Question

If $A=\left( \begin{matrix}
   1 & 3 & 0 \\
   -1 & 2 & 1 \\
   0 & 0 & 2 \\
\end{matrix} \right)$ , $B=\left( \begin{matrix}
   2 & 3 & 4 \\
   1 & 2 & 3 \\
   -1 & 1 & 2 \\
\end{matrix} \right)$ then AB is equal to?
A . $\left( \begin{matrix}
   5 & 9 & 13 \\
   -1 & 2 & 4 \\
   -1 & 2 & 4 \\
\end{matrix} \right)$
B. $\left( \begin{matrix}
   5 & 9 & 13 \\
   -1 & 2 & 4 \\
   -2 & 2 & 4 \\
\end{matrix} \right)$
C. $\left( \begin{matrix}
   1 & 2 & 4 \\
   -1 & 2 & 4 \\
   -2 & 2 & 4 \\
\end{matrix} \right)$
D. None of these

Answer 1

Hint: Given two matrices A and B of order $(3\times 3)$. We have to find AB. In order to multiply the matrices, both the matrix are of the same order. As the given matrices are of the same order, so we multiply them to get the desirable answer.

Complete step by step Solution:
Given $A=\left( \begin{matrix}
   1 & 3 & 0 \\
   -1 & 2 & 1 \\
   0 & 0 & 2 \\
\end{matrix} \right)$ and $B=\left( \begin{matrix}
   2 & 3 & 4 \\
   1 & 2 & 3 \\
   -1 & 1 & 2 \\
\end{matrix} \right)$
In matrix A, there are 3 rows and 3 columns.
Similarly in matrix B, there are 3 rows and 3 columns.
So both are of order $(3\times 3)$ matrix
The basic representation of $3\times 3$ is
$\left( \begin{matrix}
   {{a}_{11}} & {{a}_{12}} & {{a}_{13}} \\
   {{a}_{21}} & {{a}_{22}} & {{a}_{23}} \\
   {{a}_{31}} & {{a}_{32}} & {{a}_{33}} \\
\end{matrix} \right)$
Then AB will represent as $\left( \begin{matrix}
   {{a}_{11}} & {{a}_{12}} & {{a}_{13}} \\
   {{a}_{21}} & {{a}_{22}} & {{a}_{23}} \\
   {{a}_{31}} & {{a}_{32}} & {{a}_{33}} \\
\end{matrix} \right)$
AB = $\left( \begin{matrix}
   1 & 3 & 0 \\
   -1 & 2 & 1 \\
   0 & 0 & 2 \\
\end{matrix} \right)$$\left( \begin{matrix}
   2 & 3 & 4 \\
   1 & 2 & 3 \\
   -1 & 1 & 2 \\
\end{matrix} \right)$= $\left( \begin{matrix}
   {{a}_{11}} & {{a}_{12}} & {{a}_{13}} \\
   {{a}_{21}} & {{a}_{22}} & {{a}_{23}} \\
   {{a}_{31}} & {{a}_{32}} & {{a}_{33}} \\
\end{matrix} \right)$
The first digit of the first row is the sum of the multiplication of the first row of matrix A and the first column of matrix B.
That is ${{a}_{11}}=1\times 2+3\times 1+0\times -1$
Then ${{a}_{11}}=5$
Similarly, for the second digit of the first row, it is the sum of the multiplication of the first row of matrix A with the second column of matrix B.
That is ${{a}_{12}}=1\times 3+3\times 2+0\times 1$
Then ${{a}_{12}}=9$
Similarly, for the third digit of the first row, it is the sum of values of the first row of matrix A with the third column of matrix B.
That is ${{a}_{13}}=1\times 4+3\times 3+0\times 2$
Then ${{a}_{13}}=13$
Similar steps we followed to find all the values of the given matrix.
${{a}_{21}}=-1\times 2+2\times 1+1\times -1$
Then${{a}_{21}}=-1$
${{a}_{22}}=-1\times 3+2\times 2+1\times 1$
${{a}_{22}}=2$
${{a}_{23}}=-1\times 4+2\times 3+1\times 2$
Then ${{a}_{23}}=4$
Similarly, we find the third row
${{a}_{31}}=0\times 2+0\times 1+2\times -1$
Then ${{a}_{31}}=-2$
${{a}_{32}}=0\times 3+0\times 2+2\times 1$
Then ${{a}_{32}}=2$
${{a}_{33}}=0\times 4+0\times 3+2\times 2$
Then ${{a}_{33}}=4$
Substituting all the values in the AB matrix, we get
$\left( \begin{matrix}
   {{a}_{11}} & {{a}_{12}} & {{a}_{13}} \\
   {{a}_{21}} & {{a}_{22}} & {{a}_{23}} \\
   {{a}_{31}} & {{a}_{32}} & {{a}_{33}} \\
\end{matrix} \right)$= $\left( \begin{matrix}
   5 & 9 & 13 \\
   -1 & 2 & 4 \\
   -2 & 2 & 4 \\
\end{matrix} \right)$
Thus the product of matrix A and B is $\left( \begin{matrix}
   5 & 9 & 13 \\
   -1 & 2 & 4 \\
   -2 & 2 & 4 \\
\end{matrix} \right)$

Therefore, the correct option is (B).

Note: Remember that while multiplying the matrices the elements (that is ) ${{a}_{11}},{{a}_{12}}$ represents the row number of matrix 1 and column number of matrix 2. For example, ${{a}_{11}}$ tells us about the multiplication of the first row of matrix 1 with the first column of matrix 2. ${{a}_{12}}$ represents the multiplication of the first row of the matrix 1 with the second column of the matrix 2 and so on.