- #1
The attached image in post 1 of this thread doesn't specify any particular matrix, so it's not possible to determine the entries of A - 2I. You might be confusing what was in the hand-drawn sketch of the previous thread from the OP, which itself was confused.fresh_42 said:$$
A-2I =\begin{pmatrix}1&-2\\1&-2\end{pmatrix}-\begin{pmatrix}2&0\\0&2\end{pmatrix}=\begin{pmatrix}-1&-2\\1&-4\end{pmatrix}
$$
and thus ##(A-2I)^{-1}=\dfrac{1}{6}\begin{pmatrix}-4&2\\-1&-1\end{pmatrix}##
fresh_42 said:So ##A-2I## is invertible. Since we have ##A\cdot \begin{pmatrix}x\\y\end{pmatrix}=2I\cdot \begin{pmatrix}x\\y\end{pmatrix},## we get
$$
A\cdot \begin{pmatrix}x\\y\end{pmatrix}-2I\cdot \begin{pmatrix}x\\y\end{pmatrix}= (A-2I)\begin{pmatrix}x\\y\end{pmatrix}=\begin{pmatrix}0\\0\end{pmatrix}
$$
Applying ##(A-2I)^{-1}## on both sides results in
$$
(A-2I)^{-1}\cdot (A-2I)\cdot \begin{pmatrix}x\\y\end{pmatrix}= I\cdot \begin{pmatrix}x\\y\end{pmatrix} =\begin{pmatrix}x\\y\end{pmatrix} =(A-2I)^{-1}\cdot \begin{pmatrix}0\\0\end{pmatrix}=\begin{pmatrix}0\\0\end{pmatrix}
$$
Occam's razor. Why complicate things? I do not debate typos.Mark44 said:The attached image in post 1 of this thread doesn't specify any particular matrix, so it's not possible to determine the entries of A - 2I.
The OP is already sufficiently confused as evidenced in the thread title, in thinking that finding the inverse of a matrix plays any role in finding eigenvalues. Muddying up the water by tossing in a specific matrix where none was given doesn't help alleviate that confusion.fresh_42 said:Occam's razor. Why complicate things?