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vertices y=-2x^2-2x-2

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`x`²	`x`^□	log_□	√☐	^□√☐	≤	≥	□□	·	÷	`x`^◦	π
(☐)^′	`ddx`	∂∂`x`	∫	∫_□^□	lim	∑	∞	`θ`	(`f` ◦ `g`)	`f`(`x`)

☐²

x^☐

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log_□

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∫

ddx

≥	≤	·	÷	`x`^◦	(☐)	\|☐\|	(`f` ◦ `g`)	`f`(`x`)	ln	`e`^☐
(☐)^′	∂∂`x`	∫_□^□	lim	∑	sin	cos	tan	cot	csc	sec

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eigenvectors (

1	2	3
4	5	6
7	8	9

)

Solution

(

−2

), (

−2(11−3√33)

11+3√33

), (

−2(3√33+11)

−3√33+11

)

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Solution steps

One step at a time

(

1	2	3
4	5	6
7	8	9

)

Matrix eigenvectors

Let A be a square matrix, η a vector and λ a scalar that satisfy Aη=λη,
then η is an eigenvector of A and λ is the eigenvalue associated with it

To find the eigenvectors η, solve (A−λ I)η=0 for each eigenvalue λ

Find the eigenvalues for (

1	2	3
4	5	6
7	8	9

): λ=0, λ=15+3√332 , λ=15−3√332

Eigenvectors for λ=0: (

−2

)

Eigenvectors for λ=15+3√332 : (

−2(11−3√33)

11+3√33

)

Eigenvectors for λ=15−3√332 : (

−2(3√33+11)

−3√33+11

)

The eigenvectors for (

1	2	3
4	5	6
7	8	9

)

−2

), (

−2(11−3√33)

11+3√33

), (

−2(3√33+11)

−3√33+11

)

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−▭▭	<	7	8	9	÷	`AC`
+▭▭	>	4	5	6	×	☐☐☐
×▭▭	(	1	2	3	−	`x`
▭ \|▭	)	.	0	=	+	`y`

vertices y=-2x^2-2x-2

Solution

Solution steps

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