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15^{-3} \mod71

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

☐²

x^☐

√☐

^□√☐

□□

log_□

θ

∞

∫

ddx

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

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Steps Examples

Generated by AI

autovettori (

−4	−17
2	2

)

Solution

(

−3+5i

), (

−3−5i

)

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Solution steps

One step at a time

(

−4	−17
2	2

)

Matrice degli autovettori

Sia A una matrice quadrata, η un vettore e λ uno scalare tale che Aη=λη,
allora η è un autovettore di A e λ è l'autovalore associato ad esso

Per trovare gli autovettori η, risolvi (A−λ I)η=0 per ogni autovalore λ

Trova gli autovalori di (

−4	−17
2	2

): λ=−1+5i, λ=−1−5i

Autovettori per λ=−1+5i: (

−3+5i

)

Autovettori per λ=−1−5i: (

−3−5i

)

Gli autovettori per (

−4	−17
2	2

)

−3+5i

), (

−3−5i

)

Negative Power Mod Examples

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Notebook

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

15^{-3} \mod71

Solution

Solution steps

Number Line