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

☐²

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√☐

^□√☐

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log_□

θ

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ddx

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

simplify solve for inverse tangent line

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eigenvalues (

1+	+2+
3+	+4

)

Solution

λ=5+√332 , λ=5−√332

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Solution steps

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(

1+	2+
3+	4

)

Matrix eigenvalues

Let A be a square matrix, η a vector and λ a scalar that satisfy Aη=λη,
then λ is called the eigenvalue associated with the eigenvector η of A

The eigenvalues of A are the roots of the characteristic equation det(A−λ I)=0

det((

1+	+2+
3+	+4

)−λ(

1	0
0	1

)): λ²−5λ−2

Solve λ²−5λ−2=0: λ=5+√332 , λ=5−√332

The eigenvalues are:

λ=5+√332 , λ=5−√332

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1+	+2+
3+	+4

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Notebook

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

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