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

☐²

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log_□

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ddx

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

simplify solve for partial fractions long division line

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Steps Examples

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inverse (

1	2	1
6	−1	0
−1	−2	−1

)

Solution

Matrix is singular

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Solution steps

One step at a time

(

1	2	1
6	−1	0
−1	−2	−1

)⁻¹

Inverse matrix

A matrix is invertible if and only if there is a sequence of elementary row (or column) operations taking it to the identity matrix.

Cancel proportional rows by preforming R₁ ← R₁ +1 R₃

(

0	0	0
6	−1	0
−1	−2	−1

)

A matrix with a zero row is not invertible

Matrix is singular

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Chat with Symbo

היי, לשאול שאלה לגבי inverse (

1	2	1
6	−1	0
−1	−2	−1

) או בחר למטה

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Notebook

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

√`x`+2	`x`< 1
`xx`²−6`x`+8	`x`≥1

`x`²	`x`<−2
ln(`x`+5)	`x`≥−2

sin(`x`)	`x`<0
2`x`	`x`≥0

√`x`+2	`x`<1
`xx`²−6`x`+8	`x`≥1

√`x`²−4	`x`<0
`x`³−`x`	`x`≥0

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Solution

Solution steps

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