Question

what is the inverse of \[A =\left[\begin{matrix}-1&e^{2t}&0\\1&e^{2t}&0\\0&0&e^{3t}\end{matrix}\right]\]
Please, step by step

Answer 1

det A = -2 \(e^{5t}\) and cofactor of it is
\[\left[\begin{matrix}e^{5t}&-e^{3t}&0\\-e^{5t}&-e^{3t}&0\\0&0&-2e^{2t}\end{matrix}\right]\]

Answer 2

and then 1/det A *cofactor^(transpose) is
\[\frac{-1}{2e^{5t}}\left[\begin{matrix}e^{5t}&-e^{5t}&0\\-e^{3t}&-e^{3t}&0\\0&0&-2e^{2t}\end{matrix}\right]\]
\[=\left[\begin{matrix}\frac{-1}{2}&\frac{1}{2}&0\\\frac{1}{2}e^{-2t}&\frac{1}{2}e^{-2t}&0\\0&0&e^{-3t}\end{matrix}\right]\]

Answer 3

so, the last one is \(A^{-1}\), right? and to check whether it's right or wrong, \(AA^{-1}\)should be \(I\), right?

Answer 4

@e.mccormick please help me check my stuff. when checking, I don't get identity matrix

Answer 5

FYI: Transpose of the cofactor is called the adjoint.

Answer 6

FYI??

Answer 7

For Your Information

Answer 8

ok,

Answer 9

@ybarrap

Answer 10

Do you know Gauss-Jordan method for finding inverse?

Answer 11

I applied cofactor

Answer 12

Can you find inverse any way you want, or do you need to use cofactor technique?

Answer 13

any way If it helps, however, in some case I need cofactor because det A =0, right?

Answer 14

If you find detA=0, then there is no inverse.

Answer 15

i'm checking result now

Answer 16

Your inverse is correct. I also validated to get Identify.

Answer 17

I'm going to show you how did with Gauss-Jordan technique: