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(A)$3 \times 4$

(B)$3 \times 3$

(C)$4 \times 4$

(D)$4 \times 3$

Answer
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This will give us the required answer.

Also, we have a matrix B such that ${A^t}B$ and \[B{A^t}\] are both defined.

We are asked to find the order of B.

${X^t}$denotes the transpose of a matrix X and is a matrix obtained by changing the rows of matrix X into columns and the columns into rows.

Therefore, if $m \times n$ is the order of matrix X, then the order of ${X^t}$ will be $n \times m$.

Now, for a matrix Y, we say that the product ${X^t}Y$ is defined if and only if the order of Y is $m \times r$.

That is, the number of columns of the matrix ${X^t}$ should be equal to the number of rows of matrix Y.

Now, we have the order of matrix A which is $3 \times 4$. This means that the order of ${A^t}$ is $4 \times 3$.

As we are given that ${A^t}B$ is defined, we can conclude that the order of B is of the form$3 \times s$where s is some positive integer. This is because the number of columns of ${A^t}$= the number of rows of B.

Now,\[B{A^t}\]is defined$ \Rightarrow $the number of columns of B = the number of rows of ${A^t}$

No. of columns of B = s and number of rows of ${A^t}$= 4.

$ \Rightarrow s = 4$

That is, the number of columns of B = 4.

We already know that the number of rows of B = 3.

Hence B is a $3 \times 4$ matrix.

2) If A is an$m \times n$matrix and B is an $n \times r $matrix, then the order of matrix AB is $n \times r$.