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intervalo x^2+2x-1-(2x^2-3x+6)

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

☐²

x^☐

√☐

^□√☐

□□

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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vetores próprios (

1	2
0	−1

)

Solution

(

), (

−1

)

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Solution steps

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(

1	2
0	−1

)

Vetores próprios de uma matriz

Seja A uma matriz quadrada, η um vetor e λ um escalar que satisfaça Aη=λη,
então η é um vetor próprio de A e λ é o valor próprio associado a ele

Para encontrar os vetores próprios η, resolver (A−λ I)η=0 para cada valor próprio λ

Encontrar os valores próprios de (

1	2
0	−1

): λ=1, λ=−1

Vetores próprios de λ=1: (

)

Vetores próprios de λ=−1: (

−1

)

Os vetores próprios de (

1	2
0	−1

)

), (

−1

)

Fractions Add, Subtract Like Denominators Examples

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1	2
0	−1

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

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intervalo x^2+2x-1-(2x^2-3x+6)

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Solution steps

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