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AfilCa [17]
4 years ago
13

Find the characteristic polynomial of the matrix a=[−83−310].

Mathematics
1 answer:
rosijanka [135]4 years ago
5 0
Assuming the matrix is

\mathbf A=\begin{bmatrix}-8&3\\-3&10\end{bmatrix}

we have characteristic polynomial given by the equation

\det(\mathbf A-\lambda \mathbf I)=0\iff\begin{vmatrix}-8-\lambda&3\\-3&10-\lambda\end{vmatrix}=0
(-8-\lambda)(10-\lambda)-(-9)=0
\lambda^2-2\lambda-71=0
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Which answer does NOT correctly describe the graph of y = 5x + 12?
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<h2>Answer</h2>

2. The Graph Contains The Point (5, 12)

<h2>Explanation</h2>

Remember that in a linear function of the form y=mx+b, m is the slope/rate of change of the line and b is the y-intercept.

From our equation we can infer that b=12, so the y-intercept of our graph is 12 units above the origin on the point (0, 12). Therefore, choice 2 correctly describe the graph of y = 5x + 12

We can also infer that m=5, so the slope of our line is 5, which is constant, so our line has a constant rate of change. Also, since every integer can be written as a fraction with denominator 1, we can write our slope as m=\frac{5}{1}. Therefore, choices 1 and 2 correctly describe the graph of y = 5x + 12.

Now, remember that any point on the plane has coordinates (x,y), so, to check if a point is on a line, we just need to replace the x and y values in the line equation and check if the equation holds. Our point is (5, 12), so x = 5 and y = 12. Lets replace the values in our line:

y=5x+12

12=5(5)+12

12=25+12

12\neq 37

Since the equation equation doesn't hold (12 is not equal 37), we can conclude that the graph doesn't contain the point (5, 12). Therefore, choice 2 does NOT correctly describe the graph y = 5x +12.


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