Question

arătați ca det (A-2B)=0 matricea este " A= 1 2 1 -2 și B=x 1 y -1

Answer 1

$A-2B=\left[\begin{array}{ccc}1&2\\1&-2\\\end{array}\right]-2 \left[\begin{array}{ccc}x&1\\y&-1\\\end{array}\right]= \left[\begin{array}{ccc}1-2x&2-2\\1-2y&-2+2\end{array}\right]$
det (A-2B)=det$\left[\begin{array}{ccc}1-2x&0\\1-2y&0\\\end{array}\right]=(1-2x)*0-(1-2y)*0=0$, ∀x,y ∈R

Answer 2

$\displaystyle \mathtt{A=\left(\begin{array}{ccc}\mathtt1&\mathtt2\\\mathtt1&\mathtt{-2}\end{array}\right);~B= \left(\begin{array}{ccc}\mathtt x&\mathtt1\\\mathtt y&\mathtt{-1}\end{array}\right)~~~~~~~~~~~~~~~~~~~~~~~~~det(A-2B)=0}$
$\mathtt{2B=2 \cdot \left(\begin{array}{ccc}\mathtt x&\mathtt1\\\mathtt y&\mathtt{-1}\end{array}\right) = \left(\begin{array}{ccc}\mathtt{2 \cdot x}&\mathtt{2 \cdot 1}\\\mathtt{2 \cdot y}&\mathtt{2 \cdot (-1)}\end{array}\right)= \left(\begin{array}{ccc}\mathtt {2x}&\mathtt{2}\\\mathtt {2y}&\mathtt{-2}\end{array}\right)}$
$\displaystyle \mathtt{A-2B=\left(\begin{array}{ccc}\mathtt1&\mathtt2\\\mathtt1&\mathtt{-2}\end{array}\right)-\left(\begin{array}{ccc}\mathtt {2x}&\mathtt{2}\\\mathtt {2y}&\mathtt{-2}\end{array}\right)= \left(\begin{array}{ccc}\mathtt {1-2x}&\mathtt{2-2}\\\mathtt {1-2y}&\mathtt{-2-(-2)}\end{array}\right) =} \\ \\ \mathtt{=\left(\begin{array}{ccc}\mathtt {1-2x}&\mathtt{0}\\\mathtt {1-2y}&\mathtt{0}\end{array}\right)}$
$\displaystyle \mathtt{det(A-2B)= \left|\begin{array}{ccc}\mathtt {1-2x}&\mathtt{0}\\\mathtt {1-2y}&\mathtt{0}\end{array}\right| =(1-2x) \cdot 0-0 \cdot (1-2y)=0-0=0}$