Question

La punctul 2 vă rog. Dau coroana !

Answer 1

$\displaystyle \mathtt{A(a)= \left(\begin{array}{ccc}\mathtt a&\mathtt4\\\mathtt3&\mathtt{-a}\end{array}\right),a \in \mathbb{R}}\\ \\ \mathtt{a)A(1)+A(-1)=? }\\ \\ \mathtt{A(1)= \left(\begin{array}{ccc}\mathtt 1&\mathtt4\\\mathtt3&\mathtt{-1}\end{array}\right);A(-1)= \left(\begin{array}{ccc}\mathtt {-1}&\mathtt4\\\mathtt3&\mathtt1\end{array}\right)}$
$\displaystyle \mathtt{A(1)+A(-1)= \left(\begin{array}{ccc}\mathtt 1&\mathtt4\\\mathtt3&\mathtt{-1}\end{array}\right)+\left(\begin{array}{ccc}\mathtt {-1}&\mathtt4\\\mathtt3&\mathtt1\end{array}\right)=\left(\begin{array}{ccc}\mathtt {1+(-1)}&\mathtt{4+4}\\\mathtt{3+3}&\mathtt{-1+1}\end{array}\right)=} \\ \\ \mathtt{=\left(\begin{array}{ccc}\mathtt 0&\mathtt8\\\mathtt6&\mathtt0\end{array}\right) } \\ \\ \mathtt{A(1)+A(-1)=\left(\begin{array}{ccc}\mathtt 0&\mathtt8\\\mathtt6&\mathtt0\end{array}\right) }$
$\displaystyle \mathtt{b)a=3;~det\left(A^2\right)=?}\\ \\ \mathtt{A(3)=\left(\begin{array}{ccc}\mathtt 3&\mathtt4\\\mathtt3&\mathtt{-3}\end{array}\right)} \\ \\ \mathtt{A^2=A \cdot A=\left(\begin{array}{ccc}\mathtt 3&\mathtt4\\\mathtt3&\mathtt{-3}\end{array}\right)\cdot\left(\begin{array}{ccc}\mathtt 3&\mathtt4\\\mathtt3&\mathtt{-3}\end{array}\right)=}$
$\displaystyle =\mathtt{\left(\begin{array}{ccc}\mathtt {3 \cdot 3+4 \cdot 3}&\mathtt{3 \cdot 4+4 \cdot (-3)}\\\mathtt{3 \cdot 3+(-3) \cdot3}&\mathtt{3 \cdot 4+(-3) \cdot (-3)}\end{array}\right)= \left(\begin{array}{ccc}\mathtt{9+12}&\mathtt{12-12}\\\mathtt{9-9}&\mathtt{12+9}\end{array}\right)=}\\ \\ \mathtt{=\left(\begin{array}{ccc}\mathtt{21}&\mathtt0\\\mathtt0&\mathtt{21}\end{array}\right)}\\ \\ \mathtt{A^2\mathtt{=\left(\begin{array}{ccc}\mathtt{21}&\mathtt0\\\mathtt0&\mathtt{21}\end{array}\right)}}$
$\displaystyle \mathtt{det(A^2)\mathtt{=\left|\begin{array}{ccc}\mathtt{21}&\mathtt0\\\mathtt0&\mathtt{21}\end{array}\right|}=21 \cdot 21-0 \cdot 0=441}\\ \\\mathtt{det(A^2)=441}\\ \\ \mathtt{c)a=2;~A^2=-det(A)\cdot I_2}\\ \\ \mathtt{A(2)\mathtt{=\left(\begin{array}{ccc}\mathtt2&\mathtt4\\\mathtt3&\mathtt{-2}\end{array}\right)} }\\ \\ \mathtt{det(A)\mathtt{=\left|\begin{array}{ccc}\mathtt2&\mathtt4\\\mathtt3&\mathtt{-2}\end{array}\right|}=2 \cdot (-2)-4 \cdot 3=-4-12=-16}\\ \\ \mathtt{det(A)=-16 \Rightarrow -det(A)=16}$
$\displaystyle \mathtt{-det(A) \cdot I_2=16 \cdot \left(\begin{array}{ccc}\mathtt1&\mathtt0\\\mathtt0&\mathtt1\end{array}\right)}=\left(\begin{array}{ccc}\mathtt{16}&\mathtt0\\\mathtt0&\mathtt{16}\end{array}\right)$
$\mathtt{-det(A)\cdot I2= \left(\begin{array}{ccc}\mathtt{16}&\mathtt0\\\mathtt0&\mathtt{16}\end{array}\right) }}$
$\displaystyle \mathtt{A^2=A \cdot A\mathtt{=\left(\begin{array}{ccc}\mathtt2&\mathtt4\\\mathtt3&\mathtt{-2}\end{array}\right)} }\cdot \mathtt{\left(\begin{array}{ccc}\mathtt2&\mathtt4\\\mathtt3&\mathtt{-2}\end{array}\right)} }}=$
$\displaystyle \mathtt{=\left(\begin{array}{ccc}\mathtt{2 \cdot 2+4 \cdot 3}&\mathtt{2 \cdot 4+4 \cdot (-2) }\\\mathtt{3 \cdot 2+(-2) \cdot 3}&\mathtt{3 \cdot 4+(-2) \cdot (-2)}\end{array}\right)} }=\left(\begin{array}{ccc}\mathtt{4+12}&\mathtt{8-8}\\\mathtt{6-6}&\mathtt{12+4}\end{array}\right)} }=} \\ \\ \mathtt{= \left(\begin{array}{ccc}\mathtt{16}&\mathtt0\\\mathtt0&\mathtt{16}\end{array}\right)}$
$\displaystyle \mathtt{A^2= \mathtt{\left(\begin{array}{ccc}\mathtt{16}&\mathtt0\\\mathtt0&\mathtt{16}\end{array}\right);~-det(A)\cdot I_2= \left(\begin{array}{ccc}\mathtt{16}&\mathtt0\\\mathtt0&\mathtt{16}\end{array}\right) }}\\ \\ \mathtt{ \left(\begin{array}{ccc}\mathtt{16}&\mathtt0\\\mathtt0&\mathtt{16}\end{array}\right)= \left(\begin{array}{ccc}\mathtt{16}&\mathtt0\\\mathtt0&\mathtt{16}\end{array}\right) \Rightarrow A^2=-det(A) \cdot I_2}$