
If A is $2\times 3$ matrix and B is a matrix with ${{A}^{T}}B$ and $B{{A}^{T}}$ where both are well defined then find the order of the matrix $B$.
Answer
589.8k+ views
Hint: Use the knowledge of change in order of matrices in case of transpose or multiplication of matrices.
We know that any matrix $A$ with order of $m\times n$ means , it has $m$ rows and $n$ columns. The transpose of $A$ is defined as the matrix with a single right angle rotation denoted as ${{A}^{T}}$ . The order of ${{A}^{T}}$ will be $n\times m$.
Complete step-by-step solution:
It is given that the order of matrix $A$ is $2\times 3$. So its transpose ${{A}^{T}}$ will be order $3\times 2$.
Again we know that two matrices can only multiplied if and only if number of columns of the first matrix is equal to the number of rows in second matrix.
For example if the matrix $E$ is of order ${{e}_{r}}\times {{e}_{c}}$ and the matrix $F$is of order $ {{f}_{r}}\times {{f}_{c}}$, then they can only be multiplied if and only if ${{e}_{c}}={{f}_{r}}$.
It is given in the question that ${{A}^{T}}B$ is well defined which implies that number of columns of ${{A}^{T}}$ is same as number of rows of $B$ which is 2 as the order of ${{A}^{T}}$ is $3\times 2$.
Similarly it also given that $B{{A}^{T}}$ is well defined which implies number of columns of $B$ is same as number of rows of ${{A}^{T}}$ which is 3 as the order of ${{A}^{T}}$ is $3\times 2$.
So we found that $B$ has 2 rows and 3 columns. Hence the order of $B$ is $2\times 3$. \[\]
We can also take example \[A=\left[ \begin{matrix}
1 & 2 \\
3 & 4 \\
5 & 6 \\
\end{matrix} \right]\] then \[{{A}^{T}}=\left[ \begin{matrix}
1 & 3 & 5 \\
2 & 4 & 6 \\
\end{matrix} \right]\] . As ${{A}^{T}}B$ is defined the number of rows of $B$ is 2 and as $B{{A}^{T}}$ is defined the number of columns is 3.
Note: While solving this problem you need to carefully check the number of rows and columns to arrive at the correct result. And also take care of the fact that matrix multiplication is not commutative. The key concept to solve this problem is the condition for the multiplication of two matrices which needs to be fulfilled for existing multiplication.
We know that any matrix $A$ with order of $m\times n$ means , it has $m$ rows and $n$ columns. The transpose of $A$ is defined as the matrix with a single right angle rotation denoted as ${{A}^{T}}$ . The order of ${{A}^{T}}$ will be $n\times m$.
Complete step-by-step solution:
It is given that the order of matrix $A$ is $2\times 3$. So its transpose ${{A}^{T}}$ will be order $3\times 2$.
Again we know that two matrices can only multiplied if and only if number of columns of the first matrix is equal to the number of rows in second matrix.
For example if the matrix $E$ is of order ${{e}_{r}}\times {{e}_{c}}$ and the matrix $F$is of order $ {{f}_{r}}\times {{f}_{c}}$, then they can only be multiplied if and only if ${{e}_{c}}={{f}_{r}}$.
It is given in the question that ${{A}^{T}}B$ is well defined which implies that number of columns of ${{A}^{T}}$ is same as number of rows of $B$ which is 2 as the order of ${{A}^{T}}$ is $3\times 2$.
Similarly it also given that $B{{A}^{T}}$ is well defined which implies number of columns of $B$ is same as number of rows of ${{A}^{T}}$ which is 3 as the order of ${{A}^{T}}$ is $3\times 2$.
So we found that $B$ has 2 rows and 3 columns. Hence the order of $B$ is $2\times 3$. \[\]
We can also take example \[A=\left[ \begin{matrix}
1 & 2 \\
3 & 4 \\
5 & 6 \\
\end{matrix} \right]\] then \[{{A}^{T}}=\left[ \begin{matrix}
1 & 3 & 5 \\
2 & 4 & 6 \\
\end{matrix} \right]\] . As ${{A}^{T}}B$ is defined the number of rows of $B$ is 2 and as $B{{A}^{T}}$ is defined the number of columns is 3.
Note: While solving this problem you need to carefully check the number of rows and columns to arrive at the correct result. And also take care of the fact that matrix multiplication is not commutative. The key concept to solve this problem is the condition for the multiplication of two matrices which needs to be fulfilled for existing multiplication.
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