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If $\mathrm{A}=\left[\begin{array}{ll}1 & 2 \\ 4 & 1 \\ 5 & 6\end{array}\right]$ and $\mathrm{B}=\left[\begin{array}{ll}1 & 2 \\ 6 & 4 \\ 7 & 3\end{array}\right]$, then verify that
(i) $(2 \mathrm{~A}+\mathrm{B})^{\prime}=2 \mathrm{AA}+\mathrm{B}^{\prime}$.
(ii) $(\mathrm{A}-\mathrm{B})^{\prime}=\mathrm{A}^{\prime}-\mathrm{B}^{\prime}$.
(i) $(2 \mathrm{~A}+\mathrm{B})^{\prime}=2 \mathrm{AA}+\mathrm{B}^{\prime}$.
(ii) $(\mathrm{A}-\mathrm{B})^{\prime}=\mathrm{A}^{\prime}-\mathrm{B}^{\prime}$.
Solution:
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Verified Answer
We have, $\mathrm{A}=\left[\begin{array}{ll}1 & 2 \\ 4 & 1 \\ 5 & 6\end{array}\right]$ and $\mathrm{B}=\left[\begin{array}{ll}1 & 2 \\ 6 & 4 \\ 7 & 3\end{array}\right]$
(i) $\therefore \quad(2 \mathrm{~A}+\mathrm{B})=\left[\begin{array}{cc}3 & 6 \\ 14 & 6 \\ 17 & 15\end{array}\right]$
and $\quad(2 \mathrm{~A}+\mathrm{B})^{\prime}=\left[\begin{array}{ccc}3 & 14 & 17 \\ 6 & 6 & 15\end{array}\right]$
Also, $\quad 2 \mathrm{~A}^{\prime}+\mathrm{B}^{\prime}=2\left[\begin{array}{lll}1 & 4 & 5 \\ 2 & 1 & 6\end{array}\right]+\left[\begin{array}{lll}1 & 6 & 7 \\ 2 & 4 & 3\end{array}\right]$
$$
=\left[\begin{array}{ccc}
3 & 14 & 17 \\
6 & 6 & 15
\end{array}\right]=(2 \mathrm{~A}+\mathrm{B})^{\prime}
$$
Hence proved
(ii) $(\mathrm{A}-\mathrm{B})=\left[\begin{array}{cc}0 & 0 \\ -2 & -3 \\ -2 & 3\end{array}\right]$ and $(\mathrm{A}-\mathrm{B})^{\prime}=\left[\begin{array}{ccc}0 & -2 & -2 \\ 0 & -3 & 3\end{array}\right]$
Also, $\quad \mathrm{A}^{\prime}-\mathrm{B}^{\prime}=\left[\begin{array}{ccc}0 & -2 & -2 \\ 0 & -3 & 3\end{array}\right]=(\mathrm{A}-\mathrm{B})^{\prime}$
Hence proved.
(i) $\therefore \quad(2 \mathrm{~A}+\mathrm{B})=\left[\begin{array}{cc}3 & 6 \\ 14 & 6 \\ 17 & 15\end{array}\right]$
and $\quad(2 \mathrm{~A}+\mathrm{B})^{\prime}=\left[\begin{array}{ccc}3 & 14 & 17 \\ 6 & 6 & 15\end{array}\right]$
Also, $\quad 2 \mathrm{~A}^{\prime}+\mathrm{B}^{\prime}=2\left[\begin{array}{lll}1 & 4 & 5 \\ 2 & 1 & 6\end{array}\right]+\left[\begin{array}{lll}1 & 6 & 7 \\ 2 & 4 & 3\end{array}\right]$
$$
=\left[\begin{array}{ccc}
3 & 14 & 17 \\
6 & 6 & 15
\end{array}\right]=(2 \mathrm{~A}+\mathrm{B})^{\prime}
$$
Hence proved
(ii) $(\mathrm{A}-\mathrm{B})=\left[\begin{array}{cc}0 & 0 \\ -2 & -3 \\ -2 & 3\end{array}\right]$ and $(\mathrm{A}-\mathrm{B})^{\prime}=\left[\begin{array}{ccc}0 & -2 & -2 \\ 0 & -3 & 3\end{array}\right]$
Also, $\quad \mathrm{A}^{\prime}-\mathrm{B}^{\prime}=\left[\begin{array}{ccc}0 & -2 & -2 \\ 0 & -3 & 3\end{array}\right]=(\mathrm{A}-\mathrm{B})^{\prime}$
Hence proved.
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