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butalik [34]
3 years ago
6

Prove or give a counterexample:

Mathematics
1 answer:
ozzi3 years ago
4 0

Answer:

See proof below.

Step-by-step explanation:

True

For this case we need to use the following theorem "If v_1, v_2,....v_k are eigenvectors of an nxn matrix A and the associated eigenvalues \lambda_1, \lambda_2,...,\lambda_k are distinct, then v_i's are linearly independent". Now we can proof the statement like this:

Proof

Let A a nxn matrix and we can assume that A has n distinct real eingenvalues let's say \lambda_1, \lambda_2, ....,\lambda_n

From definition of eigenvector for each one \lambda_i needs to have associated an eigenvector v_i for 1 \leq i \leq n

And using the theorem from before , the n eigenvectors v_1,....,v_n are linearly independent since the \lambda_i 1\leq i \leq n are distinct so then we ensure that A is diagonalizable.

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Explain how to find the vertex of 2x^2+12x+16
Elina [12.6K]

The vertex form:

y=a(x-h)^2+k\\\\(h,\ k)-vertex

Use (a+b)^2=a^2+2ab+b^2\qquad(*)


2x^2+12x+16=2(x^2+6x)+16=2(x^2+2(x)(3))+16\\\\=2(\underbrace{x^2+2(x)(3)+3^2}_{(*)}-3^2)+16=2((x+3)^2-9)+16\\\\=2(x+3)^2+(2)(-9)+16=2(x+3)^2-18+16\\\\=\boxed{2(x-(-3))^2-2}\to h=-3,\ k=-2

<h3>Answer: (-3, -2).</h3>

Other method:

f(x)=ax^2+bx+c\\\\vertex:(h,\ k)\\\\h=\dfrac{-b}{2a},\ k=f(h)

We have

f(x)=2x^2+12x+16\\\\a=2,\ b=12,\ c=16

Substitute:

h=\dfrac{-12}{(2)(2)}=\dfrac{-12}{4}=-3\\\\k=f(-3)=2(-3)^2+12(-3)+16=2(9)-36+16=18-36+16=-2

<h3>Answer: (-3, -2).</h3>
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4 years ago
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Answer:

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Step-by-step explanation:

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7 0
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Answer:

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Step-by-step explanation:

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The angles of a linear line split equal 180

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2y is one side of it which we learned Y = 20 which makes that side 40

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