Question

Let A be the matrix \[A=\left[\begin{matrix}1 & -2\\ -2 &4\end{matrix}\right]\] decide all the 2x2 matrices B such that AB=BA=0, where 0 is the zeromatrix

Answer 1

\[\left[\begin{matrix}1 & -2 \\ -2 & 4\end{matrix}\right]\left[\begin{matrix}a &b \\ b & c\end{matrix}\right]=\left[\begin{matrix}0 &0 \\ 0& 0\end{matrix}\right]\]
\[\left[\begin{matrix}a & b \\ c & d\end{matrix}\right]\left[\begin{matrix}1 &-2\\ -2 & 4\end{matrix}\right]=\left[\begin{matrix}0 & 0\\ 0 &0\end{matrix}\right]\]

So all I can figure is that for AB=BA=0 B can be\[B=A ^{-1}=\left[\begin{matrix}4 & 2 \\ 2 & 1\end{matrix}\right]\]

Answer 2

But is there any other solution than the inverse?

Answer 3

I guess it could be \[\left[\begin{matrix}1 & -2 \\ -2 & 4\end{matrix}\right]\left[\begin{matrix}0 & 0 \\ 0& 0\end{matrix}\right]=\left[\begin{matrix}0 & 0 \\ 0& 0\end{matrix}\right]\]
\[\left[\begin{matrix}0 & 0 \\ 0& 0\end{matrix}\right]\left[\begin{matrix}1 & -2 \\ -2 & 4\end{matrix}\right]=\left[\begin{matrix}0 & 0 \\ 0& 0\end{matrix}\right]\]
but so the two solutions for AB=BA=0 should be B=A^-1 and B=0, am I right?

Answer 4

\(A\) is not invertable...so how can \(B=A^{-1}\)

Answer 5

It's not I guess, A is not invertable if AX=0 is the definition I remember right now, is that right?

Answer 6

\(Det(A)=0\) so \(A^{-1}\) does not exist

Answer 7

Ok, but is there any other solution than B equals the zerovector?

Answer 8

yes..there are infinitly many solutions

Answer 9

you have 2 of the solutions

Answer 10

So how do I show that it has infinity many solutions?

Answer 11

show that \(B=\left[\begin{matrix}4 & 2 \\ 2 & 1\end{matrix}\right]\) is a solution

Answer 12

then \(B\cdot t\) is a solution for all real numbers \(t\)

Answer 13

So B can be the same as what the inverse of A would have been if it existed?

Answer 14

it is not really the same

Answer 15

remember you have to divide by the det(A) for the shortcut way to find the inverse

Answer 16

Oh I get it, it can't be the inverse since the definition says that A^-1*A=I isn't that right?

Answer 17

I= Identity matrix

Answer 18

yes

Answer 19

Thank you for your help! :)

Answer 20

np