Question

Repede pls pana azi seara

Answer 1

$A=\left(\begin{array}{ccc}5&-1\\1&2\end{array}\right);~B=\left(\begin{array}{ccc}-1&4\\-3&1\end{array}\right)\\\\ 1)~A+B=\left(\begin{array}{ccc}5&-1\\1&2\end{array}\right)+\left(\begin{array}{ccc}-1&4\\-3&1\end{array}\right)=\left(\begin{array}{ccc}5+(-1)&-1+4\\1+(-3)&2+1\end{array}\right)=\\\\ =\left(\begin{array}{ccc}4&3\\-2&3\end{array}\right)$

$2)~A-2B=\left(\begin{array}{ccc}5&-1\\1&2\end{array}\right) -2 \cdot \left(\begin{array}{ccc}-1&4\\-3&1\end{array}\right)=\\\\ =\left(\begin{array}{ccc}5&-1\\1&2\end{array}\right)-\left(\begin{array}{ccc}2\cdot (-1)&2\cdot 4\\2 \cdot (-3)&2 \cdot 1\end{array}\right)=\left(\begin{array}{ccc}5&-1\\1&2\end{array}\right)-\left(\begin{array}{ccc}-2&8\\-6&2\end{array}\right)=\\\\=\left(\begin{array}{ccc}5-(-2)&-1-8\\1-(-6)&2-2\end{array}\right)=\left(\begin{array}{ccc}7&-9\\7&0\end{array}\right)$

$3)~A \cdot B=\left(\begin{array}{ccc}5&-1\\1&2\end{array}\right) \cdot \left(\begin{array}{ccc}-1&4\\-3&1\end{array}\right)=\\\\=\left(\begin{array}{ccc}5 \cdot (-1)+(-1) \cdot (-3)&5 \cdot 4+(-1) \cdot 1\\1 \cdot (-1)+2 \cdot (-3)&1 \cdot 4+2 \cdot 1\end{array}\right)=\left(\begin{array}{ccc}-5+3&20-1\\-1-6&4+2\end{array}\right)=\\\\=\left(\begin{array}{ccc}-2&19\\-7&6\end{array}\right)$

$4)~detA=\left|\begin{array}{ccc}5&-1\\1&2\end{array}\right|=5 \cdot 2-1 \cdot (-1)=10+1=11$

$5)~detB=\left|\begin{array}{ccc}-1&4\\-3&1\end{array}\right|=-1 \cdot 1-(-3) \cdot 4=-1+12=11 \not =0~\exists~B^{-1}$

$6)~detB=11\\\\ Transpusa~matricei~B:~^tB=\left(\begin{array}{ccc}-1&-3\\4&1\end{array}\right)\\\\ Construim~matricea~adjuncta~a~matricei~B:\\\\ b_{11}=(-1)^{1+1} \cdot 1 =1 \cdot 1=1\\\\ b_{12}=(-1)^{1+2}\cdot 4=-1 \cdot 4=-4\\\\ b_{21}=(-1)^{2+1}\cdot (-3)=-1 \cdot (-3)=3\\\\ b_{22}=(-1)^{2+2}\cdot (-1)=1 \cdot (-1)=-1\\\\ B^*=\left(\begin{array}{ccc}1&-4\\3&-1\end{array}\right)$

$\displaystyle Formula~pentru~inversa~matricei~B:~B^{-1}=\frac{1}{detB} \cdot B^*\\\\ B^{-1}=\frac{1}{11} \cdot \left(\begin{array}{ccc}1&-4\\3&-1\end{array}\right)=\left(\begin{array}{ccc}\cfrac{1}{11} &-\cfrac{4}{11} \\\cfrac{3}{11} &-\cfrac{1}{11} \end{array}\right)$

$7)~XB=A\Rightarrow X=AB^{-1}\\\\A \cdot B^{-1}=\left(\begin{array}{ccc}5&-1\\1&2\end{array}\right) \cdot\left(\begin{array}{ccc}\cfrac{1}{11} &-\cfrac{4}{11} \\\cfrac{3}{11} &-\cfrac{1}{11} \end{array}\right)=\\\\ =\left(\begin{array}{ccc}5 \cdot \cfrac{1}{11}+(-1) \cdot \cfrac{3}{11} &5 \cdot \Bigg(-\cfrac{4}{11}\Bigg)+(-1) \cdot \Bigg(-\cfrac{1}{11} \Bigg) \\1 \cdot \cfrac{1}{11}+2 \cdot \cfrac{3}{11} &1 \cdot \Bigg(-\cfrac{4}{11}\Bigg)+2 \cdot \Bigg(-\cfrac{1}{11}\Bigg) \end{array}\right)=$

$\displaystyle =\left(\begin{array}{ccc}\cfrac{2}{11} &-\cfrac{19}{11} \\\cfrac{7}{11} &-\cfrac{6}{11} \end{array}\right)=X$

$\displaystyle 8)~\left|\begin{array}{ccc}-1&1\\x+2&-3\end{array}\right|=2x-5 \Rightarrow -1 \cdot (-3)-(x+2) \cdot 1=2x-5 \Rightarrow \\\\ \Rightarrow 3-x-2=2x-5 \Rightarrow -x-2x=-5-3+2 \Rightarrow -3x=-6 \Rightarrow x=\frac{6}{3} \Rightarrow \\\\ \Rightarrow x=2$

$9)~\left|\begin{array}{ccc}-1&2&-1\\1&1&3\\1&2&1\end{array}\right|=-1 \cdot 1 \cdot 1+(-1) \cdot 1 \cdot 2+2 \cdot 3 \cdot 1 -(-1) \cdot 1 \cdot 1-\\\\~~~~~~~~~~~~~~~~~~~~~~~~~~-2 \cdot 1 \cdot 1-(-1) \cdot 3\cdot 2=-1-2+6+1-2+6=8\\--------------------------------$

$A=\left(\begin{array}{ccc}-1&2\\-1&3\end{array}\right);~B=\left(\begin{array}{ccc}1&-1\\-2&-4\end{array}\right)\\\\ 1)~A+B=\left(\begin{array}{ccc}-1&2\\-1&3\end{array}\right)+\left(\begin{array}{ccc}1&-1\\-2&-4\end{array}\right)=\left(\begin{array}{ccc}-1+1&2+(-1)\\-1+(-2)&3+(-4)\end{array}\right)=\\\\=\left(\begin{array}{ccc}0&1\\-3&-1\end{array}\right)$

$2)~A-2B=\left(\begin{array}{ccc}-1&2\\-1&3\end{array}\right)-2 \cdot \left(\begin{array}{ccc}1&-1\\-2&-4\end{array}\right)=\\\\=\left(\begin{array}{ccc}-1&2\\-1&3\end{array}\right)-\left(\begin{array}{ccc}2\cdot1&2\cdot(-1)\\2\cdot(-2)&2\cdot(-4)\end{array}\right)=\left(\begin{array}{ccc}-1&2\\-1&3\end{array}\right)-\left(\begin{array}{ccc}2&-2\\-4&-8\end{array}\right)=\\\\=\left(\begin{array}{ccc}-1-2&2-(-2)\\-1-(-4)&3-(-8)\end{array}\right)=\left(\begin{array}{ccc}-3&4\\3&11\end{array}\right)$

$3)~A \cdot B=\left(\begin{array}{ccc}-1&2\\-1&3\end{array}\right) \cdot \left(\begin{array}{ccc}1&-1\\-2&-4\end{array}\right)=\\\\=\left(\begin{array}{ccc}-1 \cdot 1+2 \cdot (-2)&-1\cdot (-1)+2\cdot(-4)\\-1 \cdot 1+3 \cdot (-2)&-1 \cdot (-1)+3\cdot(-4)\end{array}\right)=\left(\begin{array}{ccc}-1-4&1-8\\-1-6&1-12\end{array}\right)=\\\\=\left(\begin{array}{ccc}-5&-7\\-7&-11\end{array}\right)$

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

$5)~detB=\left|\begin{array}{ccc}1&-1\\-2&-4\end{array}\right|=1 \cdot (-4)-(-2)\cdot (-1)=-4-2=-6 \not=0~\exists~B^{-1}$