Question

Cine se pricepe la mate

Answer 1

$\displaystyle \mathtt{1.~~A= \left(\begin{array}{ccc}\mathtt1&\mathtt1&\mathtt{-1}\\\mathtt2&\mathtt1&\mathtt0\\\mathtt1&\mathtt{-1}&\mathtt1\end{array}\right)~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~A^{-1}=\frac{1}{det(A)}\cdot A^*}$

$\displaystyle \mathtt{det(A)=\left|\begin{array}{ccc}\mathtt1&\mathtt1&\mathtt{-1}\\\mathtt2&\mathtt1&\mathtt0\\\mathtt1&\mathtt{-1}&\mathtt1\end{array}\right|=1\cdot1\cdot1+(-1)\cdot2\cdot(-1)+1\cdot0\cdot1-}\\\\\mathtt{-(-1)\cdot1\cdot1-1\cdot2\cdot1-1\cdot0\cdot(-1)=2}$

$\displaystyle\mathtt{A=\left(\begin{array}{ccc}\mathtt1&\mathtt1&\mathtt{-1}\\\mathtt2&\mathtt1&\mathtt0\\\mathtt1&\mathtt{-1}&\mathtt1\end{array}\right)\Rightarrow A^{tr}=\left(\begin{array}{ccc}\mathtt1&\mathtt2&\mathtt1\\\mathtt1&\mathtt1&\mathtt{-1}\\\mathtt{-1}&\mathtt0&\mathtt1\end{array}\right)}$

$\displaystyle\mathtt{D_{11}=(-1)^{1+1}\cdot\left|\begin{array}{ccc}\mathtt1&\mathtt{-1}\\\mathtt0&\mathtt1\end{array}\right|=1\cdot1=1}\\\\ \mathtt{D_{12}=(-1)^{1+2}\cdot\left|\begin{array}{ccc}\mathtt1&\mathtt{-1}\\\mathtt{-1}&\mathtt1\end{array}\right|=(-1)\cdot0=0}\\\\\mathtt{D_{13}=(-1)^{1+3}\cdot \left|\begin{array}{ccc}\mathtt1&\mathtt1\\\mathtt{-1}&\mathtt0\end{array}\right|=1\cdot1=1}$
$\displaystyle\mathtt{D_{21}=(-1)^{2+1}\cdot\left|\begin{array}{ccc}\mathtt2&\mathtt1\\\mathtt0&\mathtt1\end{array}\right|=(-1)\cdot2=-2}\\\\ \mathtt{D_{22}=(-1)^{2+2}\cdot\left|\begin{array}{ccc}\mathtt1&\mathtt1\\\mathtt{-1}&\mathtt1\end{array}\right|=1\cdot2=2}\\\\ \mathtt{D_{23}=(-1)^{2+3}\cdot \left|\begin{array}{ccc}\mathtt1&\mathtt2\\\mathtt{-1}&\mathtt0\end{array}\right|=(-1)\cdot2=-2}$
$\displaystyle \mathtt{D_{31}=(-1)^{3+1}\cdot\left|\begin{array}{ccc}\mathtt2&\mathtt1\\\mathtt1&\mathtt{-1}\end{array}\right|=1\cdot (-3)=-3}\\\\\mathtt{D_{32}=(-1)^{3+2}\cdot\left|\begin{array}{ccc}\mathtt1&\mathtt1\\\mathtt1&\mathtt{-1}\end{array}\right|=(-1)\cdot (-2)=2}\\\\\mathtt{D_{33}=(-1)^{3+3}\cdot \left|\begin{array}{ccc}\mathtt1&\mathtt2\\\mathtt1&\mathtt1\end{array}\right|=1\cdot(-1)=-1}$

$\displaystyle\mathtt{A^*=\left(\begin{array}{ccc}\mathtt1&\mathtt0&\mathtt1\\\mathtt{-2}&\mathtt2&\mathtt{-2}\\\mathtt{-3}&\mathtt2&\mathtt{-1}\end{array}\right)}\\\\\mathtt{A^{-1}=\frac{1}{2}\cdot \left(\begin{array}{ccc}\mathtt1&\mathtt0&\mathtt1\\\mathtt{-2}&\mathtt2&\mathtt{-2}\\\mathtt{-3}&\mathtt2&\mathtt{-1}\end{array}\right)=\left(\begin{array}{ccc}\mathtt{ \frac{1}{2}}&\mathtt0&\mathtt{\frac{1}{2} }\\\mathtt{-1}&\mathtt1&\mathtt{-1}\\\mathtt{-\frac{3}{2}}&\mathtt1&\mathtt{- \frac{1}{2}}\end{array}\right)}$

$\displaystyle \mathtt{2.~~\left\{\begin{array}{ccc}\mathtt{x+y+2z=-1}\\\mathtt{2x-y+2z=-4}\\\mathtt{4x+y+4z=-2}\end{array}\right\Rightarrow A= \left(\begin{array}{ccc}\mathtt1&\mathtt1&\mathtt2\\\mathtt2&\mathtt{-1}&\mathtt2\\\mathtt4&\mathtt1&\mathtt4\end{array}\right)}$

$\displaystyle \mathtt{\Delta=det(A)=\left|\begin{array}{ccc}\mathtt1&\mathtt1&\mathtt2\\\mathtt2&\mathtt{-1}&\mathtt2\\\mathtt4&\mathtt1&\mathtt4\end{array}\right|=1\cdot(-1)\cdot4+2\cdot2\cdot1+1\cdot2\cdot4-}\\\\\mathtt{-2\cdot(-1)\cdot4-1\cdot2\cdot4-1\cdot2\cdot1=6}\\\\ \mathtt{\Delta=det(A)=6\ne0}$

$\displaystyle\mathtt{\Delta_x=\left|\begin{array}{ccc}\mathtt{-1}&\mathtt1&\mathtt2\\\mathtt{-4}&\mathtt{-1}&\mathtt2\\\mathtt{-2}&\mathtt1&\mathtt4\end{array}\right|=(-1)\cdot(-1)\cdot4+2\cdot(-4)\cdot1+1\cdot2\cdot(-2)-}\\\\ \mathtt{-2\cdot(-1)\cdot(-2)-1\cdot(-4)\cdot4-(-1)\cdot2\cdot1=6}\\\\ \mathtt{\Delta_x=6}$

$\displaystyle \mathtt{\Delta_y=\left|\begin{array}{ccc}\mathtt1&\mathtt{-1}&\mathtt2\\\mathtt2&\mathtt{-4}&\mathtt2\\\mathtt4&\mathtt{-2}&\mathtt4\end{array}\right|=1\cdot(-4)\cdot4+2\cdot2\cdot(-2)+(-1)\cdot2\cdot4-}\\\\\mathtt{-2\cdot(-4)\cdot4-(-1)\cdot2\cdot4-1\cdot2\cdot(-2)=12}\\\\\mathtt{\Delta_y=12}$

$\displaystyle \mathtt{\Delta_z=\left|\begin{array}{ccc}\mathtt1&\mathtt1&\mathtt{-1}\\\mathtt2&\mathtt{-1}&\mathtt{-4}\\\mathtt4&\mathtt1&\mathtt{-2}\end{array}\right|=1\cdot(-1)\cdot(-2)+(-1)\cdot2\cdot1+1\cdot(-4)\cdot4-}\\\\\mathtt{-(-1)\cdot(-1)\cdot4-1\cdot2\cdot(-2)-1\cdot(-4)\cdot1=-12}\\\\\mathtt{\Delta_z=-12}$

$\displaystyle \mathtt{x=\frac{\Delta_x}{\Delta}=\frac{6}{6}=1~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~x=1}\\\\\mathtt{y= \frac{\Delta_x}{\Delta}=\frac{12}{6}=2~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~y=2}~\\\\ \mathtt{z=\frac{\Delta_z}{\Delta}=\frac{-12}{6}=-2~~~~~~~~~~~~~~~~~~~~~~~~~~z=-2}$

$\displaystyle \mathtt{4.~a)~f(x)=3x^3-2x^2+6x}\\\\\mathtt{f'(x)=\left(3x^3-2x^2+6x\right)'=\left(3x^3\right)'-\left(2x^2\right)'+(6x)'=}\\\\ \mathtt{=3\left(x^3\right)'-2\left(x^2\right)+6x'=3\cdot3x^2-2\cdot2x+6\cdot1=9x^2-4x+6}$

$\displaystyle \mathtt{b)~f(x)=\frac{1}{x-1}}\\\\\mathtt{f'(x)=\left(\frac{1}{x-1}\right)'=\frac{1'\cdot(x-1)-1\cdot(x-1)'}{(x-1)^2}=}\\\\\mathtt{= \frac{0\cdot(x-1)-1\cdot(x'-1')}{(x-1)^2}=\frac{-1\cdot(1-0)}{(x-1)^2}=-\frac{1}{(x-1)^2}}$