Question

Multiply the matrices if possible \[A=\left[ \begin{matrix}
   40 \\
   100 \\
   50 \\
\end{matrix} \right],B=\left[ \begin{matrix}
   1000 & 500 & 5000 \\
   3000 & 1000 & 10000 \\
\end{matrix} \right]\]

Answer 1

Hint:We find the order of both the matrices by counting the number of rows and columns. We use the fact that the matrices $A,B$ can only be multiplied when numbers columns of the first matrix are equal to the number of rows of the second matrix in order of multiplication and find that we cannot take $A$ as the first matrix, so we take $B$ as the first matrix and find $BA$. \[\]

Complete step by step answer:
We know from the multiplication of matrices that two matrices $A,B$ can only be multiplied when they have compatible order. The order of a matrix is defined as $m\times n$ where $m$ is the number of rows and $n$ is the number of column. So let $A,B$ be a matrix order $m\times n$and $A,B$ be matrix of order $q\times r$. The order of $A$ and $B$ are compatible if number columns of first matrix is equal to the number of rows of second matrix in the order of multiplication. In symbols $n=q$. The order of the product matrix is $m\times r$
Let be \[A=\mathop{\left[ {{a}_{ij}} \right]}_{i=1,j=1}^{i=m,j=n}\] be the first matrix with entries ${{a}_{ij}}$ and \[B=\mathop{\left[ {{b}_{ij}} \right]}_{i=1,j=1}^{i=q,j=r}\] be the second matrix with entries${{b}_{ij}}$. We denote their product as $C=AB$. Then the entries of $C$ are given by
\[\mathop{\left[ {{c}_{ij}} \right]}_{i=1,j=1}^{i=m,j=r}={{a}_{i1}}{{b}_{1j}}+{{a}_{i2}}{{b}_{2j}}+...+{{a}_{in}}{{b}_{nj}}\]

So first find the order of the matrices. So we have the first matrix
\[A=\left[ \begin{matrix}
   40 \\
   100 \\
   50 \\
\end{matrix} \right]\]
We see that the given matrix $A$ has 3 rows and 1 column. So here we have $m=3,n=1$ and the order of the matrix is $3\times 1$. We observe the second matrix
\[B=\left[ \begin{matrix}
   1000 & 500 & 5000 \\
   3000 & 1000 & 10000 \\
\end{matrix} \right]\]
We see that the given matrix $B$ has 2 rows and 3 columns. So here we have $q=3,r=1$ and the order of the matrix is $2\times 3$. We also see that the number of column of the firs matrix $A$ is not equal to the number of rows of second matrix $B$which means $n\ne q$. So the order of matrices are not compatible for the multiplication form $AB$.\[\]
 We can take $B$ as first matrix and $A$ as second matrix. We can multiply now as $B$ has 3 columns and $A$ has 3 rows, $r=m=3$. The product matrix $BA$ will have the order $q\times n=2\times 1$. So we multiply row by column to have
\[\begin{align}
  & BA=\left[ \begin{matrix}
   1000 & 500 & 5000 \\
   3000 & 1000 & 10000 \\
\end{matrix} \right]\times \left[ \begin{matrix}
   40 \\
   100 \\
   50 \\
\end{matrix} \right] \\
 & =\left[ \begin{matrix}
   1000\times 40+500\times 100+5000\times 50 \\
   3000\times 40+1000\times 100+10000\times 50 \\
\end{matrix} \right] \\
 & =\left[ \begin{matrix}
   340000 \\
   1170000 \\
\end{matrix} \right] \\
\end{align}\]

Note:
We note that the matrix multiplication is not commutative which means if the matrix products exist $AB\ne BA$. We found in this problem that $AB$ does not exist. If both $AB$ and $BA$ exists then both $A,B$ are square matrices of the same order.
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