-9}\\\mathtt{-48-24}&\mathtt{40-(-12)}\end{array}\right)=\left(\begin{array}{ccc}\mathtt{34}&\mathtt{-27}\\\mathtt{-72}&\mathtt{52}\end{array}\right)}$

$\displaystyle \mathtt{2.~~~x*y=3xy+12(x+y)+44}\\ \\ \mathtt{a)~(-4)*4=?}\\ \\ \mathtt{(-4)*4=3\cdot(-4)\cdot4+12(-4+4)+44=-12\cdot4+12\cdot0+44=}\\ \\ \mathtt{=-48+0+44=-4}\\ \\ \mathtt{b)~3(x+4)(y+4)-4=3xy+12x+12y+48-4=}\\ \\ \mathtt{=3xy+12x+12y+44=3xy+12(x+y)+44=x*y}$

$\displaystyle\mathtt{c)~x*x=5}\\\\\mathtt{3x\cdot x+12(x+x)+44=5}\\\\\mathtt{3x^2+12\cdot2x+44-5=0}\\ \\ \mathtt{3x^2+24x+39=0\big|:3}\\\\\mathtt{x^2+8x+13=0}\\\\ \mathtt{a=1,~b=8,~c=13}\\\\\mathtt{\Delta=b^2-4ac=8^2-4\cdot1\cdot13=64-52=12\ \textgreater \ 0}\\\\\mathtt{x_1=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-8-\sqrt{12}}{2\cdot1}=\frac{-8-2\sqrt{3}}{2}=\frac{2(-4-\sqrt{3})}{2}=-4-\sqrt{3}}\\\\ \mathtt{x_2=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-8+\sqrt{12}}{2\cdot1}= \frac{-8+2\sqrt{3}}{2}=\frac{2(-4+\sqrt{3})}{2}=-4+\sqrt{3}}$