
If A is a matrix of order $m\times n$and B is a matrix such that AB’ and B’A are both defined, then the order of matrix B is
A. $m\times m$
B. $n\times n$
C. $n\times m$
D. $m\times n$
Answer
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Hint: To solve this question, we have to know the condition for the product of two matrices to exist. The product of 2 matrices A, B which is A$\times $B is defined when the number of columns of the first matrix which is A is equal to the number of rows of the second matrix which is B. Using this property, we can get the rows and columns of the matrix B as required in the question.
Complete step-by-step solution:
In the notation of order of a matrix M of order m$\times $n, m is the number of rows and n is the number of columns. Let C, D be two matrices of orders $a\times b$ and $x\times y$ respectively.
For the product C$\times $D to be defined, we have to apply the condition for the product to be defined which is the number of columns of C is equal to the number of rows of D. Mathematically it is b = x$\to \left( 1 \right)$.
The important point that we have to note here is that there is no condition for the number of rows of first matrix C and the number of columns of second matrix D.
In the question, it is given that the matrix A is ordered m$\times $n. Let the order of B be k$\times $l.
For a matrix A of order x$\times $y, the transpose A’ will be of the order y$\times $x.
Let us consider the first product in the question which is AB’. The order of B’ will be l$\times $k. For AB’ to be defined, from the equation-1, we get that
$\therefore $ l = n$\to \left( 2 \right)$
Let us consider the first product in the question which is B’A. The order of B’ will be l$\times $k. For B’A to be defined, from the equation-1, we get that
$\therefore $ k = m$\to \left( 3 \right)$
From equations- 2 and 3 we get the order of B as m$\times $n.
$\therefore $ The order of matrix B is m$\times $n. The answer is option D.
Note: In the answer, we got the inference that the order of matrix A and B are the same for the condition that AB’ and B’A be defined. So we can infer from this result that for two matrices C and D, the products CD and DC are defined when the conditions D is of the order of transpose of C and vice versa which is C is of the order of transpose of D should be satisfied.
Complete step-by-step solution:
In the notation of order of a matrix M of order m$\times $n, m is the number of rows and n is the number of columns. Let C, D be two matrices of orders $a\times b$ and $x\times y$ respectively.
For the product C$\times $D to be defined, we have to apply the condition for the product to be defined which is the number of columns of C is equal to the number of rows of D. Mathematically it is b = x$\to \left( 1 \right)$.
The important point that we have to note here is that there is no condition for the number of rows of first matrix C and the number of columns of second matrix D.
In the question, it is given that the matrix A is ordered m$\times $n. Let the order of B be k$\times $l.
For a matrix A of order x$\times $y, the transpose A’ will be of the order y$\times $x.
Let us consider the first product in the question which is AB’. The order of B’ will be l$\times $k. For AB’ to be defined, from the equation-1, we get that
$\therefore $ l = n$\to \left( 2 \right)$
Let us consider the first product in the question which is B’A. The order of B’ will be l$\times $k. For B’A to be defined, from the equation-1, we get that
$\therefore $ k = m$\to \left( 3 \right)$
From equations- 2 and 3 we get the order of B as m$\times $n.
$\therefore $ The order of matrix B is m$\times $n. The answer is option D.
Note: In the answer, we got the inference that the order of matrix A and B are the same for the condition that AB’ and B’A be defined. So we can infer from this result that for two matrices C and D, the products CD and DC are defined when the conditions D is of the order of transpose of C and vice versa which is C is of the order of transpose of D should be satisfied.
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