Answer
The product is, $\left[ \begin{array}{*{35}{r}}
2 & -1 \\
13 & 1 \\
\end{array} \right]$.
Work Step by Step
Consider the given matrices
$\begin{align}
& A=\left[ \begin{array}{*{35}{r}}
1 & -1 & 0 \\
2 & 1 & 3 \\
\end{array} \right] \\
& B=\left[ \begin{array}{*{35}{r}}
4 & -1 \\
2 & 0 \\
1 & 1 \\
\end{array} \right]
\end{align}$
Therefore, the order of matrix $A$ is $2\times 3$ and $B$ is $3\times 2$; therefore, the product of the matrices is defined and the order of the matrix will be $2\times 2$
$\begin{align}
& AB=\left[ \begin{array}{*{35}{r}}
1 & -1 & 0 \\
2 & 1 & 3 \\
\end{array} \right]\left[ \begin{array}{*{35}{r}}
4 & -1 \\
2 & 0 \\
1 & 1 \\
\end{array} \right] \\
& =\left[ \begin{array}{*{35}{r}}
\left( 1 \right)\left( 4 \right)+\left( -1 \right)\left( 2 \right)+\left( 0 \right)\left( 1 \right) & \left( 1 \right)\left( -1 \right)+\left( -1 \right)\left( 0 \right)+\left( 0 \right)\left( 1 \right) \\
\left( 2 \right)\left( 4 \right)+\left( 1 \right)\left( 2 \right)+\left( 3 \right)\left( 1 \right) & \left( 2 \right)\left( -1 \right)+\left( 1 \right)\left( 0 \right)+\left( 3 \right)\left( 1 \right) \\
\end{array} \right] \\
& =\left[ \begin{array}{*{35}{r}}
4-2+0 & -1+0+0 \\
8+2+3 & -2+0+3 \\
\end{array} \right] \\
& =\left[ \begin{array}{*{35}{r}}
2 & -1 \\
13 & 1 \\
\end{array} \right]
\end{align}$
The product of the matrices is $\left[ \begin{array}{*{35}{r}}
2 & -1 \\
13 & 1 \\
\end{array} \right]$.
