Question

If
$A=\left( \begin{matrix}
   1 & 2 & -1 \\
   3 & 0 & 2 \\
   4 & 5 & 0 \\
\end{matrix} \right)$,
$B=\left( \begin{matrix}
   1 & 0 & 0 \\
   2 & 1 & 0 \\
   0 & 1 & 3 \\
\end{matrix} \right)$ then $AB$ is
A. $\left[ \begin{matrix}
   5 & 1 & -3 \\
   3 & 2 & 6 \\
   14 & 5 & 0 \\
\end{matrix} \right]$
B. $\left[ \begin{matrix}
   11 & 4 & 3 \\
   1 & 2 & 3 \\
   0 & 3 & 3 \\
\end{matrix} \right]$
C. $\left[ \begin{matrix}
   1 & 8 & 4 \\
   2 & 9 & 6 \\
   0 & 2 & 0 \\
\end{matrix} \right]$
D. $\left[ \begin{matrix}
   0 & 1 & 2 \\
   5 & 4 & 3 \\
   1 & 8 & 2 \\
\end{matrix} \right]$

Answer 1

Hint: In this Question, we have to find the product of two matrices, Matrix A and Matrix B. we have to multiply each element of a row of matrix A with each element of a column of matrix B. Then adding all the pairs will give us the final answer.

Formula Used: A matrix is denoted by $A=\left[ {{a}_{ij}} \right]$ where $i$ represents rows and $j$ represents columns
Then, the multiplication of two $3\times 3$ matrices is as follows:
If $A=\left[ {{a}_{ij}} \right]$ is an $m\times n$ matrix and $B=\left[ {{b}_{ij}} \right]$ is $n\times p$ matrix, the product $[AB]$ becomes an $m\times p$ matrix
Consider matrix
$A=\left( \begin{matrix}
   {{a}_{11}} & {{a}_{12}} & {{a}_{13}} \\
   {{a}_{21}} & {{a}_{22}} & {{a}_{23}} \\
   {{a}_{31}} & {{a}_{32}} & {{a}_{33}} \\
\end{matrix} \right)$
and matrix
$B=\left( \begin{matrix}
   {{b}_{11}} & {{b}_{12}} & {{b}_{13}} \\
   {{b}_{21}} & {{b}_{22}} & {{b}_{23}} \\
   {{b}_{31}} & {{b}_{32}} & {{b}_{33}} \\
\end{matrix} \right)$
Then the product of these two matrices is calculated as
$\left( \begin{matrix}
   {{a}_{11}}{{b}_{11}}+{{a}_{12}}{{b}_{21}}+{{a}_{13}}{{b}_{31}} & {{a}_{11}}{{b}_{12}}+{{a}_{12}}{{b}_{22}}+{{a}_{13}}{{b}_{32}} & {{a}_{11}}{{b}_{13}}+{{a}_{12}}{{b}_{23}}+{{a}_{13}}{{b}_{33}} \\
   {{a}_{21}}{{b}_{11}}+{{a}_{22}}{{b}_{21}}+{{a}_{23}}{{b}_{31}} & {{a}_{21}}{{b}_{12}}+{{a}_{22}}{{b}_{22}}+{{a}_{23}}{{b}_{32}} & {{a}_{21}}{{b}_{13}}+{{a}_{22}}{{b}_{23}}+{{a}_{23}}{{b}_{33}} \\
   {{a}_{31}}{{b}_{11}}+{{a}_{32}}{{b}_{21}}+{{a}_{33}}{{b}_{31}} & {{a}_{31}}{{b}_{12}}+{{a}_{32}}{{b}_{22}}+{{a}_{33}}{{b}_{32}} & {{a}_{31}}{{b}_{13}}+{{a}_{32}}{{b}_{23}}+{{a}_{33}}{{b}_{33}} \\
\end{matrix} \right)$

Complete step by step solution: The given two matrices are
$A=\left( \begin{matrix}
   1 & 2 & -1 \\
   3 & 0 & 2 \\
   4 & 5 & 0 \\
\end{matrix} \right)$
and
$B=\left( \begin{matrix}
   1 & 0 & 0 \\
   2 & 1 & 0 \\
   0 & 1 & 3 \\
\end{matrix} \right)$
For finding the product of these two matrices we have to multiply each element of row of Matrix A to each element of column of Matrix B.
$\begin{align}
  & AB=\left( \begin{matrix}
   1\times 1+2\times 2+(-1)(0) & 1\times 0+2\times 1+(-1)(1) & 1\times 0+2\times 0+(-1)(3) \\
   3\times 1+0\times 2+2\times 0 & 3\times 0+0\times 1+2\times 1 & 3\times 0+0\times 0+2\times 3 \\
   4\times 1+5\times 2+0\times 0 & 4\times 0+5\times 1+0\times 0 & 4\times 0+5\times 0+0\times 3 \\
\end{matrix} \right) \\
 & \text{ }=\left( \begin{matrix}
   1+4+0 & 0+2-1 & 0+0-3 \\
   3+0+0 & 0+0+2 & 0+0+6 \\
   4+10+0 & 0+5+0 & 0+0+0 \\
\end{matrix} \right) \\
 & \text{ }=\left( \begin{matrix}
   5 & 1 & -3 \\
   3 & 2 & 6 \\
   14 & 5 & 0 \\
\end{matrix} \right) \\
\end{align}$
Hence, this is the required solution

Option ‘A’ is correct

Note: Here, we may go wrong with the multiplication of 3\times 3 matrices and can exchange rows to columns or columns to rows. We have to keep in mind that we have to multiply the row of Matrix A by the column of matrix B.