Question

Ofer toate punctele ​

Answer 1

Răspuns:

Explicație pas cu pas:

Answer 2

$1.~~A(x)=\left(\begin{array}{ccc}x+2&3\\3&x+2\end{array}\right),~x\in\mathbb{R},~B=\left(\begin{array}{ccc}-1&5\\5&-1\end{array}\right),~I_2=\left(\begin{array}{ccc}1&0\\0&1\end{array}\right)$

$a)~det(A(2)-I_2)=?\\\\A(2)=\left(\begin{array}{ccc}2+2&3\\3&2+2\end{array}\right)=\left(\begin{array}{ccc}4&3\\3&4\end{array}\right)\\\\A(2)-I_2=\left(\begin{array}{ccc}4&3\\3&4\end{array}\right)-\left(\begin{array}{ccc}1&0\\0&1\end{array}\right)=\left(\begin{array}{ccc}4-1&3-0\\3-0&4-1\end{array}\right)=\left(\begin{array}{ccc}3&3\\3&3\end{array}\right)\\\\det(A(2)-I_2)=\left|\begin{array}{ccc}3&3\\3&3\end{array}\right|=3 \cdot 3-3\cdot 3=9-9=0$

$b)~det(A(a))=0,~a\in\mathbb{R}~~~~~~~~~~~~~~~~~A(a)=\left(\begin{array}{ccc}a+2&3\\3&a+2\end{array}\right)\\\\det(A(a))=\left|\begin{array}{ccc}a+2&3\\3&a+2\end{array}\right|=(a+2)(a+2)-3\cdot 3=\\\\=a^2+2a+2a+4-9=a^2+4a-5\\\\det(A(a))=0\Rightarrow a^2+4a-5=0\\\\ \Delta=4^2-4\cdot 1 \cdot (-5)=16+20=36\\\\a_1=\cfrac{-4-\sqrt{36} }{2 \cdot 1} =\cfrac{-4-6}{2} =\cfrac{-10}{2} =-5\\\\a_2=\cfrac{-4+\sqrt{36} }{2 \cdot 1} =\cfrac{-4+6}{2} =\cfrac{2}{2}=1$

$c)~X\in M_2(\mathbb{R})~~~~~~~~~~~~~~~~~~A(-1)\cdot X=B\Rightarrow X=(A(-1))^{-1}B\\\\A(-1)=\left(\begin{array}{ccc}-1+2&3\\3&-1+2\end{array}\right)=\left(\begin{array}{ccc}1&3\\3&1\end{array}\right)\\\\det(A(-1))=\left|\begin{array}{ccc}1&3\\3&1\end{array}\right|=1\cdot1-3\cdot3=1-9=-8 \not=0 \Rightarrow \\~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\Rightarrow Matricea~este~inversabila.$

$Transpusa~matricei~A(-1):~^tA(-1)=\left(\begin{array}{ccc}1&3\\3&1\end{array}\right)\\\\Construim~matricea~adjuncta~a~matricei~A(-1):\\\\a_{11}=(-1)^{1+1}\cdot 1=1 \cdot 1=1\\\\a_{12}=(-1)^{1+2}\cdot 3=(-1)\cdot 3=-3\\\\a_{21}=(-1)^{2+1}\cdot 3=(-1) \cdot 3=-3\\\\a_{22}=(-1)^{2+2}\cdot 1=1 \cdot 1=1$

$Matricea~adjuncta:~(A(-1))^*=\left(\begin{array}{ccc}1&-3\\-3&1\end{array}\right)\\\\ Aplicam~formula:~(A(-1))^{-1}=\cfrac{1}{det(A(-1))}\cdot (A(-1))^* \\\\(A(-1))^{-1}=\Bigg(-\cfrac{1}{8}\Bigg) \cdot \left(\begin{array}{ccc}1&-3\\-3&1\end{array}\right)=$

$=\left(\begin{array}{ccc}\Bigg(-\cfrac{1}{8}\Bigg)\cdot 1 &\Bigg(-\cfrac{1}{8}\Bigg)\cdot (-3) \\\Bigg(-\cfrac{1}{8}\Bigg) \cdot (-3) &\Bigg(-\cfrac{1}{8}\Bigg)\cdot 1 \end{array}\right)=\left(\begin{array}{ccc}-\cfrac{1}{8} &\cfrac{3}{8} \\\cfrac{3}{8} &-\cfrac{1}{8} \end{array}\right)$

$(A(-1))^{-1}B=\left(\begin{array}{ccc}-\cfrac{1}{8} &\cfrac{3}{8} \\\cfrac{3}{8} &-\cfrac{1}{8} \end{array}\right) \cdot \left(\begin{array}{ccc}-1&5\\5&-1\end{array}\right)=\\\\=\left(\begin{array}{ccc}\Bigg(-\cfrac{1}{8} \Bigg) \cdot(-1)+\cfrac{3}{8}\cdot 5 &\Bigg(-\cfrac{1}{8}\Bigg)\cdot 5+\cfrac{3}{8}\cdot (-1) \\\cfrac{3}{8}\cdot (-1)+\Bigg(-\cfrac{1}{8}\Bigg) \cdot 5 &\cfrac{3}{8} \cdot 5+\Bigg(-\cfrac{1}{8}\Bigg) \cdot(-1) \end{array}\right)=$

$\left(\begin{array}{ccc}\cfrac{1}{8} +\cfrac{15}{8} &-\cfrac{5}{8} -\cfrac{3}{8} \\-\cfrac{3}{8}-\cfrac{5}{8} &\cfrac{15}{8}+\cfrac{1}{8} \end{array}\right)=\left(\begin{array}{ccc}\cfrac{16}{8} &-\cfrac{8}{8} \\-\cfrac{8}{8} &\cfrac{16}{8} \end{array}\right)=\left(\begin{array}{ccc}2&-1\\-1&2\end{array}\right)=X$

$X=\left(\begin{array}{ccc}2&-1\\-1&2\end{array}\right)$