If A is a matrix of order $m × n$ and $B$ is a matrix such that $AB^T$ and $B^TA$ are both well-defined matrices, then order of matrix B is

Matrix A has order $m × n$ (m rows, n columns).

Matrix B exists such that $AB^T$ and $B^T A$ are both well-defined.

For matrix multiplication to be possible, the number of columns in the first matrix must equal the number of rows in the second matrix.

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Let matrix B have order $p × q$ (p rows, q columns).

Then $B^T$ (transpose of B) will have order $q × p$.

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For $AB^T$ to be well-defined:

A is $(m × n)$

$B^T$ is $(q × p)$

Therefore, $n = q$

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For $B^T A$ to be well-defined:

$B^T$ is $(q × p)$

A is $(m × n)$

Therefore, $p = m$

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From the two conditions:

$p = m$

$q = n$

Since B has order $p × q$:

Order of matrix B $= m × n$