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Let $A$ and $B$ be square matrices of the order $3 \times 3$. Is $(\mathrm{AB})^2=\mathrm{A}^2 \mathrm{~B}^2$ ? Give reasons.
Solution:
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Verified Answer
Since, $\mathrm{A}$ and $\mathrm{B}$ are square matrices of order $3 \times 3$
$$
\begin{array}{rlr}
\therefore \mathrm{AB}^2 & =\mathrm{AB} \cdot \mathrm{AB} & \\
& =\mathrm{ABAB} & \\
& =\mathrm{AABB} & \\
& =\mathrm{A}^2 \mathrm{~B}^2 &
\end{array}
$$
So, $\mathrm{AB}^2=\mathrm{A}^2 \mathrm{~B}^2$ is true when $\mathrm{AB}=\mathrm{BA}$.
$$
\begin{array}{rlr}
\therefore \mathrm{AB}^2 & =\mathrm{AB} \cdot \mathrm{AB} & \\
& =\mathrm{ABAB} & \\
& =\mathrm{AABB} & \\
& =\mathrm{A}^2 \mathrm{~B}^2 &
\end{array}
$$
So, $\mathrm{AB}^2=\mathrm{A}^2 \mathrm{~B}^2$ is true when $\mathrm{AB}=\mathrm{BA}$.
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