# Solve this problem

Question:

$\left[\begin{array}{c}1 \\ -1 \\ 2\end{array}\right]\left[\begin{array}{lll}2 & 1 & -1\end{array}\right]=$ _________

Solution:

The order of matrix $\left[\begin{array}{c}1 \\ -1 \\ 2\end{array}\right]$ is $3 \times 1$ and the order of matrix $\left[\begin{array}{lll}2 & 1 & -1\end{array}\right]$ is $1 \times 3$.

So, $\left[\begin{array}{c}1 \\ -1 \\ 2\end{array}\right]\left[\begin{array}{lll}2 & 1 & -1\end{array}\right]$ is a matrix of order $3 \times 3$.

$\left[\begin{array}{c}1 \\ -1 \\ 2\end{array}\right]\left[\begin{array}{lll}2 & 1 & -1\end{array}\right]=\left[\begin{array}{ccc}1 \times 2 & 1 \times 1 & 1 \times(-1) \\ (-1) \times 2 & (-1) \times 1 & (-1) \times(-1) \\ 2 \times 2 & 2 \times 1 & 2 \times(-1)\end{array}\right]=\left[\begin{array}{ccc}2 & 1 & -1 \\ -2 & -1 & 1 \\ 4 & 2 & -2\end{array}\right]$

$\left[\begin{array}{c}1 \\ -1 \\ 2\end{array}\right]\left[\begin{array}{lll}2 & 1 & -1\end{array}\right]=$ $\left[\begin{array}{ccc}2 & 1 & -1 \\ -2 & -1 & 1 \\ 4 & 2 & -2\end{array}\right]$