Question

Fie matricea A= ( 2 4 )
3 -2
Demonstrati ca A la a doua = - det(A) I indice 2 .

Answer 1

$\displaystyle \mathtt{A= \left(\begin{array}{ccc}\mathtt2&\mathtt4\\\mathtt3&\mathtt{-2}\end{array}\right)}\\ \\ \mathtt{A^2=-det(A) \cdot I_2} \\ \\ \mathtt{det(A)= \left|\begin{array}{ccc}\mathtt2&\mathtt4\\\mathtt3&\mathtt{-2}\end{array}\right|=2 \cdot (-2)-4 \cdot 3=-4-12=-16}\\ \\ \mathtt{det(A)=-16 \Rightarrow -det(A)=16}$
$\displaystyle \mathtt{-det(A) \cdot I_2=16 \cdot \left(\begin{array}{ccc}\mathtt1&\mathtt0\\\mathtt0&\mathtt1\end{array}\right) = \left(\begin{array}{ccc}\mathtt{16 \cdot 1}&\mathtt{16 \cdot 0}\\\mathtt{16 \cdot 0}&\mathtt{16 \cdot 1}\end{array}\right)= \left(\begin{array}{ccc}\mathtt{16}&\mathtt0\\\mathtt0&\mathtt{16}\end{array}\right)}\\ \\ \mathtt{-det(A) \cdot I_2= \left(\begin{array}{ccc}\mathtt{16}&\mathtt0\\\mathtt0&\mathtt{16}\end{array}\right) }$
$\mathtt{A^2=A \cdot A= \left(\begin{array}{ccc}\mathtt2&\mathtt4\\\mathtt3&\mathtt{-2}\end{array}\right) \cdot \left(\begin{array}{ccc}\mathtt2&\mathtt4\\\mathtt3&\mathtt{-2}\end{array}\right) =}$
$\displaystyle \mathtt{= \left(\begin{array}{ccc}\mathtt{2 \cdot 2+4 \cdot 3}&\mathtt{2 \cdot 4+4 \cdot (-2)}\\\mathtt{3 \cdot 2+(-2) \cdot 3}&\mathtt{3\cdot 4+(-2) \cdot (-2)}\end{array}\right)= \left(\begin{array}{ccc}\mathtt{4+12}&\mathtt{8-8}\\\mathtt{6-6}&\mathtt{12+4}\end{array}\right)=}\\ \\ \mathtt{= \left(\begin{array}{ccc}\mathtt{16}&\mathtt0\\\mathtt0&\mathtt{16}\end{array}\right) } \\ \\ \mathtt{A^2=\left(\begin{array}{ccc}\mathtt{16}&\mathtt0\\\mathtt0&\mathtt{16}\end{array}\right) }$
$\mathtt{\Rightarrow A^2=-det(A) \cdot I_2}$