Question

In Problems 25– 30, find the product.

$\left[\begin{array}{ccc}4& -2& 3\\ 0& 1& 2\\ -1& 0& 1\end{array}\right]\left[\begin{array}{cc}2& 6\\ 1& -1\\ 0& 2\end{array}\right]$

Short answer

The product of the matrices $\left[\begin{array}{ccc}4& -2& 3\\ 0& 1& 2\\ -1& 0& 1\end{array}\right]\left[\begin{array}{cc}2& 6\\ 1& -1\\ 0& 2\end{array}\right]$is $\left[\begin{array}{cc}6& 32\\ 1& 3\\ -2& -4\end{array}\right]$

Step-by-step solution

Step 1. Given Information

The matrices:

$\left[\begin{array}{ccc}4& -2& 3\\ 0& 1& 2\\ -1& 0& 1\end{array}\right]\left[\begin{array}{cc}2& 6\\ 1& -1\\ 0& 2\end{array}\right]$

Step 2. Observe the matrix

The first matrix is $3\times 3$matrix and the second matrix is $3\times 2$, so the resultant matrix will be of the form $3\times 2$ matrix

Step 3. Find the product

$$\begin{gathered} \left[\begin{array}{ccc}4& -2& 3\\ 0& 1& 2\\ -1& 0& 1\end{array}\right]\left[\begin{array}{cc}2& 6\\ 1& -1\\ 0& 2\end{array}\right]=\left[\begin{array}{cc}4\left(2\right)+-2\left(1\right)+3\left(0\right)& 4\left(6\right)+-2(-1)+3\left(2\right)\\ 0\left(2\right)+1\left(1\right)+2\left(0\right)& 0\left(6\right)+1(-1)+2\left(2\right)\\ -1\left(2\right)+0\left(1\right)+1\left(0\right)& -1\left(6\right)+0(-1)+1\left(2\right)\end{array}\right] \\ =\left[\begin{array}{cc}6& 32\\ 1& 3\\ -2& -4\end{array}\right] \end{gathered}$$