Question
Let $A =\left[\begin{array}{ll}4 & -2 \\ 6 & -3\end{array}\right], B =\left[\begin{array}{cc}0 & 2 \\ 1 & -1\end{array}\right]$ and $C =\left[\begin{array}{cc}-2 & 3 \\ 1 & -1\end{array}\right]$. Find $A^2 - A + BC$.
✓
Answer
$A=\left[\begin{array}{ll}4 & -2 \\ 6 & -3\end{array}\right], B=\left[\begin{array}{cc}0 & 2 \\ 1 & -1\end{array}\right] \text { and } C=\left[\begin{array}{cc}-2 & 3 \\ 1 & -1\end{array}\right] . $
$ \therefore A^2=\left[\begin{array}{ll}4 & -2 \\ 6 & -3\end{array}\right]\left[\begin{array}{ll}4 & -2 \\ 6 & -3\end{array}\right] $
$ =\left[\begin{array}{cc}16-12 & -8+6 \\ 24-18 & -12+9\end{array}\right]$
$BC =\left[\begin{array}{cc}0 & 2 \\ 1 & -1\end{array}\right]\left[\begin{array}{cc}-2 & 3 \\ 1 & -1\end{array}\right]=\left[\begin{array}{cc}0+2 & 0-2 \\ -2-1 & 3+1\end{array}\right] $
$ =\left[\begin{array}{cc}2 & -2 \\ -3 & 4\end{array}\right]$
Now $A ^2- A + BC$
$ =\left[\begin{array}{ll} 4 & -2 \\ 6 & -3 \end{array}\right]-\left[\begin{array}{cc} 4 & -2 \\ 6 & -3
\end{array}\right]+\left[\begin{array}{cc} 2 & -2 \\ -3 & 4 \end{array}\right] $
$=\left[\begin{array}{ll} 0 & 0 \\ 0 & 0 \end{array}\right]+\left[\begin{array}{cc} 2 & -2 \\ -3 & 4 \end{array}\right]=\left[\begin{array}{cc} 2 & -2 \\-3 & 4 \end{array}\right] $
$ \therefore A^2=\left[\begin{array}{ll}4 & -2 \\ 6 & -3\end{array}\right]\left[\begin{array}{ll}4 & -2 \\ 6 & -3\end{array}\right] $
$ =\left[\begin{array}{cc}16-12 & -8+6 \\ 24-18 & -12+9\end{array}\right]$
$BC =\left[\begin{array}{cc}0 & 2 \\ 1 & -1\end{array}\right]\left[\begin{array}{cc}-2 & 3 \\ 1 & -1\end{array}\right]=\left[\begin{array}{cc}0+2 & 0-2 \\ -2-1 & 3+1\end{array}\right] $
$ =\left[\begin{array}{cc}2 & -2 \\ -3 & 4\end{array}\right]$
Now $A ^2- A + BC$
$ =\left[\begin{array}{ll} 4 & -2 \\ 6 & -3 \end{array}\right]-\left[\begin{array}{cc} 4 & -2 \\ 6 & -3
\end{array}\right]+\left[\begin{array}{cc} 2 & -2 \\ -3 & 4 \end{array}\right] $
$=\left[\begin{array}{ll} 0 & 0 \\ 0 & 0 \end{array}\right]+\left[\begin{array}{cc} 2 & -2 \\ -3 & 4 \end{array}\right]=\left[\begin{array}{cc} 2 & -2 \\-3 & 4 \end{array}\right] $
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