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

☐²

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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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lu decomposition (

1	2	3	4
5	6	7	8
3	2	3	2
3	1	7	8

)

Solution

L=(

1	0	0	0
5	1	0	0
3	1	1	0
3	54	4	1

), U=(

1	2	3	4
0	−4	−8	−12
0	0	2	2
0	0	0	3

)

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Solution steps

One step at a time

(

1	2	3	4
5	6	7	8
3	2	3	2
3	1	7	8

)

Multiply by the identity matrix

1	0	0	0
0	1	0	0
0	0	1	0
0	0	0	1

)(

1	2	3	4
5	6	7	8
3	2	3	2
3	1	7	8

)

Use Gaussian Elimination

1	0	0	0
5	1	0	0
3	1	1	0
3	54	4	1

)(

1	2	3	4
0	−4	−8	−12
0	0	2	2
0	0	0	3

)

Therefore

L=(

1	0	0	0
5	1	0	0
3	1	1	0
3	54	4	1

), U=(

1	2	3	4
0	−4	−8	−12
0	0	2	2
0	0	0	3

)

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Hi, ask a question about lu decomposition (

1	2	3	4
5	6	7	8
3	2	3	2
3	1	7	8

)

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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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