amplitude of 2sec(x)

Topic

Full pad

`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

implicit derivative tangent volume laplace fourier

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Steps Examples

Generated by AI

e^{(1 2
0 −1
)}

Solution

(

`e`	`e`²−1`e`
0	1`e`

)

Show Steps

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Solution steps

One step at a time

e^{(1 2
0 −1
)}

Diagonalize (

1	2
0	−1

): (

1	−1
0	1

)(

1	0
0	−1

)(

1	1
0	1

)

For diagonalizable matrix A=PDP⁻¹ the matrix exponential is e^A=Pe^DP⁻¹

1	−1
0	1

)e^{(1 0
0 −1
)}(

1	1
0	1

)

e^{(1 0
0 −1
)}=(

`e`	0
0	1`e`

)

1	−1
0	1

)(

`e`	0
0	1`e`

)(

1	1
0	1

)

(

1	−1
0	1

)(

`e`	0
0	1`e`

)(

1	1
0	1

)=(

`e`	`e`²−1`e`
0	1`e`

)

`e`	`e`²−1`e`
0	1`e`

)

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