Question

Matrix Question

Answer 1

show that \[X(t) = \left[\begin{matrix}e^{2t} & e^{3t} \\ -3e^{2t} & -2e^{3t}\end{matrix}\right]\]
is a solution to\[X'(t) = \left[\begin{matrix}1 & -1 \\ 2 &4\end{matrix}\right]X(t)\]

Answer 2

so i took the derivative of X(t) and got\[\left[\begin{matrix}2e^{2t} &3e^{3t} \\-2e^{2t} & -6e^{3t}\end{matrix}\right]\]...

so i multiply X(t) with the regular matrix there and i don't get the X'(t) i just calculated... so i'm confused

Answer 3

are you sure you copied down all the right signs? no typos?

Answer 4

derivative of -3e^(2t) = -6e^2t

Answer 5

yes, there is a typo

X(t) should be
\[\left[\begin{matrix}e^{2t} & e^{3t} \\ -e^{2t} & -2e^{3t}\end{matrix}\right]\]

Answer 6

how to we make a space in latek?
sothatitsnotlikethis

Answer 7

yes, sorry about that.

Answer 8

my derivative should still be correct

Answer 9

space in latex `\quad` for large space, `\:` for small spaces...

Answer 10

thanks you two:)

Answer 11

You should get the right answer now with the fixed typo :)

Answer 12

^

Answer 13

sorry i only typed in the matrix wrong, i worked it out with the correct values. the derivative didn't match my answer though :(