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parity 1/(x^2)

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собственные значения (

1	2	1
6	−1	0
−1	−2	−1

)

Solution

λ=0, λ=3, λ=−4

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(

1	2	1
6	−1	0
−1	−2	−1

)

Собственные значения матрицы

Пусть A будет квадратной матрицей, η вектором и λ скаляром, которые удовлетворяют Aη=λη,
тогда λ называется собственным значением, связанным с собственным вектором η из A

собственные значения A являются корнями характеристического уравнения det(A−λ I)=0

det((

1	2	1
6	−1	0
−1	−2	−1

)−λ(

1	0	0
0	1	0
0	0	1

)): −λ³−λ²+12λ

Решить −λ³−λ²+12λ=0: λ=0, λ=3, λ=−4

Собственные значения:

λ=0, λ=3, λ=−4

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

parity 1/(x^2)

Solution

Solution steps

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