Question
Let $A=\left[\begin{array}{ll}1 & 0 \\ 2 & 1\end{array}\right], B=\left[\begin{array}{cc}2 & 3 \\ -1 & 0\end{array}\right]$. Find $A^2+A B+B^2$
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Answer
$A=\left[\begin{array}{ll}1 & 0 \\ 2 & 1\end{array}\right], B=\left[\begin{array}{cc}2 & 3 \\ -1 & 0\end{array}\right]$
$A^2=A \times A=\left[\begin{array}{ll}1 & 0 \\ 2 & 1\end{array}\right] \times\left[\begin{array}{ll}1 & 0 \\ 2 & 1\end{array}\right]$
$=\left[\begin{array}{ll}1 \times 1+0 \times 2 & 1 \times 0+0 \times 1 \\ 2 \times 1+1 \times 2 & 2 \times 0+1 \times 1\end{array}\right]=\left[\begin{array}{ll}1 & 0 \\ 4 & 1\end{array}\right]$
$A B=A \times B=\left[\begin{array}{ll}1 & 0 \\ 2 & 1\end{array}\right] \times\left[\begin{array}{cc}2 & 3 \\ -1 & 0\end{array}\right] $
$=\left[\begin{array}{ll}1 \times 2+0 \times(-1) & 1 \times 3+0 \times 0 \\ 2 \times 2+1 \times(-1) & 2 \times 3+1 \times 0\end{array}\right] =\left[\begin{array}{ll}2 & 3 \\ 3 & 6\end{array}\right]$
$B^2=B \times B=\left[\begin{array}{cc}2 & 3 \\ -1 & 0\end{array}\right] \times\left[\begin{array}{cc}2 & 3 \\ -1 & 0\end{array}\right] $
$ =\left[\begin{array}{cc}2 \times 2+3 \times(-1) & 2 \times 3+3 \times 0 \\ (-1) \times 2+0 \times(-1) & -1 \times 3+0 \times 0\end{array}\right] =\left[\begin{array}{cc}1 & 6 \\ -2 & -3\end{array}\right] $
$\therefore A^2+A B+B^2=\left[\begin{array}{ll}1 & 0 \\ 4 & 1\end{array}\right]+\left[\begin{array}{ll}2 & 3 \\ 3 & 6\end{array}\right]+\left[\begin{array}{cc}1 & 6 \\ -2 & -3\end{array}\right] =\left[\begin{array}{ll}4 & 9 \\ 5 & 4\end{array}\right]$
$A^2=A \times A=\left[\begin{array}{ll}1 & 0 \\ 2 & 1\end{array}\right] \times\left[\begin{array}{ll}1 & 0 \\ 2 & 1\end{array}\right]$
$=\left[\begin{array}{ll}1 \times 1+0 \times 2 & 1 \times 0+0 \times 1 \\ 2 \times 1+1 \times 2 & 2 \times 0+1 \times 1\end{array}\right]=\left[\begin{array}{ll}1 & 0 \\ 4 & 1\end{array}\right]$
$A B=A \times B=\left[\begin{array}{ll}1 & 0 \\ 2 & 1\end{array}\right] \times\left[\begin{array}{cc}2 & 3 \\ -1 & 0\end{array}\right] $
$=\left[\begin{array}{ll}1 \times 2+0 \times(-1) & 1 \times 3+0 \times 0 \\ 2 \times 2+1 \times(-1) & 2 \times 3+1 \times 0\end{array}\right] =\left[\begin{array}{ll}2 & 3 \\ 3 & 6\end{array}\right]$
$B^2=B \times B=\left[\begin{array}{cc}2 & 3 \\ -1 & 0\end{array}\right] \times\left[\begin{array}{cc}2 & 3 \\ -1 & 0\end{array}\right] $
$ =\left[\begin{array}{cc}2 \times 2+3 \times(-1) & 2 \times 3+3 \times 0 \\ (-1) \times 2+0 \times(-1) & -1 \times 3+0 \times 0\end{array}\right] =\left[\begin{array}{cc}1 & 6 \\ -2 & -3\end{array}\right] $
$\therefore A^2+A B+B^2=\left[\begin{array}{ll}1 & 0 \\ 4 & 1\end{array}\right]+\left[\begin{array}{ll}2 & 3 \\ 3 & 6\end{array}\right]+\left[\begin{array}{cc}1 & 6 \\ -2 & -3\end{array}\right] =\left[\begin{array}{ll}4 & 9 \\ 5 & 4\end{array}\right]$
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