Question

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Answer 1

$1)~A=\left(\begin{array}{ccc}2&2&3\\1&-1&0\\-1&2&1\end{array}\right);~~~~B=\left(\begin{array}{ccc}1&-4&-3\\1&-5&-3\\-1&6&4\end{array}\right)\\\\Matricea~A~este~inversa~matricei~B~daca~A\cdot B=B\cdot A=I_3\\\\A\cdot B=\left(\begin{array}{ccc}2&2&3\\1&-1&0\\-1&2&1\end{array}\right)\cdot\left(\begin{array}{ccc}1&-4&-3\\1&-5&-3\\-1&6&4\end{array}\right)=$

$=\tiny\left(\begin{array}{ccc}2\cdot1+2\cdot1+3\cdot(-1)&2\cdot(-4)+2\cdot(-5)+3\cdot6&2\cdot(-3)+2\cdot(-3)+3\cdot4\\1\cdot1+(-1)\cdot1+0\cdot(-1)&1\cdot(-4)+(-1)\cdot(-5)+0\cdot6&1\cdot(-3)+(-1)\cdot(-3)+0\cdot4\\-1\cdot1+2\cdot1+1\cdot(-1)&-1\cdot(-4)+2\cdot(-5)+1\cdot6&-1\cdot(-3)+2\cdot(-3)+1\cdot4\end{array}\right)=$

$=\normalsize\left(\begin{array}{ccc}2+2-3&-8-10+18&-6-6+12\\1-1+0&-4+5+0&-3+3+0\\-1+2-1&4-10+6&3-6+4\end{array}\right)=\left(\begin{array}{ccc}1&0&0\\0&1&0\\0&0&1\end{array}\right)=I_3\\\\\\B\cdot A=\left(\begin{array}{ccc}1&-4&-3\\1&-5&-3\\-1&6&4\end{array}\right)\cdot\left(\begin{array}{ccc}2&2&3\\1&-1&0\\-1&2&1\end{array}\right)=$

$=\tiny\left(\begin{array}{ccc}1\cdot2+(-4)\cdot1+(-3)\cdot(-1)&1\cdot2+(-4)\cdot(-1)+(-3)\cdot2&1\cdot3+(-4)\cdot0+(-3)\cdot1\\1\cdot2+(-5)\cdot1+(-3)\cdot(-1)&1\cdot2+(-5)\cdot(-1)+(-3)\cdot2&1\cdot3+(-5)\cdot0+(-3)\cdot1\\-1\cdot2+6\cdot1+4\cdot(-1)&-1\cdot2+6\cdot(-1)+4\cdot2&-1\cdot3+6\cdot0+4\cdot1\end{array}\right)=\\\\\\=\normalsize\left(\begin{array}{ccc}2-4+3&2+4-6&3+0-3\\2-5+3&2+5-6&3+0-3\\-2+6-4&-2-6+8&-3+0+4\end{array}\right)=\left(\begin{array}{ccc}1&0&0\\0&1&0\\0&0&1\end{array}\right)=I_3$

$A\cdot B=B\cdot A=I_3\Rightarrow A~este~inversa~matricei~B$

$~A=\left(\begin{array}{ccc}-3&-5\\-1&2\end{array}\right),~~~~B=\left(\begin{array}{ccc}1&1&1\\4&3&2\\1&2&3\end{array}\right)\\\\Matricea~A~este~inversabila~daca~detA\not=0\\\\detA=\left|\begin{array}{ccc}-3&-5\\-1&2\end{array}\right|=-3\cdot2-(-5)\cdot(-1)=-6-5=-11\\\\detA=-11\not=0\Rightarrow Matricea~A~este~inversabila$

$Matricea~B~est~inversabila~daca~detB\not=0\\\\detB=\left|\begin{array}{ccc}1&1&1\\4&3&2\\1&2&3\end{array}\right|=1\cdot 3 \cdot 3+1\cdot2\cdot1+1\cdot4\cdot2-1\cdot3\cdot1-1\cdot2\cdot2-\\~~~~~~~~~~~~~~~~~~~~~~~~~~~-1\cdot4\cdot3=9+2+8-3-4-12=0\\\\detB=0\Rightarrow Matricea~B~nu~este~inversabila$

$3)~a\in\mathbb{R};~~~A=\left(\begin{array}{ccc}1&a\\3&6\end{array}\right);~~~B=\left(\begin{array}{ccc}3&1&-1\\1&2&a\\2&-1&1\end{array}\right)\\\\Matricea~A~este~inversabila~daca~detA\not=0\\\\detA=\left|\begin{array}{ccc}1&a\\3&6\end{array}\right|=1\cdot6-a\cdot3=6-3a\\\\6-3a=0\Rightarrow -3a=-6\Rightarrow a=\dfrac{6}{3}\Rightarrow a=2\\\\detA\not=0\Rightarrow a\not=2\Rightarrow a\in\mathbb{R}~ \backslash~ \{2\}$

$Matricea~B~este~inversabila~daca~detB\not=0\\\\detB=\left|\begin{array}{ccc}3&1&-1\\1&2&a\\2&-1&1\end{array}\right|=3\cdot2\cdot1+1\cdot a\cdot2+(-1)\cdot1\cdot(-1)-(-1)\cdot2\cdot2-\\~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~-3\cdot a\cdot(-1)-1\cdot1\cdot1=6+2a+1+4+3a-1=\\\\~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~=5a+10\\\\5a+10=0\Rightarrow 5a=-10\Rightarrow a=-\dfrac{10}{5} \Rightarrow a=-2\\\\detB\not=0\Rightarrow a\not=-2\Rightarrow a\in\mathbb{R}~\backslash~\{-2\}$