points of d'intersection 1/y

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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

simplify solve for expand factor rationalize

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diagonaliser (

−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

)

Diagonalisation d'une matrice

Une matrice A peut être diagonalisées'il existe une matrice inversible P et une matrice diagonale D tel que A=PDP⁻¹

Trouver les valeurs propres pour (

−4	−17
2	2

): λ=−1+5i, λ=−1−5i

La matrice diagonale D est composée des valeurs propres :

D=(

−1+5`i`	0
0	−1−5`i`

)

Trouver les vecteurs propres pour (

−4	−17
2	2

): (

−3+5i

), (

−3−5i

)

Les vecteurs propres correspondant aux valeurs propres en Dcomposent les colonnes de P :

P=(

−3+5`i`	−3−5`i`
2	2

)

Trouver P⁻¹: (

−`i`10	14 −`i`320
`i`10	14 +`i`320

)

Vérifier que A=PDP⁻¹

Résoudre 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

)

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−4	−17
2	2

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

points of d'intersection 1/y

Solution

Solution steps

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