Question
Find the matrix $A,$ If $B =\left[\begin{array}{ll}2 & 1 \\ 0 & 1\end{array}\right]$ and $B^2=B+\frac{1}{2} A$
✓
Answer
$B^2=B+\frac{1}{2} A$
$A=2\left(B^2-B\right)$
$\begin{array}{l}B^2=\left[\begin{array}{ll}2 & 1 \\ 0 & 1\end{array}\right]\left[\begin{array}{ll}2 & 1 \\ 0 & 1\end{array}\right] \end{array} $
$ =\left[\begin{array}{llll}4 & +0 & 2 & +1 \\ 0 & +0 & 0 & +1\end{array}\right] $
$=\left[\begin{array}{ll}4 & 3 \\ 0 & 1\end{array}\right] $
$\begin{array}{l}=B^2-B=\left[\begin{array}{ll}4 & 3 \\ 0 & 1\end{array}\right]-\left[\begin{array}{ll}2 & 1 \\ 0 & 1\end{array}\right] \end{array} $
$ =\left[\begin{array}{ll}2 & 2 \\ 0 & 0\end{array}\right]$
$A=2\left(B^2-B\right)$
$=2\left[\begin{array}{ll}2 & 2 \\ 0 & 0\end{array}\right]$
$=\left[\begin{array}{ll}4 & 4 \\ 0 & 0\end{array}\right]$
$A=2\left(B^2-B\right)$
$\begin{array}{l}B^2=\left[\begin{array}{ll}2 & 1 \\ 0 & 1\end{array}\right]\left[\begin{array}{ll}2 & 1 \\ 0 & 1\end{array}\right] \end{array} $
$ =\left[\begin{array}{llll}4 & +0 & 2 & +1 \\ 0 & +0 & 0 & +1\end{array}\right] $
$=\left[\begin{array}{ll}4 & 3 \\ 0 & 1\end{array}\right] $
$\begin{array}{l}=B^2-B=\left[\begin{array}{ll}4 & 3 \\ 0 & 1\end{array}\right]-\left[\begin{array}{ll}2 & 1 \\ 0 & 1\end{array}\right] \end{array} $
$ =\left[\begin{array}{ll}2 & 2 \\ 0 & 0\end{array}\right]$
$A=2\left(B^2-B\right)$
$=2\left[\begin{array}{ll}2 & 2 \\ 0 & 0\end{array}\right]$
$=\left[\begin{array}{ll}4 & 4 \\ 0 & 0\end{array}\right]$
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