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If \( \mathrm{A} \) is a matrix of order \( \mathrm{m} \times \mathrm{n} \) and \( \mathrm{B} \) is a matrix such that \( \mathrm{AB}^{\prime} \) and \( \mathrm{B}^{\prime} \mathrm{A} \) are both defined, the
order of the matrix \( B \) is
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order of the matrix \( B \) is
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The correct answer is:
\( \mathrm{m} \times \mathrm{n} \)
Given that, matrix $\mathrm{A}$ is $\mathrm{m} \times \mathrm{n}$
Let matrix $\mathrm{B}$ is $x \times y$.
So, $\mathrm{AB}^{\prime}$ is defined then, $\mathrm{n}=\mathrm{y}$
Similarly, $\mathrm{B}$ 'A is defined then, $\mathrm{x}=\mathrm{m}$
Therefore, order of matrix $\mathrm{B}$ is $\mathrm{m} \times \mathrm{n}$.
Let matrix $\mathrm{B}$ is $x \times y$.
So, $\mathrm{AB}^{\prime}$ is defined then, $\mathrm{n}=\mathrm{y}$
Similarly, $\mathrm{B}$ 'A is defined then, $\mathrm{x}=\mathrm{m}$
Therefore, order of matrix $\mathrm{B}$ is $\mathrm{m} \times \mathrm{n}$.
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