Question

Compute the products Ax in Exercises 16 through 19

using paper and pencil (if the products are defined).

16.

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

Short answer

The dot product$A{x}^{\to }$ of $\left[\begin{array}{cc}0& 1\\ 3& 2\end{array}\right]\left[\begin{array}{c}2\\ -3\end{array}\right]$is$\left[\begin{array}{c}-3\\ 0\end{array}\right]$

Step-by-step solution

Step 1:

Find $A\overline{x}$in terms of rows of A.

$\left[\begin{array}{cc}0& 1\\ 3& 2\end{array}\right].\left[\begin{array}{c}2\\ -3\end{array}\right]$

$$\begin{gathered} \Rightarrow \left[\begin{array}{c}\left(0\right)\left(2\right)+\left(1\right)\left(-3\right)\\ \left(3\right)\left(2\right)+\left(2\right)\left(-3\right)\end{array}\right] \\ \Rightarrow \left[\begin{array}{c}0-3\\ 6-6\end{array}\right] \\ \Rightarrow \left[\begin{array}{c}-3\\ 0\end{array}\right] \end{gathered}$$

Step 2:

Find $A\overline{x}$ in terms of columns of A.

role="math" localid="1659339166632" $$\begin{gathered} \left[\begin{array}{cc}0& 1\\ 3& 2\end{array}\right].\left[\begin{array}{c}2\\ -3\end{array}\right] \\ \Rightarrow 2\left[\begin{array}{c}0\\ 3\end{array}\right]-3\left[\begin{array}{c}1\\ 2\end{array}\right] \\ \Rightarrow \left[\begin{array}{c}0\\ 6\end{array}\right]-\left[\begin{array}{c}3\\ 6\end{array}\right] \\ \Rightarrow \left[\begin{array}{c}-3\\ 0\end{array}\right] \end{gathered}$$