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11) find the product. $-2\begin{bmatrix}1&2\\3&-1\\5&0end{bmatrix}$ opt…

Question

  1. find the product.

$-2\begin{bmatrix}1&2\\3&-1\\5&0end{bmatrix}$
options:
$\begin{bmatrix}2&1\\-1&3\\0&5end{bmatrix}$
$\begin{bmatrix}2&4\\6&-2\\10&0end{bmatrix}$
$\begin{bmatrix}-1&0\\1&-3\\-7&-2end{bmatrix}$
$\begin{bmatrix}-2&-4\\-6&2\\-10&0end{bmatrix}$

  1. choose the best answer.

which is the equation of an ellipse?
options:
$\frac{(x - 5)^2}{16}+\frac{(y + 2)^2}{9}=1$
$\frac{(x - 5)^2}{16}+\frac{(y + 2)}{9}=1$
$\frac{(x - 5)^2}{16}-\frac{(y + 2)^2}{9}=1$
$(x - 5)^2+(y + 2)^2=25$

Explanation:

Step1: Multiply scalar by each element

Multiply -2 by every entry in the matrix:

$$-2 \times LATEXBLOCK1 = LATEXBLOCK2 $$

Step2: Calculate each product

Compute the value of each entry:

$$\begin{bmatrix} -2 & -4 \\ -6 & 2 \\ -10 & 0 \end{bmatrix}$$

Step3: Identify ellipse standard form

Recall ellipse equation: $\frac{(x-h)^2}{a^2}+\frac{(y-k)^2}{b^2}=1$. Match to options.

Answer:

Question 11:

$$\begin{bmatrix} -2 & -4 \\ -6 & 2 \\ -10 & 0 \end{bmatrix}$$

(the fourth option)

Question 12:

$\boldsymbol{\frac{(x - 5)^2}{16} + \frac{(y + 2)^2}{9} = 1}$ (the first option)