Question

Exercitiile 15 si 16 rog frumos..

Answer 1

$15.~~A=\left(\begin{array}{ccc}2&1\\-1&-3\end{array}\right);~~~I_2=\left(\begin{array}{ccc}1&0\\0&1\end{array}\right);~~~B(a)=\left(\begin{array}{ccc}a&a\\3&6\end{array}\right);~~~a\in\mathbb{R}\\\\a)~detA=-5\\\\detA=\left|\begin{array}{ccc}2&1\\-1&-3\end{array}\right|=2\cdot(-3)-1\cdot(-1)=-6+1=-5$

$b)~B(1)\cdot B(-1)+3A=4B(a);~~~~~a=?\\\\B(1)=\left(\begin{array}{ccc}1&1\\3&6\end{array}\right);~~~~~B(-1)=\left(\begin{array}{ccc}-1&-1\\3&6\end{array}\right)\\\\3A=3 \cdot \left(\begin{array}{ccc}2&1\\-1&-3\end{array}\right)=\left(\begin{array}{ccc}3\cdot 2&3\cdot1\\3\cdot(-1)&3\cdot (-3)\end{array}\right)=\left(\begin{array}{ccc}6&3\\-3&-9\end{array}\right)$

$4B(a)=4 \cdot \left(\begin{array}{ccc}a&a\\3&6\end{array}\right)=\left(\begin{array}{ccc}4a&4a\\4\cdot 3&4\cdot6\end{array}\right)=\left(\begin{array}{ccc}4a&4a\\12&24\end{array}\right)$

$B(1)\cdot B(-1)=\left(\begin{array}{ccc}1&1\\3&6\end{array}\right) \cdot \left(\begin{array}{ccc}-1&-1\\3&6\end{array}\right)=\\\\=\left(\begin{array}{ccc}1\cdot(-1)+1\cdot3&1\cdot(-1)+1\cdot 6\\3\cdot(-1)+6\cdot 3&3\cdot(-1)+6\cdot6\end{array}\right)=\left(\begin{array}{ccc}-1+3&-1+6\\-3+18&-3+36\end{array}\right)=\\\\=\left(\begin{array}{ccc}2&5\\15&33\end{array}\right)$

$B(1)\cdot B(-1)+3A=\left(\begin{array}{ccc}2&5\\15&33\end{array}\right)+\left(\begin{array}{ccc}6&3\\-3&-9\end{array}\right)=\\\\=\left(\begin{array}{ccc}2+6&5+3\\15+(-3)&33+(-9)\end{array}\right)=\left(\begin{array}{ccc}8&8\\12&24\end{array}\right)$

$B(1)\cdot B(-1)+3A=4B(a) \Rightarrow \left(\begin{array}{ccc}8&8\\12&24\end{array}\right)=\left(\begin{array}{ccc}4a&4a\\12&24\end{array}\right)\Rightarrow \\\\\Rightarrow 8=4a\Rightarrow a=\cfrac{8}{4} \Rightarrow a=2$

$c)~X\in M_2(R)~~~~~~~~~~~~~X\cdot\underbrace{(A-2I_2)}=\underbrace{B(0)}~~~~~~~~~XC=B\Rightarrow X=BC^{-1}\\~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~C~~~~~~~~~~~B\\\\I_2=\left(\begin{array}{ccc}1&0\\0&1\end{array}\right)\Rightarrow 2I_2=2\cdot \left(\begin{array}{ccc}1&0\\0&1\end{array}\right)=\left(\begin{array}{ccc}2\cdot 1&2\cdot0\\2\cdot0&2\cdot 1\end{array}\right)=\left(\begin{array}{ccc}2&0\\0&2\end{array}\right)\\\\B(0)=\left(\begin{array}{ccc}0&0\\3&6\end{array}\right)=B$

$A-2I_2=\left(\begin{array}{ccc}2&1\\-1&-3\end{array}\right)-\left(\begin{array}{ccc}2&0\\0&2\end{array}\right)=\left(\begin{array}{ccc}2-2&1-0\\-1-0&-3-2\end{array}\right)=\\\\=\left(\begin{array}{ccc}0&1\\-1&-5\end{array}\right)=C$

$detC=\left|\begin{array}{ccc}0&1\\-1&-5\end{array}\right|=0\cdot (-5)-1\cdot(-1)=0+1=1 \Rightarrow \\~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\Rightarrow Matricea~este~inversabila.\\\\ Transpusa~matricei~C:~^tC=\left(\begin{array}{ccc}0&-1\\1&-5\end{array}\right)\\\\ Construim~matricea~adjuncta~a~matricei~C:\\\\a_{11}=(-1)^{1+1}\cdot(-5)=-5\\\\a_{12}=(-1)^{1+2}\cdot1=-1\\\\a_{21}=(-1)^{2+1}\cdot(-1)=1\\\\a_{22}=(-1)^{2+2}\cdot 0=0$

$Matricea~adjuncta~este:~~C^*=\left(\begin{array}{ccc}-5&-1\\1&0\end{array}\right)$

$Aplicam~formula:~C^{-1}=\cfrac{1}{detC} \cdot C^*\\\\C^{-1}=\cfrac{1}{1} \cdot \left(\begin{array}{ccc}-5&-1\\1&0\end{array}\right)=\left(\begin{array}{ccc}-5&-1\\1&0\end{array}\right)$

$BC^{-1}=\left(\begin{array}{ccc}0&0\\3&6\end{array}\right) \cdot \left(\begin{array}{ccc}-5&-1\\1&0\end{array}\right)=\\\\=\left(\begin{array}{ccc}0 \cdot (-5)+0\cdot 1&0\cdot (-1)+0\cdot 0\\3\cdot (-5)+6\cdot 1&3\cdot(-1)+6\cdot0\end{array}\right)=\left(\begin{array}{ccc}0+0&0+0\\-15+6&-3+0\end{array}\right)=\\\\=\left(\begin{array}{ccc}0&0\\-9&-3\end{array}\right) \Rightarrow X=\left(\begin{array}{ccc}0&0\\-9&-3\end{array}\right)$

$16.~~A=\left(\begin{array}{ccc}3&-4\\-2&3\end{array}\right);~~~B=\left(\begin{array}{ccc}-1&x\\1&-1\end{array}\right);~~~x\in\mathbb{R}$

$a)~detA=1\\\\detA=\left|\begin{array}{ccc}3&-4\\-2&3\end{array}\right|=3\cdot3-(-4)\cdot(-2)=9-8=1$

$b)~B \cdot B=A;~~~~~~x=?\\\\B\cdot B=\left(\begin{array}{ccc}-1&x\\1&-1\end{array}\right) \cdot \left(\begin{array}{ccc}-1&x\\1&-1\end{array}\right)=\\\\=\left(\begin{array}{ccc}(-1)\cdot (-1)+x\cdot 1&(-1)\cdot x+x \cdot (-1)\\1\cdot(-1)+(-1)\cdot 1&1\cdot x+(-1)\cdot (-1)\end{array}\right)=\\\\=\left(\begin{array}{ccc}1+x&-x-x\\-1-1&x+1\end{array}\right)=\left(\begin{array}{ccc}1+x&-2x\\-2&x+1\end{array}\right)$

$B\cdot B=A \Rightarrow \left(\begin{array}{ccc}1+x&-2x\\-2&x+1\end{array}\right)=\left(\begin{array}{ccc}3&-4\\-2&3\end{array}\right) \Rightarrow \\\\ \Rightarrow \left\{\begin{array}{ccc}1+x=3\Rightarrow x=3-1\Rightarrow x=2\\-2x=-4\Rightarrow x=\cfrac{4}{2} \Rightarrow x=2\\-2=-2\\x+1=3\Rightarrow x=3-1\Rightarrow x=2\end{array}\right~~~~~~~~~~~~~~~~~~~~~~~~~~~x=2$