Question
Let $A=\left[\begin{array}{cc}2 & 1 \\ 0 & -2\end{array}\right], B=\left[\begin{array}{cc}4 & 1 \\ -3 & -2\end{array}\right]$ and $C=\left[\begin{array}{ll}-3 & 2 \\ -1 & 4\end{array}\right]$. Find $A^2 + AC -5B$.
✓
Answer
Given : $A =\left[\begin{array}{cc}2 & 1 \\ 0 & -2\end{array}\right], B =\left[\begin{array}{cc}4 & 1 \\ -3 & -2\end{array}\right]$ and $C =\left[\begin{array}{ll}-3 & 2 \\ -1 & 4\end{array}\right]$
Now,
$ A^2=\left[\begin{array}{cc} 2 & 1 \\ 0 & -2 \end{array}\right]\left[\begin{array}{cc} 2 & 1 \\
0 & -2 \end{array}\right]=\left[\begin{array}{ll} 4+0 & 2-2 \\ 0+0 & 0+4 \end{array}\right]=\left[\begin{array}{ll}
4 & 0 \\ 0 & 4 \end{array}\right] $
$5 B=\left[\begin{array}{cc} 20 & 5 \\ -15 & -10 \end{array}\right]$
$A C=\left[\begin{array}{cc}2 & 1 \\ 0 & -2\end{array}\right]\left[\begin{array}{ll}-3 & 2 \\ -1 & 4\end{array}\right]=\left[\begin{array}{cc}-6-1 & 4+4 \\ 0+2 & 0-8\end{array}\right]=\left[\begin{array}{cc}-7 & 8 \\ 2 & -8\end{array}\right] $
$ \therefore A^2+A C-5 B=\left[\begin{array}{ll}4 & 0 \\ 0 & 4\end{array}\right]+\left[\begin{array}{cc}-7 & 8 \\ 2 & -8\end{array}\right]-\left[\begin{array}{cc}20 & 5 \\ -15 & -10\end{array}\right] $
$ =\left[\begin{array}{cc}4-7-20 & 0+8-5 \\ 0+2+15 & 4-8+10\end{array}\right] $
$ =\left[\begin{array}{cc}-23 & 3 \\ 17 & 6\end{array}\right] .$
Now,
$ A^2=\left[\begin{array}{cc} 2 & 1 \\ 0 & -2 \end{array}\right]\left[\begin{array}{cc} 2 & 1 \\
0 & -2 \end{array}\right]=\left[\begin{array}{ll} 4+0 & 2-2 \\ 0+0 & 0+4 \end{array}\right]=\left[\begin{array}{ll}
4 & 0 \\ 0 & 4 \end{array}\right] $
$5 B=\left[\begin{array}{cc} 20 & 5 \\ -15 & -10 \end{array}\right]$
$A C=\left[\begin{array}{cc}2 & 1 \\ 0 & -2\end{array}\right]\left[\begin{array}{ll}-3 & 2 \\ -1 & 4\end{array}\right]=\left[\begin{array}{cc}-6-1 & 4+4 \\ 0+2 & 0-8\end{array}\right]=\left[\begin{array}{cc}-7 & 8 \\ 2 & -8\end{array}\right] $
$ \therefore A^2+A C-5 B=\left[\begin{array}{ll}4 & 0 \\ 0 & 4\end{array}\right]+\left[\begin{array}{cc}-7 & 8 \\ 2 & -8\end{array}\right]-\left[\begin{array}{cc}20 & 5 \\ -15 & -10\end{array}\right] $
$ =\left[\begin{array}{cc}4-7-20 & 0+8-5 \\ 0+2+15 & 4-8+10\end{array}\right] $
$ =\left[\begin{array}{cc}-23 & 3 \\ 17 & 6\end{array}\right] .$
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