Question

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Answer 1

$1)~A=\left(\begin{array}{ccc}-1&2&1\\3&0&2\\4&-1&2\end{array}\right);~~~~~B=\left(\begin{array}{ccc}-2&5&-4\\-2&6&-5\\3&-7&6\end{array}\right)\\\\Matricea~A~este~inversa~matricei~B~daca~A\cdot B=B\cdot A=I_3\\\\A\cdot B=\left(\begin{array}{ccc}-1&2&1\\3&0&2\\4&-1&2\end{array}\right)\cdot\left(\begin{array}{ccc}-2&5&-4\\-2&6&-5\\3&-7&6\end{array}\right)=$

$=\tiny{\left(\begin{array}{ccc}-1\cdot(-2)+2\cdot(-2)+1\cdot3&-1\cdot5+2\cdot6+1\cdot(-7)&-1\cdot(-4)+2\cdot(-5)+1\cdot6\\3\cdot(-2)+0\cdot(-2)+2\cdot3&3\cdot5+0\cdot5+2\cdot(-7)&3\cdot(-4)+0\cdot(-5)+2\cdot6\\4\cdot(-2)+(-1)\cdot(-2)+2\cdot3&4\cdot5+(-1)\cdot6+2\cdot(-7)&4\cdot(-4)+(-1)\cdot(-5)+2\cdot6\end{array}\right) }}=$

$=\normalsize{\left(\begin{array}{ccc}2-4+3&-5+12-7&4-10+6\\-6+0+6&15+0-14&-12+0+12\\-8+2+6&20-6-14&-16+5+12\end{array}\right) }=\left(\begin{array}{ccc}1&0&0\\0&1&0\\0&0&1\end{array}\right)=I_3$

$B\cdot A=\left(\begin{array}{ccc}-2&5&-4\\-2&6&-5\\3&-7&6\end{array}\right)\cdot\left(\begin{array}{ccc}-1&2&1\\3&0&2\\4&-1&2\end{array}\right) =\\\\\\=\scriptsize{\left(\begin{array}{ccc}-2\cdot(-1)+5\cdot3+(-4)\cdot4&-2\cdot2+5\cdot0+(-4)\cdot(-1)&-2\cdot1+5\cdot2+(-4)\cdot2\\-2\cdot(-1)+6\cdot3+(-5)\cdot4&-2\cdot2+6\cdot0+(-5)\cdot(-1)&-2\cdot1+6\cdot2+(-5)\cdot2\\3\cdot(-1)+(-7)\cdot3+6\cdot4&3\cdot2+(-7)\cdot0+6\cdot(-1)&3\cdot1+(-7)\cdot2+6\cdot2\end{array}\right)}=$

$= \normalsize{\left(\begin{array}{ccc}2+15-16&-4+0+4&-2+10-8\\2+18-20&-4+0+5&-2+12-10\\-3-21+24&6+0-6&3-14+12\end{array}\right)=\left(\begin{array}{ccc}1&0&0\\0&1&0\\0&0&1\end{array}\right) }=I_3$

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

$2)~A=\left(\begin{array}{ccc}2&-5\\3&-7\end{array}\right);~~~B=\left(\begin{array}{ccc}3&-2&0\\0&2&2\\1&-2&-3\end{array}\right)\\\\Matricea~A~este~inversabila~daca~detA\not=0\\\\\ detA=\left|\begin{array}{ccc}2&-5\\3&-7\end{array}\right|=2\cdot(-7)-(-5)\cdot3=-14+15=1\\\\detA=1\not=0\Rightarrow Matricea~A~este~inversabila.$

$Matricea~B~este~inversabila~daca~detB\not=0\\\\detB=\left|\begin{array}{ccc}3&-2&0\\0&2&2\\1&-2&-3\end{array}\right|=3\cdot2\cdot(-3)+(-2)\cdot2\cdot1+0\cdot0\cdot(-2)-0\cdot2\cdot1-\\~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~-3\cdot2\cdot(-2)-(-2)\cdot0\cdot(-3)=\\\\~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~=-18-4+0-0+12-0=-10\\\\detB=-10\not=0\Rightarrow Matricea~B~este~inversabila.$

$3)~a\in\mathbb{R};~~~~A=\left(\begin{array}{ccc}2&a\\3&-6\end{array}\right) ;~~~~B=\left(\begin{array}{ccc}a&a+1&2\\1&1&-3\\0&a&1\end{array}\right)\\\\Matricea~A~este~inversabila~daca~detA\not=0\\\\detA=\left|\begin{array}{ccc}2&a\\3&-6\end{array}\right|=2\cdot(-6)-a\cdot3=-12-3a\\\\-12-3a=0\Rightarrow -3a=12\Rightarrow a=-\dfrac{12}{3}\Rightarrow a=-4\\\\detA\not=0\Rightarrow a\not=-4\Rightarrow a\in \mathbb{R} ~\backslash~\{-4\}$

$Matricea~B~este~inversabila~daca~detB\not=0\\\\detB=\left|\begin{array}{ccc}a&a+1&2\\1&1&-3\\0&a&1\end{array}\right|=a\cdot1\cdot1+(a+1)\cdot(-3)\cdot0+2\cdot1 \cdot a-\\~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~-2\cdot1\cdot0-a\cdot(-3)\cdot a-(a+1)\cdot1\cdot1=\\\\~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~=a+0+2a-0+3a^2-a-1=3a^2+2a-1$

$3a^2+2a-1=0\\\\\Delta=2^2-4\cdot3\cdot(-1)=4+12=16 > 0\\\\a_1=\dfrac{-2-\sqrt{16} }{2\cdot3} =\dfrac{-2-4}{6}=\dfrac{-6}{6} =-1\\\\a_2=\dfrac{-2+\sqrt{16} }{2\cdot3} =\dfrac{-2+4}{6} =\dfrac{2}{6} =\dfrac{1}{3} \\\\detB\not=0\Rightarrow a_1\not=-1\\~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\Bigg|\Rightarrow a\in\mathbb{R} ~\backslash~ \bigg\{-1;~\dfrac{1}{3}\bigg\} \\~~~~~~~~~~~~~\Rightarrow a_2\not=\dfrac{1}{3}$