Question

Se considera matricea A(a) = (a 4 )
(3 -a)
@€ R
a) Demonstrati ca pentru a =2, A la puterea a 2-a =-determinant (A) •i2

Answer 1

a=2 => A(a)=( 2    4)
                       ( 3  -2)

A^2= A*A= (2   4) *(2   4)= (16  0)
                    (3  -2)  (3  -2)   (0   16)

det(A(2))= 2    4 = -4 -12 = -16
                   3  -2

det (A(2))* I2= -16* (1  0) = ( -16  0)   =>  A^2 = det(A(a))*i2
                                  (0  1)    (0  -16)

Answer 2

$\displaystyle \mathtt{A(a)= \left(\begin{array}{ccc}\mathtt a&\mathtt4\\\mathtt3&\mathtt{-a}\end{array}\right)~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~A^2=-det(A) \cdot I_2}\\ \\ \mathtt{a=2 \Rightarrow A(2)= \left(\begin{array}{ccc}\mathtt 2&\mathtt4\\\mathtt3&\mathtt{-2}\end{array}\right)}$
$\displaystyle \mathtt{A^2= \left(\begin{array}{ccc}\mathtt 2&\mathtt4\\\mathtt3&\mathtt{-2}\end{array}\right) \cdot \left(\begin{array}{ccc}\mathtt 2&\mathtt4\\\mathtt3&\mathtt{-2}\end{array}\right) }= \\ \\ =\mathtt{\left(\begin{array}{ccc}\mathtt {2 \cdot 2+4 \cdot 3}&\mathtt{2 \cdot 4+4 \cdot (-2)}\\\mathtt{3 \cdot 2+(-2) \cdot 3}&\mathtt{3 \cdot 4+(-2) \cdot (-2)}\end{array}\right)= \left(\begin{array}{ccc}\mathtt {4+12}&\mathtt{8-8}\\\mathtt{6-6}&\mathtt{12+4}\end{array}\right)=}$
$\displaystyke \mathtt{= \left(\begin{array}{ccc}\mathtt {16}&\mathtt0\\\mathtt0&\mathtt{16}\end{array}\right)}$
$\displaystyle \mathtt{det(A)=\left|\begin{array}{ccc}\mathtt {2}&\mathtt4\\\mathtt3&\mathtt{-2}\end{array}\right|=2 \cdot (-2)-4 \cdot 3=-4-12=-16}\\ \\ \mathtt{det(A)=-16 \Rightarrow -det(A)=16}}$
$\displaystyle \mathtt{-det(A) \cdot I_2=16 \cdot \left(\begin{array}{ccc}\mathtt {1}&\mathtt0\\\mathtt0&\mathtt{1}\end{array}\right)=\left(\begin{array}{ccc}\mathtt {16 \cdot 1}&\mathtt16 \cdot0\\\mathtt16\cdot0&\mathtt{16\cdot1}\end{array}\right)=\left(\begin{array}{ccc}\mathtt {16}&\mathtt0\\\mathtt0&\mathtt{16}\end{array}\right) }\\ \\ \mathtt{\Rightarrow A^2=-det(A) \cdot I_2}$