Question

If \[A\] and \[B\] are square matrices of the same order, then
A. \[{(AB)^\prime } = {A^\prime }{B^\prime }\]
B. \[{(AB)^\prime } = {B^\prime }{A^\prime }\]
C. \[AB = O\] if \[|A| = 0\] or \[|B| = 0\]
D.\[(AB = O)\]; If \[A = I\] or \[B = I\]

Answer 1

Hint:A square matrix is one with no rows that are equal to the number of columns in the matrix. The square matrix has an order of \[n \times n\] where n can be any natural number. We can take two arbitrary square matrices, A and B, and then use matrix multiplication to enlarge the LHS.
Formula Used:
Matrix property:
\[\sum\limits_{r = 1}^n {{a_{jr}}} {b_{ri}}\]

Complete Step-by-Step Solution:We have been provided in the question that,
\[A\] and \[B\] are of the same order of square matrix
And we are to find any of the given possibilities would be true.
Let us assume that, \[A = {\left[ {{a_{ij}}} \right]_{\rm{m}}} \times n\] and \[B = {\left[ {{b_{ij}}} \right]_{n \times p}}\] be two matrices.
Then,
We can consider that \[AB\] is a \[m \times p\] matrix.
Therefore, from the above it is understood that, \[{(AB)^\prime }\] is a \[p \times m\] matrix.
Since, it is known that \[{A^\prime }\] and \[{B^\prime }\] are \[n \times m\] and \[p \times n\] matrices.
Therefore, we can have as \[{B^\prime }{A^\prime }\] is a \[p \times m\] matrix.
Thus, from the above statements we came to a conclusion that the two matrices \[{(AB)^\prime }\] and \[{B^\prime }{A^\prime }\] are of the same order such that \[{\left( {{{(AB)}^\prime }} \right)_{ij}} = {(AB)_{ij}}\]
\[ = \sum\limits_{r = 1}^n {{a_{jr}}} {b_{ri}}\]
\[ = \sum\limits_{r = 1}^n {{b_{ri}}} {a_{jr}}\]
\[ = \sum\limits_{r = 1}^n {{{\left( {{B^\prime }} \right)}_{ir}}} {\left( {{A^\prime }} \right)_{rj}}\]
Thus, the answer would be obvious.
Therefore, if \[A\] and \[B\] are square matrices of the same order, then \[{(AB)^\prime } = {B^\prime }{A^\prime }\]
Hence, the option B is correct

Note: To resolve these types of questions, students should constantly remember the determinant and matrix property, which states that if we have any square matrix of order n, then \[|KA| = {K^n}\left| A \right|\] Many students frequently forget this property of determinants and matrices, and as a result, they are unable to solve the question.