invers (x/(-2))^3

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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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eigenvektoren (

6	−1
2	3

)

Solution

(

), (

)

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Solution steps

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(

6	−1
2	3

)

Matrix Eigenvektoren

Lass A eine Quadratmatrix, η ein Vektor und λ ein Skalar sein, welches diese Bedingung erfüllt: Aη=λη,
dann ist η ein Eigenvektor von A und λ ist ein Eigenwert, der damit verbunden ist.

Um die Eigenvektoren η zu finden, löse (A−λ I)η=0 für jeden Eigenwert λ

Ermittle die Eigenwerte von (

6	−1
2	3

): λ=5, λ=4

Eigenvektoren für λ=5: (

)

Eigenvektoren für λ=4: (

)

Die Eigenvektoren für (

6	−1
2	3

)

), (

)

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6	−1
2	3

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

invers (x/(-2))^3

Solution

Solution steps

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