Here's the question you clicked on:
DLS
Matrix help!
\[If~A=\left[\begin{matrix}\alpha & 0 \\ 1 & 1\end{matrix}\right] \] \[\And~B=\left[\begin{matrix}1 & 5 \\ 0 & 1\end{matrix}\right]\] such that \[A^2=B\] Then \[\alpha=?\]
Options: A)1 B)-1 C)4 D)None of these
\[\LARGE \left[\begin{matrix}\alpha^2 & 0 \\ \alpha+1& 1\end{matrix}\right]=\left[\begin{matrix}1 & 0 \\ 5 & 1\end{matrix}\right]\] I am getting this after equating A^2=B I am getting all the 3 options A,B,C. But it is a single choice question.
@agent0smith @dmezzullo @electrokid
the lower left entry should be in B is 0. so you have a^2 =1 and a+1=0
are you sure about the B matrix? A*A is [ a^2 0 ] [ a+1 1 ] which cannot match B as given
which version of B is the one given? the original or the one you posted later?
\[~B=\left[\begin{matrix}1 & 5 \\ 0 & 1\end{matrix}\right] \] \[\LARGE \left[\begin{matrix}\alpha^2 & 0 \\ \alpha+1& 1\end{matrix}\right]=\left[\begin{matrix}1 & 0 \\ 5 & 1\end{matrix}\right]\] which is B...
so alpha^2 = 1 and alpha+1 = 5...??? that doesn't seem to work :/
as you noticed, there is no unique solution for alpha...
A,B,C all 3 are correct but single choice .-.
If you're sure that second matrix is right, then there's no solution. The first matrix would give alpha is -1.
I think you should choose option D
yes it is right :| the solution says none of these since this is an absurd case :O