Question

calculate this reflection of the triangle:
\\(\

$$\begin{bmatrix}1&0\\0& - 1\\end{bmatrix}$$

\

$$\begin{bmatrix}0&3&0\\0&0&2\\end{bmatrix}$$

=\

$$\begin{bmatrix}0&a&b\\0&c&d\\end{bmatrix}$$

a =
b =
c =
d =
(0, 2)
(0, 0)
(3, 0)

Explanation:

Step1: Matrix - multiplication rule

For two matrices $A =

$$\begin{bmatrix}1&0\\0& - 1\end{bmatrix}$$

$ and $B=

$$\begin{bmatrix}0&3&0\\0&0&2\end{bmatrix}$$

$, the element in the $i -$th row and $j -$th column of the product matrix $AB$ is the dot - product of the $i -$th row of $A$ and the $j -$th column of $B$.

Step2: Calculate $a$

$a$ is the element in the first row and second column of the product matrix. The first row of $A$ is

$$\begin{bmatrix}1&0\end{bmatrix}$$

and the second column of $B$ is

$$\begin{bmatrix}3\\0\end{bmatrix}$$

. So $a=(1\times3)+(0\times0)=3$.

Step3: Calculate $b$

$b$ is the element in the first row and third column of the product matrix. The first row of $A$ is

$$\begin{bmatrix}1&0\end{bmatrix}$$

and the third column of $B$ is

$$\begin{bmatrix}0\\2\end{bmatrix}$$

. So $b=(1\times0)+(0\times2)=0$.

Step4: Calculate $c$

$c$ is the element in the second row and second column of the product matrix. The second row of $A$ is

$$\begin{bmatrix}0& - 1\end{bmatrix}$$

and the second column of $B$ is

$$\begin{bmatrix}3\\0\end{bmatrix}$$

. So $c=(0\times3)+(-1\times0)=0$.

Step5: Calculate $d$

$d$ is the element in the second row and third column of the product matrix. The second row of $A$ is

$$\begin{bmatrix}0& - 1\end{bmatrix}$$

and the third column of $B$ is

$$\begin{bmatrix}0\\2\end{bmatrix}$$

. So $d=(0\times0)+(-1\times2)=-2$.

Answer:

$a = 3$
$b = 0$
$c = 0$
$d=-2$