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vértices ((x+3)^2}{25}-\frac{(y-4)^2)/9 =1

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

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Steps Examples

Generated by AI

diagonalizar (

−4	−17
2	2

)

Solution

P=(

−3+5`i`	−3−5`i`
2	2

), D=(

−1+5`i`	0
0	−1−5`i`

), P⁻¹=(

−`i`10	14 −`i`320
`i`10	14 +`i`320

)

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Solution steps

One step at a time

(

−4	−17
2	2

)

Diagonalização de uma matriz

Uma matriz A é diagonalizável se existe uma matriz inversível P e uma matriz diagonal D tal que A=PDP⁻¹

Encontrar os valores próprios de (

−4	−17
2	2

): λ=−1+5i, λ=−1−5i

A matriz diagonal D é composta pelos valores próprios:

D=(

−1+5`i`	0
0	−1−5`i`

)

Encontrar os vetores próprios de (

−4	−17
2	2

): (

−3+5i

), (

−3−5i

)

Os vetores próprios correspondentes aos valores próprios mn D compõem as colunas de P :

P=(

−3+5`i`	−3−5`i`
2	2

)

Encontrar P⁻¹: (

−`i`10	14 −`i`320
`i`10	14 +`i`320

)

Verificar que A=PDP⁻¹

Resolver PDP⁻¹: (

−4	−17
2	2

)

P=(

−3+5`i`	−3−5`i`
2	2

), D=(

−1+5`i`	0
0	−1−5`i`

), P⁻¹=(

−`i`10	14 −`i`320
`i`10	14 +`i`320

)

Negative Exponent Examples

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−4	−17
2	2

) ou escolha abaixo

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

vértices ((x+3)^2}{25}-\frac{(y-4)^2)/9 =1

Solution

Solution steps

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